Chapter 6 Permutations and Combinations (Q & A)

Solution:

The expression $n!$ is read as "n factorial" and it represents the product of all positive integers up to n.

The formula for n-factorial is:

$n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1$

For the given question, we need to evaluate $5!$.

Using the formula, we can expand $5!$ as follows:

$5! = 5 \times 4 \times 3 \times 2 \times 1$

Now, we perform the multiplication:

$5! = (5 \times 4) \times 3 \times 2 \times 1$

$5! = 20 \times 3 \times 2 \times 1$

$5! = (20 \times 3) \times 2 \times 1$

$5! = 60 \times 2 \times 1$

$5! = (60 \times 2) \times 1$

$5! = 120 \times 1$

$5! = 120$

Thus, the value of $5!$ is 120.

Comparing this result with the given options, we find that it matches option (C).

Hence, the correct option is (C) 120.

Solution:

The value of zero factorial, denoted as $0!$, is defined by mathematical convention.

By definition, the value of $0!$ is 1.

$0! = 1$

Explanation:

This definition is consistent with the recursive property of factorials. The factorial of a non-negative integer $n$, denoted by $n!$, is defined as:

$n! = n \times (n-1)!$

Let's use this relation for $n=1$:

$1! = 1 \times (1-1)!$

$1! = 1 \times 0!$

We know that $1! = 1$. Substituting this into the equation:

$1 = 1 \times 0!$

To solve for $0!$, we can divide both sides by 1:

$0! = \frac{1}{1}$

$0! = 1$

Alternate Solution (Combinatorial Argument):

The factorial $n!$ represents the number of ways to arrange (permute) $n$ distinct objects.

For example, $3! = 6$ is the number of ways to arrange 3 objects.

Following this logic, $0!$ is the number of ways to arrange 0 objects.

There is exactly one way to arrange zero objects: the empty arrangement. You do nothing, and that is one distinct outcome.

Therefore, based on the combinatorial interpretation, the value of $0!$ is 1.

From the explanation, we can conclude that the value of $0!$ is 1.

Comparing this result with the given options, we find that it matches option (B).

Hence, the correct option is (B) 1.

Solution:

We need to evaluate the expression $\frac{8!}{6!}$.

The factorial of a non-negative integer $n$, denoted by $n!$, is the product of all positive integers less than or equal to $n$.

$n! = n \times (n-1) \times (n-2) \times \dots \times 1$

A useful property of factorials is that we can express a larger factorial in terms of a smaller one. For example, $n! = n \times (n-1)!$.

We can apply this property to expand $8!$ in terms of $6!$:

$8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$

This can also be written as:

$8! = 8 \times 7 \times (6 \times 5 \times 4 \times 3 \times 2 \times 1)$

$8! = 8 \times 7 \times 6!$

Now, we substitute this expression for $8!$ into the original fraction:

$\frac{8!}{6!} = \frac{8 \times 7 \times 6!}{6!}$

We can cancel the common term $6!$ from the numerator and the denominator:

$\frac{8!}{6!} = \frac{8 \times 7 \times \cancel{6!}}{\cancel{6!}}$

This leaves us with:

$8 \times 7$

Performing the multiplication:

$8 \times 7 = 56$

Thus, the value of $\frac{8!}{6!}$ is 56.

Comparing this result with the given options, we find that it matches option (A).

Hence, the correct option is (A) 56.

Solution:

We are given the equation:

We need to find the value of the positive integer 'n'. To do this, we can calculate the factorial values of integers starting from 1 until we get 720.

The factorial of a number is the product of all positive integers up to that number.

• $1! = 1$
• $2! = 2 \times 1 = 2$
• $3! = 3 \times 2 \times 1 = 6$
• $4! = 4 \times 3 \times 2 \times 1 = 24$
• $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$
• $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$

We can see that $6!$ is equal to 720.

Therefore, if $n! = 720$, then $n = 6$.

Alternate Solution:

We can find the value of 'n' by successively dividing 720 by consecutive integers starting from 2, until we reach 1.

Start with 720.

Divide by 2: $\frac{720}{2} = 360$

Divide the result by 3: $\frac{360}{3} = 120$

Divide the result by 4: $\frac{120}{4} = 30$

Divide the result by 5: $\frac{30}{5} = 6$

Divide the result by 6: $\frac{6}{6} = 1$

Since we have performed division by consecutive integers from 2 up to 6, this means:

$720 = 2 \times 3 \times 4 \times 5 \times 6$

We can add $1 \times$ to the product as it does not change the value:

$720 = 1 \times 2 \times 3 \times 4 \times 5 \times 6$

This is the definition of $6!$.

$720 = 6!$

So, the value of $n$ is 6.

Comparing our result with the given options, we find that it matches option (B).

Hence, the correct option is (B) 6.

Solution:

We are asked to express the product $10 \times 9 \times 8 \times 7$ in factorial notation.

Recall the definition of the factorial of a number $n$, denoted as $n!$:

$n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1$

The given expression is $10 \times 9 \times 8 \times 7$.

This expression looks like the beginning of the expansion of $10!$. Let's write out $10!$:

$10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$

We can see that our expression $10 \times 9 \times 8 \times 7$ is a part of $10!$. The missing part is $6 \times 5 \times 4 \times 3 \times 2 \times 1$, which is precisely $6!$.

So, we can rewrite $10!$ as:

$10! = (10 \times 9 \times 8 \times 7) \times (6!)$

To find an expression for $10 \times 9 \times 8 \times 7$, we can rearrange this equation by dividing both sides by $6!$:

$10 \times 9 \times 8 \times 7 = \frac{10!}{6!}$

Alternate Method:

We start with the expression $10 \times 9 \times 8 \times 7$.

To convert this into a factorial form, we can multiply and divide by the factorial of the number that is one less than the last number in the product (which is 7). So we multiply and divide by $(7-1)! = 6!$.

$10 \times 9 \times 8 \times 7 = \frac{10 \times 9 \times 8 \times 7 \times 6!}{6!}$

The numerator, $10 \times 9 \times 8 \times 7 \times 6!$, is the complete expansion of $10!$.

So, the expression becomes:

$\frac{10!}{6!}$

Therefore, the expression $10 \times 9 \times 8 \times 7$ written in terms of factorials is $\frac{10!}{6!}$.

Comparing this result with the given options, we find that it matches option (A).

Hence, the correct option is (A) $\frac{10!}{6!}$.

Analysis of Assertion and Reason:

We need to evaluate both the Assertion (A) and the Reason (R) to determine the correct option.

Assertion (A): $5! + 4! = 9!$

To check if the assertion is true, we will calculate the values of the Left Hand Side (LHS) and the Right Hand Side (RHS) of the equation.

Left Hand Side (LHS) = $5! + 4!$

First, we calculate the values of the individual factorials:

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

$4! = 4 \times 3 \times 2 \times 1 = 24$

Now, we add them:

LHS = $120 + 24 = 144$

Right Hand Side (RHS) = $9!$

Now, we calculate the value of $9!$:

$9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$

$9! = 362,880$

Comparison:

LHS = 144 and RHS = 362,880.

Since $144 \neq 362,880$, the assertion $5! + 4! = 9!$ is false.

Reason (R): Factorial represents the product of integers from 1 to the number.

This statement gives the definition of a factorial. For any positive integer $n$, the factorial of $n$, denoted by $n!$, is defined as:

$n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1$

This is the product of all positive integers from 1 to $n$. Therefore, the Reason (R) is a true statement.

Conclusion:

We have determined that Assertion (A) is false and Reason (R) is true.

Let's look at the options:

(A) Both A and R are true and R is the correct explanation of A. (Incorrect, A is false)

(B) Both A and R are true but R is not the correct explanation of A. (Incorrect, A is false)

(C) A is false but R is true. (Correct)

(D) Both A and R are false. (Incorrect, R is true)

Hence, the correct option is (C) A is false but R is true.

Solution:

We need to evaluate the sum: $\frac{1}{8!} + \frac{1}{9!} + \frac{1}{10!}$

To add these fractions, we first need to find a common denominator. The denominators are $8!$, $9!$, and $10!$. The least common multiple (LCM) of these is the largest factorial among them, which is $10!$.

We will convert each fraction to an equivalent fraction with the denominator $10!$.

Step 1: Convert the first term $\frac{1}{8!}$

We know that $10! = 10 \times 9 \times 8!$. So, we multiply the numerator and denominator of $\frac{1}{8!}$ by $10 \times 9 = 90$.

$\frac{1}{8!} = \frac{1 \times (10 \times 9)}{8! \times (10 \times 9)} = \frac{90}{10!}$

Step 2: Convert the second term $\frac{1}{9!}$

We know that $10! = 10 \times 9!$. So, we multiply the numerator and denominator of $\frac{1}{9!}$ by $10$.

$\frac{1}{9!} = \frac{1 \times 10}{9! \times 10} = \frac{10}{10!}$

Step 3: The third term $\frac{1}{10!}$

This term already has the common denominator.

Step 4: Add the fractions

Now we can add the three equivalent fractions:

$\frac{1}{8!} + \frac{1}{9!} + \frac{1}{10!} = \frac{90}{10!} + \frac{10}{10!} + \frac{1}{10!}$

Since the denominators are the same, we add the numerators:

$\frac{90 + 10 + 1}{10!} = \frac{101}{10!}$

Thus, the value of the expression is $\frac{101}{10!}$.

Comparing this result with the given options, we find that it matches option (B).

Hence, the correct option is (B) $\frac{101}{10!}$.

Solution:

The question asks to complete the definition of a factorial for a non-negative integer.

The definition of the factorial of a positive integer $n$ is given by:

$n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1$

The definition is extended to include the non-negative integer 0. This is a special case that is defined by mathematical convention.

By definition, the value of zero factorial, denoted as $0!$, is 1.

$0! = 1$

Justification for the definition:

This definition is not arbitrary; it is chosen to be consistent with other mathematical properties, particularly the recursive relationship for factorials:

$n! = n \times (n-1)!$

If we let $n=1$, this relationship becomes:

$1! = 1 \times (1-1)!$

$1! = 1 \times 0!$

Since we know that $1! = 1$, we can substitute this value into the equation:

$1 = 1 \times 0!$

Dividing both sides by 1 gives:

$0! = 1$

Therefore, to maintain the consistency of the recursive factorial formula, $0!$ must be defined as 1.

Completing the statement in the question:

The factorial of a non-negative integer $n$, denoted by $n!$, is the product of all positive integers less than or equal to $n$, with the special case that $0! = 1$.

Comparing this with the given options, we find that the correct value to fill in the blank is 1, which corresponds to option (B).

Hence, the correct option is (B) 1.

Solution:

We need to evaluate the expression $(3!)^2$.

This expression requires us to first calculate the value of the term inside the parentheses, which is $3!$, and then square the result.

Step 1: Calculate the value of $3!$

The factorial of a number $n$, denoted as $n!$, is the product of all positive integers from 1 to $n$.

So, for $3!$, we have:

$3! = 3 \times 2 \times 1$

$3! = 6$

Step 2: Square the result from Step 1

Now we substitute the value of $3!$ back into the original expression:

$(3!)^2 = (6)^2$

Squaring 6 means multiplying it by itself:

$(6)^2 = 6 \times 6 = 36$

Thus, the value of $(3!)^2$ is 36.

Comparing this result with the given options:

(A) 6

(B) 9

(C) 36

(D) 216

The calculated value matches option (C).

Hence, the correct option is (C) 36.

Given:

The equation involving a positive integer $n$ is:

To Find:

The value of $n$.

Solution:

The given equation is $\frac{n!}{2!(n-2)!} = 28$.

First, let's simplify the factorial expression. We can expand $n!$ as:

$n! = n \times (n-1) \times (n-2) \times \dots \times 1$

This can be written in terms of $(n-2)!$ as:

$n! = n \times (n-1) \times (n-2)!$

Also, we know that $2! = 2 \times 1 = 2$.

Now, substitute these expressions back into the equation (i):

$\frac{n \times (n-1) \times (n-2)!}{2 \times (n-2)!} = 28$

We can cancel the common term $(n-2)!$ from the numerator and the denominator, provided $n \geq 2$.

$\frac{n \times (n-1)}{2} = 28$

Now, we solve for $n$. Multiply both sides by 2:

$n(n-1) = 28 \times 2$

$n(n-1) = 56$

Expand the left side to form a quadratic equation:

$n^2 - n = 56$

$n^2 - n - 56 = 0$

We can solve this quadratic equation by factoring. We need to find two numbers that multiply to -56 and add up to -1. These numbers are -8 and 7.

$(n-8)(n+7) = 0$

This gives two possible solutions for $n$:

$n - 8 = 0 \implies n = 8$

$n + 7 = 0 \implies n = -7$

Since the question states that $n$ is a positive integer, we discard the negative solution $n = -7$.

Thus, the value of $n$ is 8.

Alternate Solution (Testing the Options):

The given expression $\frac{n!}{2!(n-2)!}$ is the formula for combinations, $\binom{n}{2}$, which simplifies to $\frac{n(n-1)}{2}$. We need to find which option for $n$ makes this expression equal to 28.

Let's test each option:

(A) n = 7:

$\frac{7(7-1)}{2} = \frac{7 \times 6}{2} = \frac{42}{2} = 21$. This is not 28.

(B) n = 8:

$\frac{8(8-1)}{2} = \frac{8 \times 7}{2} = \frac{56}{2} = 28$. This matches the given value.

(C) n = 9:

$\frac{9(9-1)}{2} = \frac{9 \times 8}{2} = \frac{72}{2} = 36$. This is not 28.

(D) n = 10:

$\frac{10(10-1)}{2} = \frac{10 \times 9}{2} = \frac{90}{2} = 45$. This is not 28.

By testing the options, we find that $n=8$ is the correct solution.

Both methods confirm that the value of $n$ is 8. Comparing this with the options, it matches option (B).

Hence, the correct option is (B) 8.

Solution:

The case study requires the student to calculate the value of $6!$ (read as "6 factorial").

The factorial of a positive integer $n$, denoted by $n!$, is the product of all positive integers from 1 up to $n$.

The formula for n-factorial is:

$n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1$

To calculate $6!$, we apply this formula for $n=6$:

$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$

Now, we perform the multiplication:

$6 \times 5 = 30$

$30 \times 4 = 120$

$120 \times 3 = 360$

$360 \times 2 = 720$

$720 \times 1 = 720$

So, the value of $6!$ is 720.

Thus, the correct value for the student's calculation is 720.

Comparing this result with the given options:

(A) 720

(B) 360

(C) 120

(D) 60

The calculated value matches option (A).

Hence, the correct option is (A) 720.

Solution:

The problem asks us to simplify the expression $\frac{n!}{(n-1)!}$ for the specific value $n=5$.

First, let's simplify the general expression $\frac{n!}{(n-1)!}$.

We know the recursive definition of a factorial is $n! = n \times (n-1)!$. This allows us to express a factorial in terms of the factorial of the next smaller integer.

We can substitute this definition for $n!$ into the numerator of the given expression:

$\frac{n!}{(n-1)!} = \frac{n \times (n-1)!}{(n-1)!}$

Now, we can cancel the common term $(n-1)!$ from both the numerator and the denominator:

$\frac{n \times \cancel{(n-1)!}}{\cancel{(n-1)!}} = n$

So, the general expression simplifies to just $n$.

The question requires us to find the value for $n=5$. Substituting this into our simplified result, we get:

Value = 5

Alternate Solution:

We can first substitute $n=5$ into the expression and then evaluate it.

The expression becomes:

$\frac{5!}{(5-1)!} = \frac{5!}{4!}$

Now, we can expand the factorial in the numerator until we reach the factorial in the denominator:

$5! = 5 \times 4!$

Substituting this back into the fraction:

$\frac{5 \times 4!}{4!}$

By canceling the $4!$ term, we get:

$\frac{5 \times \cancel{4!}}{\cancel{4!}} = 5$

Both methods confirm that the simplified value is 5.

Comparing this with the given options, it matches option (A).

Hence, the correct option is (A) 5.

Solution:

The question asks to identify the fundamental concept used for calculating the total number of possible outcomes when a series of choices are made.

Let's analyze the given options:

• (A) Factorial calculation: A factorial, $n!$, is used to calculate the number of ways to arrange (permute) a set of $n$ distinct objects. While this is a type of counting, it's a specific case and not the general principle for any series of choices.

• (B) Fundamental Principle of Counting: This principle (also known as the multiplication rule) states that if an event can occur in $m$ ways, and following it, a second event can occur in $n$ ways, then the total number of ways the two events can occur in sequence is $m \times n$. This principle directly addresses how to calculate the total number of outcomes from a series of choices. For example, if you have 3 shirts and 2 pairs of pants, the total number of outfits is $3 \times 2 = 6$. This perfectly matches the description in the question.

• (C) Combination calculation: A combination is a way of selecting items from a collection, such that the order of selection does not matter. It answers the question "How many ways can I choose $k$ items from a set of $n$ items?". This is about selection, not necessarily a sequence of choices leading to a final outcome.

• (D) Average calculation: This is a statistical method to find a central value of a set of numbers and is unrelated to counting the number of possible outcomes in a series of events.

Based on the analysis, the Fundamental Principle of Counting is the general rule for determining the total number of outcomes when making a series of choices.

Hence, the correct option is (B) Fundamental Principle of Counting.

Given:

A customer is choosing a meal from a menu with the following options:

• Number of choices for a starter = 3
• Number of choices for a main course = 5
• Number of choices for a dessert = 2

The customer must choose one item from each category.

To Find:

The total number of different ways a customer can choose a complete meal.

Solution:

This problem involves a sequence of independent choices. We can use the Fundamental Principle of Counting (also known as the multiplication principle) to find the total number of possible outcomes.

The principle states that if an event can occur in 'm' ways, and a second independent event can occur in 'n' ways, and a third independent event can occur in 'p' ways, then the total number of ways that all events can occur in sequence is the product of the number of ways for each event.

The selection of a meal consists of three independent events:

1. Choosing a starter: There are 3 ways to do this.
2. Choosing a main course: There are 5 ways to do this.
3. Choosing a dessert: There are 2 ways to do this.

To find the total number of different meal combinations, we multiply the number of choices for each part of the meal:

Total number of ways = (Number of starters) $\times$ (Number of main courses) $\times$ (Number of desserts)

Total number of ways = $3 \times 5 \times 2$

Let's perform the multiplication:

Total number of ways = $15 \times 2$

Total number of ways = $30$

Therefore, there are 30 different ways a customer can choose a meal consisting of one starter, one main course, and one dessert.

Comparing this result with the given options:

(A) 10

(B) 15

(C) 30

(D) 60

The calculated value matches option (C).

Hence, the correct option is (C) 30.

Given:

The set of available digits is {1, 2, 3, 4, 5}.

We need to form 2-digit numbers.

The condition is that repetition of digits is allowed.

To Find:

The total number of 2-digit numbers that can be formed under the given conditions.

Solution:

To form a 2-digit number, we need to fill two places: the tens place and the units place.

We can use the Fundamental Principle of Counting to solve this problem.

Step 1: Filling the tens place

The digit for the tens place can be chosen from any of the available digits: {1, 2, 3, 4, 5}.

So, the number of ways to fill the tens place is 5.

Step 2: Filling the units place

The digit for the units place can also be chosen from the available digits: {1, 2, 3, 4, 5}.

Since repetition of digits is allowed, the choice for the units place is independent of the choice for the tens place. We still have 5 options.

So, the number of ways to fill the units place is 5.

Step 3: Calculating the total number of 2-digit numbers

According to the Fundamental Principle of Counting, the total number of 2-digit numbers is the product of the number of choices for each place.

Total number of ways = (Number of choices for tens place) $\times$ (Number of choices for units place)

Total number of ways = $5 \times 5$

Total number of ways = $25$

Therefore, 25 different 2-digit numbers can be formed.

Comparing this result with the given options:

(A) 10

(B) 20

(C) 25

(D) 50

The calculated value matches option (C).

Hence, the correct option is (C) 25.

Given:

The set of available digits is {1, 2, 3, 4, 5}.

We need to form 3-digit numbers.

The condition is that repetition of digits is NOT allowed.

To Find:

The total number of 3-digit numbers that can be formed under the given conditions.

Solution:

To form a 3-digit number, we need to fill three places: the hundreds place, the tens place, and the units place. We will use the Fundamental Principle of Counting.

Step 1: Filling the hundreds place

We can choose any of the 5 available digits {1, 2, 3, 4, 5} for the hundreds place.

Number of choices for the hundreds place = 5.

Step 2: Filling the tens place

Since repetition of digits is not allowed, the digit used for the hundreds place cannot be used again. This leaves us with $5 - 1 = 4$ remaining digits.

Number of choices for the tens place = 4.

Step 3: Filling the units place

Now, two distinct digits have been used for the hundreds and tens places. This leaves us with $5 - 2 = 3$ remaining digits.

Number of choices for the units place = 3.

Step 4: Calculating the total number of 3-digit numbers

According to the Fundamental Principle of Counting, the total number of 3-digit numbers is the product of the number of choices for each place.

Total number of ways = (Choices for hundreds place) $\times$ (Choices for tens place) $\times$ (Choices for units place)

Total number of ways = $5 \times 4 \times 3$

Total number of ways = $20 \times 3$

Total number of ways = $60$

Alternate Solution (Using Permutations):

This problem is equivalent to finding the number of arrangements (permutations) of 5 distinct items taken 3 at a time. The formula for permutations is:

$P(n, r) = \frac{n!}{(n-r)!}$

Here, $n=5$ (the number of available digits) and $r=3$ (the number of digits in the number to be formed).

$P(5, 3) = \frac{5!}{(5-3)!} = \frac{5!}{2!}$

Expanding the factorials:

$P(5, 3) = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = \frac{120}{2} = 60$

Alternatively, we can simplify the fraction before multiplying:

$P(5, 3) = \frac{5 \times 4 \times 3 \times \cancel{2!}}{\cancel{2!}} = 5 \times 4 \times 3 = 60$

Both methods show that 60 different 3-digit numbers can be formed.

Comparing this result with the given options, it matches option (A).

Hence, the correct option is (A) 60.

Given:

• Telephone numbers consist of 7 digits.
• The first two digits are fixed as 25.
• The remaining 5 digits can be any digit from 0 to 9.
• Repetition of digits is allowed for the remaining 5 places.

To Find:

The total number of possible telephone numbers that can be formed.

Solution:

A 7-digit telephone number can be represented by 7 places:

_ _ _ _ _ _ _

We are given that the first two digits are fixed. The first digit must be 2, and the second digit must be 5.

So, the number of choices for the first place is 1 (only the digit 2).

The number of choices for the second place is 1 (only the digit 5).

The structure of the number is: 2 5 _ _ _ _ _

We need to determine the number of possibilities for the remaining 5 places (from the 3rd to the 7th digit).

The digits available for these places are {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, which means there are 10 possible choices for each place.

We can use the Fundamental Principle of Counting to find the total number of combinations.

• Choices for the 3rd digit: 10 (any from 0-9)
• Choices for the 4th digit: 10 (repetition is allowed)
• Choices for the 5th digit: 10 (repetition is allowed)
• Choices for the 6th digit: 10 (repetition is allowed)
• Choices for the 7th digit: 10 (repetition is allowed)

The total number of possible telephone numbers is the product of the number of choices for each of the 5 variable positions.

Total possible numbers = $10 \times 10 \times 10 \times 10 \times 10$

This can be written in exponential form as:

Total possible numbers = $10^5$

We can also write this as 100,000.

Therefore, there are $10^5$ possible telephone numbers.

Comparing this result with the given options:

(A) $10^5$

(B) $10^7$

(C) $8^5$

(D) $2 \times 10^5$

The calculated value matches option (A).

Hence, the correct option is (A) $10^5$.

Analysis of Assertion and Reason:

We need to evaluate both the Assertion (A) and the Reason (R) and determine the relationship between them.

Assertion (A): The number of ways to flip a coin and roll a standard six-sided die is $2 \times 6 = 12$.

Let's break down the process into two independent events:

1. Event 1: Flipping a coin.

   A standard coin has two possible outcomes: Heads (H) or Tails (T). So, there are 2 ways for this event to occur.

2. Event 2: Rolling a standard six-sided die.

   A standard die has six faces, numbered 1, 2, 3, 4, 5, 6. So, there are 6 possible outcomes for this event.

To find the total number of combined outcomes when both events happen, we multiply the number of outcomes for each event. This is an application of the Fundamental Principle of Counting.

Total number of ways = (Number of outcomes for coin flip) $\times$ (Number of outcomes for die roll)

Total number of ways = $2 \times 6 = 12$.

The statement in the assertion is correct. Thus, Assertion (A) is true.

Reason (R): The Fundamental Principle of Counting states that if one event can occur in $m$ ways and another event can occur in $n$ ways, then both events can occur in $m \times n$ ways.

This statement is the precise definition of the Fundamental Principle of Counting (also known as the multiplication principle). Therefore, the Reason (R) is a true statement.

Conclusion:

We have established that both Assertion (A) and Reason (R) are true. Now we must check if R is the correct explanation for A.

The calculation in Assertion (A) is a direct application of the principle described in Reason (R). The first event (flipping a coin) has $m=2$ outcomes, and the second event (rolling a die) has $n=6$ outcomes. The total number of ways is $m \times n = 2 \times 6 = 12$.

Therefore, Reason (R) is indeed the correct explanation for Assertion (A).

This corresponds to option (A).

Hence, the correct option is (A) Both A and R are true and R is the correct explanation of A.

Given:

• There are 5 different coloured flags available.
• A signal is generated by arranging two flags, one above the other.

The arrangement "one above the other" implies that the order of the flags matters (e.g., Red above Blue is a different signal from Blue above Red). Also, since the flags are distinct, once a flag is used in one position, it cannot be used in the other position (no repetition).

To Find:

The total number of different signals that can be generated.

Solution:

We can solve this using the Fundamental Principle of Counting. A signal consists of two positions that need to be filled with flags:

1. The top flag position.
2. The bottom flag position.

Step 1: Filling the top flag position

We have 5 different coloured flags to choose from for the top position.

Number of choices for the top position = 5.

Step 2: Filling the bottom flag position

After one flag has been chosen for the top position, it cannot be used again. This leaves us with $5 - 1 = 4$ remaining flags.

Number of choices for the bottom position = 4.

Step 3: Calculating the total number of signals

The total number of different signals is the product of the number of choices for each position.

Total number of signals = (Choices for top position) $\times$ (Choices for bottom position)

Total number of signals = $5 \times 4$

Total number of signals = $20$

Alternate Solution (Using Permutations):

This problem is about finding the number of arrangements (permutations) of 2 flags chosen from a set of 5. The order matters, so we use the permutation formula:

$P(n, r) = \frac{n!}{(n-r)!}$

Here, $n=5$ (the total number of different flags) and $r=2$ (the number of flags used in each signal).

$P(5, 2) = \frac{5!}{(5-2)!} = \frac{5!}{3!}$

We can expand the numerator to simplify:

$P(5, 2) = \frac{5 \times 4 \times 3!}{3!} = \frac{5 \times 4 \times \cancel{3!}}{\cancel{3!}}$

$P(5, 2) = 5 \times 4 = 20$

Both methods show that 20 different signals can be generated.

Comparing this result with the given options, it matches option (C).

Hence, the correct option is (C) 20.

Solution:

The question asks to complete the definition of the Fundamental Principle of Counting, which is also known as the Multiplication Principle.

This principle is used to find the total number of outcomes when two or more events occur in sequence.

The principle states:

If a first event can occur in $m$ different ways, and after it has occurred, a second event can occur in $n$ different ways, then the total number of ways that the two events can occur in the specified order is the product of the number of ways for each event.

Total number of ways = $m \times n$

For example, if a person has 3 different shirts and 2 different pairs of pants, the total number of different outfits they can wear is found by multiplying the number of choices for each item:

Total outfits = (Number of shirts) $\times$ (Number of pants) = $3 \times 2 = 6$.

The sum $m+n$ would correspond to the Addition Principle, which applies when a choice is made between two mutually exclusive sets of options (e.g., choosing one item from a set of $m$ OR one item from a set of $n$).

Based on the definition, the statement should be completed as:

If an event can occur in $m$ different ways, and if following this event, another event can occur in $n$ different ways, then the total number of ways for both events to occur in the given order is $m \times n$.

Comparing this with the given options, it matches option (C).

Hence, the correct option is (C) $m \times n$.

Given:

• The set of available digits is {1, 2, 3, 4, 5, 6}.
• We need to form 3-digit numbers.
• The numbers must be even.
• Repetition of digits is allowed.

To Find:

The total number of even 3-digit numbers that can be formed.

Solution:

To solve this, we will use the Fundamental Principle of Counting. A 3-digit number has three places: the hundreds place, the tens place, and the units place.

The main constraint is that the number must be even. A number is even if its units digit is an even number. When there is a constraint on a particular digit's place, it's best to fill that place first.

Step 1: Filling the units place

The number must be even, so the digit in the units place must be an even digit. From the given set {1, 2, 3, 4, 5, 6}, the even digits are {2, 4, 6}.

So, there are 3 possible choices for the units place.

Step 2: Filling the hundreds place

The problem states that repetition of digits is allowed. This means we can use any of the available digits for the hundreds place, regardless of what was chosen for the units place.

The available digits are {1, 2, 3, 4, 5, 6}.

So, there are 6 possible choices for the hundreds place.

Step 3: Filling the tens place

Similar to the hundreds place, repetition is allowed, so we can use any of the 6 available digits for the tens place.

So, there are 6 possible choices for the tens place.

Step 4: Calculating the total number of even 3-digit numbers

According to the Fundamental Principle of Counting, the total number of possible numbers is the product of the number of choices for each place.

Total numbers = (Choices for hundreds place) $\times$ (Choices for tens place) $\times$ (Choices for units place)

Total numbers = $6 \times 6 \times 3$

Total numbers = $36 \times 3$

Total numbers = $108$

Therefore, 108 even 3-digit numbers can be formed using the given digits with repetition allowed.

Comparing this result with the given options, it matches option (A).

Hence, the correct option is (A) 108.

Solution:

The problem asks for the number of ways to arrange 5 distinct books on a shelf. This is a problem of finding the total number of permutations of 5 items.

The number of ways to arrange 'n' distinct objects in a sequence is given by n-factorial, denoted as $n!$.

In this case, we have $n = 5$ distinct books.

So, the total number of arrangements is $5!$.

Let's calculate the value of $5!$:

$5! = 5 \times 4 \times 3 \times 2 \times 1$

$5! = 20 \times 3 \times 2 \times 1$

$5! = 60 \times 2 \times 1$

$5! = 120 \times 1$

$5! = 120$

Alternate Solution (Using the Fundamental Principle of Counting):

We can think of this as filling 5 positions on the shelf with 5 distinct books.

• For the first position, we have 5 choices of books.
• After placing one book, we have 4 books left. For the second position, we have 4 choices.
• For the third position, we have 3 choices.
• For the fourth position, we have 2 choices.
• For the last position, we have 1 choice.

The total number of arrangements is the product of the number of choices for each position:

Total arrangements = $5 \times 4 \times 3 \times 2 \times 1 = 120$

Both methods show that there are 120 ways to arrange the 5 distinct books.

Comparing this result with the given options:

(A) 5

(B) 25

(C) 120

(D) 625

The calculated value matches option (C).

Hence, the correct option is (C) 120.

Solution:

The question asks for the standard formula for calculating the number of permutations of $n$ distinct objects taken $r$ at a time.

A permutation is an arrangement of objects in a specific order. The number of permutations of $n$ distinct objects taken $r$ at a time, denoted by ${}_nP_r$ or $P(n, r)$, represents the number of different ways to choose and arrange $r$ objects from a set of $n$ distinct objects.

The derivation of the formula using the Fundamental Principle of Counting is as follows:

• There are $n$ choices for the 1st position.
• There are $(n-1)$ choices for the 2nd position.
• ...
• There are $(n-r+1)$ choices for the $r$-th position.

So, ${}_nP_r = n \times (n-1) \times \dots \times (n-r+1)$.

To express this in a compact factorial form, we multiply and divide by $(n-r)!$:

${}_nP_r = \frac{n \times (n-1) \times \dots \times (n-r+1) \times (n-r)!}{(n-r)!}$

The numerator is now the full expansion of $n!$.

Thus, the formula is:

${}_nP_r = \frac{n!}{(n-r)!}$

Let's analyze the given options:

• (A) ${}_nP_r = \frac{n!}{(n-r)!}$: This is the correct standard formula for permutations.
• (B) ${}_nP_r = \frac{n!}{r!(n-r)!}$: This is the formula for combinations (${}_nC_r$), where the order of selection does not matter.
• (C) ${}_nP_r = \frac{(n-r)!}{n!}$: This is the reciprocal of the correct formula and is incorrect.
• (D) ${}_nP_r = n!$: This is the formula for the number of permutations of $n$ objects taken all $n$ at a time (i.e., when $r=n$), not the general formula for taking $r$ at a time.

Therefore, the correct formula is the one presented in option (A).

Hence, the correct option is (A) ${}_nP_r = \frac{n!}{(n-r)!}$.

Solution:

We need to evaluate the expression ${}_7P_3$, which represents the number of permutations of 7 distinct objects taken 3 at a time.

The formula for the number of permutations of $n$ distinct objects taken $r$ at a time is:

${}_nP_r = \frac{n!}{(n-r)!}$

In this problem, we have $n=7$ and $r=3$.

Substituting these values into the formula:

${}_7P_3 = \frac{7!}{(7-3)!}$

${}_7P_3 = \frac{7!}{4!}$

Now, we can expand the numerator $7!$ as $7 \times 6 \times 5 \times 4!$ to simplify the expression:

${}_7P_3 = \frac{7 \times 6 \times 5 \times 4!}{4!}$

We can cancel the common term $4!$ from the numerator and the denominator:

${}_7P_3 = 7 \times 6 \times 5$

Performing the multiplication:

${}_7P_3 = 42 \times 5$

${}_7P_3 = 210$

Alternate Solution (Using the Counting Principle):

We need to arrange 3 items chosen from 7. This is equivalent to filling 3 positions:

• For the first position, we have 7 choices.
• Since we cannot repeat items, for the second position, we have 6 choices remaining.
• For the third position, we have 5 choices remaining.

The total number of arrangements is the product of the choices for each position:

Total arrangements = $7 \times 6 \times 5 = 210$

Both methods yield the same result.

Comparing our result with the given options:

(A) 35

(B) 210

(C) 840

(D) 5040

The calculated value matches option (B).

Hence, the correct option is (B) 210.

Solution:

We need to find the number of different ways the letters of the word "INDIA" can be arranged. This is a problem of permutations with repetitions because the letter 'I' appears more than once.

Step 1: Analyze the letters in the word "INDIA"

• The total number of letters in the word is 5. So, $n = 5$.
• Let's count the frequency of each letter:
  • 'I' appears 2 times
  • 'N' appears 1 time
  • 'D' appears 1 time
  • 'A' appears 1 time

Step 2: Use the formula for permutations with repetitions

The formula for the number of distinct permutations of $n$ objects where there are $p_1$ identical items of one kind, $p_2$ of a second kind, ..., and $p_k$ of a k-th kind is:

$\frac{n!}{p_1! \times p_2! \times \dots \times p_k!}$

Step 3: Apply the formula to our problem

In the word "INDIA":

• $n = 5$ (total letters)
• The letter 'I' is repeated 2 times, so we have one repeating group with $p_1 = 2$.

The number of distinct arrangements is:

Number of arrangements = $\frac{n!}{p_1!} = \frac{5!}{2!}$

Step 4: Calculate the value

First, we evaluate the factorials:

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

$2! = 2 \times 1 = 2$

Now, we perform the division:

Number of arrangements = $\frac{120}{2} = 60$

Therefore, there are 60 different ways to arrange the letters of the word "INDIA".

Comparing this result with the given options:

(A) 120

(B) 60

(C) 30

(D) 24

The calculated value matches option (B).

Hence, the correct option is (B) 60.

Solution:

We need to find the number of different arrangements (permutations) of the letters of the word "ALLEHABAD". This is a problem of permutations with repetitions, as some letters appear more than once.

Step 1: Analyze the letters in the word "ALLEHABAD"

First, we count the total number of letters and the frequency of each repeating letter in the word.

The word is A-L-L-E-H-A-B-A-D.

• Total number of letters (n) = 9.
• The letter 'A' appears 3 times.
• The letter 'L' appears 2 times.
• The other letters (E, H, B, D) appear once.

Step 2: Use the formula for permutations with repetitions

The formula for the number of distinct permutations of $n$ objects where there are $p_1$ identical items of one kind, $p_2$ of a second kind, etc., is:

Number of arrangements = $\frac{n!}{p_1! \times p_2! \times \dots}$

Step 3: Apply the formula to the given word

Using the counts from Step 1, the correct formula for the word "ALLEHABAD" is:

Number of arrangements = $\frac{9!}{3! \times 2!}$

Note on the Question:

The correctly derived expression, $\frac{9!}{3! \times 2!}$, is not present in the given options. This suggests there might be a typographical error in the question's word or its options.

A common version of this problem uses the standard spelling of the city, "ALLAHABAD". Let's analyze this word:

• Word: A-L-L-A-H-A-B-A-D
• Total letters (n) = 9
• 'A' appears 4 times.
• 'L' appears 2 times.

For the word "ALLAHABAD", the correct formula would be:

Number of arrangements = $\frac{9!}{4! \times 2!}$

This expression is also not among the options.

Given the choices, there is a possibility that the question intends for the solver to account only for the most frequent repeating letter in "ALLAHABAD", which is 'A' (4 times), and to ignore the repetition of 'L'. While this is mathematically incorrect, it would lead to one of the options.

Under this flawed assumption (using "ALLAHABAD" and ignoring the 'L's), the expression would be:

Number of arrangements $\approx \frac{9!}{4!}$

This matches option (B). Due to the clear error in the question's construction, this is the most likely intended, albeit incorrect, answer.

Hence, based on the assumption of a common type of error in the problem statement, the intended option is (B) $\frac{9!}{4!}$.

Solution:

The problem asks for the number of ways 6 distinct students can be seated in a row. This is a classic permutation problem, as the order in which the students are seated matters.

The number of ways to arrange $n$ distinct objects is given by n-factorial, denoted as $n!$.

In this case, we have $n = 6$ students.

So, the total number of seating arrangements is $6!$.

Now, let's calculate the value of $6!$:

$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$

$6! = 30 \times 4 \times 3 \times 2 \times 1$

$6! = 120 \times 3 \times 2 \times 1$

$6! = 360 \times 2 \times 1$

$6! = 720 \times 1$

$6! = 720$

Alternate Solution (Using the Fundamental Principle of Counting):

We can think of this as filling 6 seats in a row.

• For the first seat, there are 6 possible students who can sit.
• Once the first seat is filled, there are 5 remaining students for the second seat.
• For the third seat, there are 4 remaining students.
• For the fourth seat, there are 3 remaining students.
• For the fifth seat, there are 2 remaining students.
• For the last seat, there is only 1 student left.

The total number of ways is the product of the number of choices for each seat:

Total arrangements = $6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$

Both methods show that there are 720 ways to seat the 6 students.

Comparing this result with the given options:

(A) 6

(B) 36

(C) 720

(D) 6! + 6

The calculated value matches option (C).

Hence, the correct option is (C) 720.

Analysis of Assertion and Reason:

We need to evaluate both the Assertion (A) and the Reason (R) and determine their relationship.

Assertion (A): Permutations are used when the order of arrangement matters.

This statement is the fundamental definition of a permutation. A permutation is an arrangement of a set of objects in a specific sequence. For example, the arrangement 'ABC' is a different permutation from 'BCA', even though they use the same letters. This is contrasted with combinations, where the order does not matter. Therefore, Assertion (A) is true.

Reason (R): The formula ${}_nP_r$ accounts for the different possible orders of the selected items.

The formula for permutations, ${}_nP_r = \frac{n!}{(n-r)!}$, is derived by considering the number of choices for each position in an arrangement of $r$ items. The calculation $n \times (n-1) \times \dots \times (n-r+1)$ inherently counts each unique sequence as a distinct outcome. For instance, selecting A then B is counted separately from selecting B then A. Thus, the formula is specifically designed to count ordered arrangements. Therefore, Reason (R) is true.

Conclusion:

Both Assertion (A) and Reason (R) are true statements.

Furthermore, Reason (R) provides the mathematical justification for the concept described in Assertion (A). Assertion (A) states a conceptual fact (permutations are about order), and Reason (R) explains that the mathematical tool used for permutations (${}_nP_r$) is structured precisely to implement this concept of order.

Therefore, Reason (R) is the correct explanation for Assertion (A).

This corresponds to option (A).

Hence, the correct option is (A) Both A and R are true and R is the correct explanation of A.

Given:

• There are 8 runners participating in a race.
• Medals are awarded for three distinct places: first, second, and third.

This means the order in which the runners finish matters. For example, Runner A getting first and Runner B getting second is a different outcome from Runner B getting first and Runner A getting second.

To Find:

The total number of different ways the first, second, and third place medals can be distributed among the 8 runners.

Solution (Using the Fundamental Principle of Counting):

We can think of this as filling three positions (1st, 2nd, and 3rd place) from the available 8 runners.

Step 1: Filling the first place

Any of the 8 runners can win the first place medal.

Number of choices for first place = 8.

Step 2: Filling the second place

Once the first place winner is decided, that runner is no longer available for second place. This leaves $8 - 1 = 7$ runners.

Number of choices for second place = 7.

Step 3: Filling the third place

After the first and second places are filled, there are $8 - 2 = 6$ runners remaining.

Number of choices for third place = 6.

Step 4: Calculating the total number of ways

The total number of different ways to distribute the medals is the product of the number of choices for each place:

Total ways = (Choices for 1st) $\times$ (Choices for 2nd) $\times$ (Choices for 3rd)

Total ways = $8 \times 7 \times 6$

Total ways = $56 \times 6$

Total ways = $336$

Alternate Solution (Using Permutations):

Since the order of the top three runners matters, this is a permutation problem. We need to find the number of permutations of 8 objects taken 3 at a time, denoted as ${}_8P_3$.

The formula for permutations is ${}_nP_r = \frac{n!}{(n-r)!}$.

Here, $n = 8$ (the total number of runners) and $r = 3$ (the number of medal positions).

${}_8P_3 = \frac{8!}{(8-3)!} = \frac{8!}{5!}$

Expanding the numerator:

${}_8P_3 = \frac{8 \times 7 \times 6 \times 5!}{5!}$

Canceling the $5!$ term:

${}_8P_3 = 8 \times 7 \times 6 = 336$

Both methods confirm that there are 336 different ways to distribute the medals.

Comparing this with the given options, it matches option (D).

Hence, the correct option is (D) 336.

Analysis of the Case Study Problem:

The problem presents a specific scenario or condition and then asks a question based on it.

Given Condition:

• There are 8 runners in total.
• A specific group of 5 runners (the Indian runners) finish in the top 5 positions.

To Find:

The number of different ways the top 5 positions can be filled, given the condition.

Solution:

The problem states that the top 5 positions are filled by the 5 Indian runners. This condition restricts the set of runners we are considering for the top 5 places. Instead of choosing from all 8 runners, we are only concerned with the arrangement of the 5 specific Indian runners within the top 5 spots.

The task is to find the number of ways to arrange these 5 distinct runners in the 5 distinct positions (first, second, third, fourth, and fifth place).

This is a permutation of 5 distinct objects taken all 5 at a time, which is calculated as $5!$.

Let's calculate the value of $5!$:

$5! = 5 \times 4 \times 3 \times 2 \times 1$

$5! = 120$

The question asks for the expression that represents this calculation. The number of ways to arrange 5 items is $5!$.

Analysis of the Options:

• (A) ${}_8P_5$: This would be the answer if the question were "How many ways can the top 5 positions be filled by any of the 8 runners?", without the condition about the Indian runners.
• (B) ${}_8C_5$: This calculates the number of ways to choose a group of 5 runners, but it does not account for the order of finishing, which is essential for race results.
• (C) 5!: This correctly represents the number of ways to arrange the 5 specific Indian runners in the top 5 positions, which is the scenario described by the problem's condition.
• (D) 8!: This represents the total number of ways all 8 runners can be arranged, not just the top 5.

Given the condition in the problem, the correct approach is to find the number of arrangements for the 5 specified runners.

Hence, the correct option is (C) 5!.

Solution:

The question asks to complete a fundamental statement about permutations.

The phrase "the number of permutations of $n$ distinct objects" is standard terminology for the number of ways to arrange all $n$ of those distinct objects in a sequence.

This is a specific case of the general permutation formula, ${}_nP_r$, where we are choosing to arrange all $r=n$ objects from the available set of $n$.

So, the correct notation for this operation is ${}_nP_n$.

Verification using the formula:

The formula for permutations is ${}_nP_r = \frac{n!}{(n-r)!}$.

When we are arranging all $n$ objects, we set $r=n$. Substituting this into the formula gives:

${}_nP_n = \frac{n!}{(n-n)!}$

${}_nP_n = \frac{n!}{0!}$

Since by definition $0! = 1$, the expression becomes:

${}_nP_n = \frac{n!}{1} = n!$

This confirms that ${}_nP_n$ is the correct symbolic representation for the number of permutations of $n$ distinct objects, which equals $n!$.

Analysis of Options:

• (A) ${}_nP_1$: This represents arranging 1 object out of $n$, which equals $n$. Incorrect.
• (B) ${}_nP_n$: This represents arranging all $n$ objects out of $n$, which equals $n!$. This is the correct representation.
• (C) ${}_nC_n$: This represents choosing (without order) all $n$ objects from $n$, which equals 1. Incorrect.
• (D) $n$: This is just the number of objects, not the number of ways to arrange them. Incorrect.

Therefore, the statement is correctly completed as: "The number of permutations of $n$ distinct objects is ${}_nP_n$."

Hence, the correct option is (B) ${}_nP_n$.

Given:

• The word is "ROSE".
• The letters available are R, O, S, E. All 4 letters are distinct.
• We need to form 4-letter words.
• Each letter must be used exactly once.

To Find:

The total number of 4-letter words that can be formed.

Solution:

The problem asks for the number of ways to arrange the 4 distinct letters of the word "ROSE". Since the order of letters matters in forming a word, this is a permutation problem.

The number of ways to arrange $n$ distinct objects is given by n-factorial, denoted as $n!$.

In this case, we have $n = 4$ distinct letters.

So, the total number of arrangements (words) is $4!$.

Let's calculate the value of $4!$:

$4! = 4 \times 3 \times 2 \times 1$

$4! = 12 \times 2 \times 1$

$4! = 24 \times 1$

$4! = 24$

Alternate Solution (Using the Fundamental Principle of Counting):

We can think of this as filling four positions to form a 4-letter word.

• For the first position, we have 4 choices (R, O, S, or E).
• After filling the first position, we have 3 letters remaining. So, for the second position, we have 3 choices.
• After filling the first two positions, we have 2 letters remaining. For the third position, we have 2 choices.
• For the last position, we have only 1 letter remaining, so we have 1 choice.

The total number of words is the product of the number of choices for each position:

Total words = $4 \times 3 \times 2 \times 1 = 24$

Both methods show that 24 different 4-letter words can be formed.

Comparing this result with the given options:

(A) 4

(B) 12

(C) 24

(D) 48

The calculated value matches option (C).

Hence, the correct option is (C) 24.

Solution:

The question asks for the standard formula for calculating the number of combinations of $n$ distinct objects taken $r$ at a time.

A combination is a selection of items from a set where the order of selection does not matter. The number of combinations of $n$ distinct objects taken $r$ at a time is denoted by ${}_nC_r$, $C(n, r)$, or $\binom{n}{r}$.

The formula is derived by taking the number of permutations, ${}_nP_r = \frac{n!}{(n-r)!}$, and dividing by the number of ways to arrange the $r$ selected items, which is $r!$. This division removes the effect of order.

${}_nC_r = \frac{{}_nP_r}{r!} = \frac{n!/((n-r)!)}{r!} = \frac{n!}{r!(n-r)!}$

So, the standard formula for combinations is:

${}_nC_r = \frac{n!}{r!(n-r)!}$

Analysis of the Options:

• (A) ${}_nC_r = \frac{n!}{(n-r)!}$: This is the formula for permutations (${}_nP_r$), where order matters. Incorrect.
• (B) ${}_nC_r = \frac{n!}{r!(n-r)!}$: This is the correct standard formula for combinations.
• (C) ${}_nC_r = \frac{(n-r)!}{n!}$: This is the reciprocal of the permutation formula and is incorrect.
• (D) ${}_nC_r = n!$: This is the formula for the number of ways to arrange all $n$ distinct objects (${}_nP_n$). Incorrect.

Therefore, the correct formula for combinations is the one presented in option (B).

Hence, the correct option is (B) ${}_nC_r = \frac{n!}{r!(n-r)!}$.

Solution:

We need to evaluate the expression ${}_5C_2$, which represents the number of ways to choose 2 items from a set of 5 distinct items, where the order of selection does not matter.

The formula for the number of combinations of $n$ distinct objects taken $r$ at a time is:

${}_nC_r = \frac{n!}{r!(n-r)!}$

In this problem, we have $n=5$ and $r=2$.

Substituting these values into the formula:

${}_5C_2 = \frac{5!}{2!(5-2)!}$

${}_5C_2 = \frac{5!}{2!3!}$

Now, we can expand the factorials to calculate the value. It's often easier to expand the largest factorial in the numerator until it matches the largest factorial in the denominator.

${}_5C_2 = \frac{5 \times 4 \times 3!}{2! \times 3!}$

We can cancel the common term $3!$ from the numerator and the denominator:

${}_5C_2 = \frac{5 \times 4}{2!}$

Since $2! = 2 \times 1 = 2$, the expression becomes:

${}_5C_2 = \frac{5 \times 4}{2} = \frac{20}{2}$

${}_5C_2 = 10$

Therefore, the value of ${}_5C_2$ is 10.

Comparing this result with the given options:

(A) 10

(B) 20

(C) 12

(D) 5

The calculated value matches option (A).

Hence, the correct option is (A) 10.

Solution:

The problem asks for the number of ways to select a committee of 3 members from a group of 7 people. When forming a committee, the order in which the members are selected does not matter. For instance, selecting the group {Alice, Bob, Charlie} is the same as selecting {Charlie, Alice, Bob}. Therefore, this is a problem of combinations, not permutations.

We need to find the number of combinations of 7 items taken 3 at a time, which is denoted by ${}_7C_3$.

Calculation:

The formula for combinations is:

In this problem, we have:

• $n = 7$ (the total number of people to choose from)
• $r = 3$ (the number of members in the committee)

Substituting these values into the formula:

${}_7C_3 = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!}$

Now, we expand the factorials to simplify the expression:

${}_7C_3 = \frac{7 \times 6 \times 5 \times 4!}{3! \times 4!}$

We can cancel out the $4!$ term from the numerator and denominator:

${}_7C_3 = \frac{7 \times 6 \times 5}{3!}$

We know that $3! = 3 \times 2 \times 1 = 6$. So the expression becomes:

${}_7C_3 = \frac{7 \times 6 \times 5}{6}$

We can cancel out the 6:

${}_7C_3 = \frac{7 \times \cancel{6} \times 5}{\cancel{6}} = 7 \times 5 = 35$

Therefore, there are 35 different ways to select the committee.

Comparing this result with the given options:

(A) 21

(B) 35

(C) 210

(D) 7!

The calculated value matches option (B).

Hence, the correct option is (B) 35.

Solution:

We need to identify the correct property of combinations among the given options.

The formula for combinations is ${}_nC_r = \frac{n!}{r!(n-r)!}$. Let's analyze each option using this formula or by a conceptual argument.

(A) ${}_nC_r = {}_nC_{n-r}$

Let's evaluate the right-hand side using the formula:

${}_nC_{n-r} = \frac{n!}{(n-r)!(n - (n-r))!}$

${}_nC_{n-r} = \frac{n!}{(n-r)!(n - n + r)!}$

${}_nC_{n-r} = \frac{n!}{(n-r)!r!}$

This is identical to the formula for ${}_nC_r$. So, the statement ${}_nC_r = {}_nC_{n-r}$ is correct.

Conceptually, this property means that the number of ways to choose $r$ items from a set of $n$ is the same as the number of ways to choose the $(n-r)$ items to leave behind.

(B) ${}_nC_0 = 0$

Let's evaluate ${}_nC_0$ using the formula:

${}_nC_0 = \frac{n!}{0!(n-0)!} = \frac{n!}{0!n!}$

Since $0!=1$, we have:

${}_nC_0 = \frac{n!}{1 \times n!} = \frac{\cancel{n!}}{\cancel{n!}} = 1$

The number of ways to choose zero items is 1 (the empty set). Therefore, the statement ${}_nC_0 = 0$ is incorrect.

(C) ${}_nC_n = n$

Let's evaluate ${}_nC_n$ using the formula:

${}_nC_n = \frac{n!}{n!(n-n)!} = \frac{n!}{n!0!}$

Since $0!=1$, we have:

${}_nC_n = \frac{n!}{n! \times 1} = \frac{\cancel{n!}}{\cancel{n!}} = 1$

The number of ways to choose all $n$ items from a set of $n$ is 1 (the whole set). Therefore, the statement ${}_nC_n = n$ is incorrect.

(D) ${}_nC_r = {}_nC_{r+1}$

This is not a general property. It is only true for a specific value of $r$ if $n$ is odd (specifically when $r = (n-1)/2$). For example, ${}_5C_2 = 10$ and ${}_5C_3 = 10$. However, ${}_5C_1 = 5$ and ${}_5C_2 = 10$, so ${}_5C_1 \neq {}_5C_2$. Thus, this is not a general property and is incorrect.

Based on the analysis, the only correct property among the choices is the one in option (A).

Hence, the correct option is (A) ${}_nC_r = {}_nC_{n-r}$.

Solution:

The problem asks for the number of ways to choose 2 books from a collection of 9 distinct books. When "choosing" a set of items, the order in which they are selected does not matter. Therefore, this is a combination problem.

We need to find the number of combinations of 9 distinct items taken 2 at a time, which is denoted by ${}_9C_2$.

Calculation:

The formula for combinations is:

In this problem, we have:

• $n = 9$ (the total number of distinct books)
• $r = 2$ (the number of books to choose)

Substituting these values into the formula:

${}_9C_2 = \frac{9!}{2!(9-2)!} = \frac{9!}{2!7!}$

Now, we expand the largest factorial in the numerator until it matches the largest factorial in the denominator:

${}_9C_2 = \frac{9 \times 8 \times 7!}{2! \times 7!}$

We can cancel out the $7!$ term:

${}_9C_2 = \frac{9 \times 8}{2!}$

We know that $2! = 2 \times 1 = 2$. So the expression becomes:

${}_9C_2 = \frac{9 \times 8}{2} = \frac{72}{2}$

${}_9C_2 = 36$

Therefore, there are 36 different ways to choose 2 books from the collection of 9.

Comparing this result with the given options:

(A) 18

(B) 36

(C) 72

(D) 81

The calculated value matches option (B).

Hence, the correct option is (B) 36.

Analysis of Assertion and Reason:

We need to evaluate both the Assertion (A) and the Reason (R) and determine their relationship.

Assertion (A): Combinations are used when the order of selection does NOT matter.

This statement is the fundamental definition of a combination. A combination refers to the selection of items from a collection, such that the order of selection is irrelevant. For example, choosing a committee of two people, {Alice, Bob}, is the same combination as choosing {Bob, Alice}. This is the key distinction between combinations and permutations. Therefore, Assertion (A) is true.

Reason (R): The formula for combinations divides the number of permutations by the factorial of the number of items selected, effectively removing the impact of order.

Let's examine the relationship between the formulas for permutations and combinations:

• The number of permutations (ordered arrangements) of $n$ items taken $r$ at a time is ${}_nP_r = \frac{n!}{(n-r)!}$.
• The number of combinations (unordered selections) of $n$ items taken $r$ at a time is ${}_nC_r = \frac{n!}{r!(n-r)!}$.

We can see that ${}_nC_r = \frac{{}_nP_r}{r!}$.

For any set of $r$ items that are chosen, there are $r!$ ways to arrange them. The permutation formula ${}_nP_r$ counts all these $r!$ arrangements as distinct outcomes. By dividing ${}_nP_r$ by $r!$, the combination formula groups all these different orderings into a single selection, thus removing the impact of order. Therefore, Reason (R) is a true statement that correctly describes the mathematical relationship and its effect.

Conclusion:

Both Assertion (A) and Reason (R) are true statements.

Furthermore, Reason (R) provides the precise mathematical explanation for why Assertion (A) is true. Assertion (A) states the concept, and Reason (R) explains how the formula achieves that concept. The act of dividing the number of permutations by $r!$ is exactly what makes the result independent of order.

Therefore, Reason (R) is the correct explanation for Assertion (A).

This corresponds to option (A).

Hence, the correct option is (A) Both A and R are true and R is the correct explanation of A.

Given:

• A box contains 5 red balls and 3 blue balls.
• All balls of the same color are assumed to be distinct for the purpose of selection.

Task:

To find the number of ways to select 2 red balls from the box.

Solution:

The problem asks us to select 2 red balls. This means we are only concerned with the 5 red balls available in the box. The blue balls are irrelevant for this specific task.

When we "select" or "choose" items from a group, the order in which we pick them does not matter. For example, picking Red Ball 1 and then Red Ball 2 is the same selection as picking Red Ball 2 and then Red Ball 1. Therefore, this is a combination problem.

We need to find the number of ways to choose 2 items from a set of 5. This is represented by the combination notation ${}_5C_2$.

Analysis of the Options:

• (A) ${}_5P_2$: This is the formula for permutations. It would be correct if the order of selection mattered (e.g., "pick a first red ball and a second red ball"). This is not what the question asks.
• (B) ${}_5C_2$: This correctly represents the number of ways to choose (in any order) 2 red balls from the available 5 red balls.
• (C) ${}_5C_0 \times {}_3C_2$: This represents choosing 0 red balls AND 2 blue balls. This is incorrect.
• (D) ${}_5C_2 + {}_3C_0$: This represents the sum of ways to choose 2 red balls and the ways to choose 0 blue balls. While the first term is correct, the addition is not part of the required calculation for the stated task. The expression should simply be ${}_5C_2$.

The correct mathematical expression that describes the number of ways to select 2 red balls from the 5 available is ${}_5C_2$.

Calculating the value (for confirmation):

${}_5C_2 = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10$ ways.

The question asks for the correct representation, which is ${}_5C_2$.

Hence, the correct option is (B) ${}_5C_2$.

Given:

• A box contains 5 red balls and 3 blue balls.

Task:

To find the number of ways to select one red ball AND one blue ball.

Solution:

This problem involves two independent events occurring together:

1. Selecting 1 red ball from the 5 available red balls.
2. Selecting 1 blue ball from the 3 available blue balls.

We need to find the number of ways for each event and then use the Fundamental Principle of Counting (Multiplication Principle) to find the total number of ways for both events to happen.

Step 1: Calculate the number of ways to select one red ball.

The order of selection does not matter, so we use combinations. We need to choose 1 red ball from 5.

Number of ways = ${}_5C_1 = \frac{5!}{1!(5-1)!} = \frac{5!}{1!4!} = 5$.

Step 2: Calculate the number of ways to select one blue ball.

Similarly, we use combinations to choose 1 blue ball from 3.

Number of ways = ${}_3C_1 = \frac{3!}{1!(3-1)!} = \frac{3!}{1!2!} = 3$.

Step 3: Combine the results.

According to the Multiplication Principle, the total number of ways to select one red ball AND one blue ball is the product of the number of ways for each selection.

Total ways = (Ways to select 1 red ball) $\times$ (Ways to select 1 blue ball)

Total ways = ${}_5C_1 \times {}_3C_1$

Calculating the final value: Total ways = $5 \times 3 = 15$.

The question asks for the correct expression representing this calculation.

Analysis of the Options:

• (A) ${}_5C_1 + {}_3C_1$: This would be the number of ways to select either one red ball OR one blue ball. This is incorrect.
• (B) ${}_5P_1 \times {}_3P_1$: This uses permutations, which implies order matters. In a selection problem, order is irrelevant. Incorrect formulation.
• (C) ${}_5C_1 \times {}_3C_1$: This correctly uses combinations for each selection and the multiplication principle to find the total number of ways for the combined event. This is correct.
• (D) ${}_8C_2$: This calculates the total ways to choose any 2 balls from the 8 available, which includes pairs of two red balls and pairs of two blue balls. This does not specifically address the condition of one of each color.

The correct expression is ${}_5C_1 \times {}_3C_1$.

Hence, the correct option is (C) ${}_5C_1 \times {}_3C_1$.

Solution:

The question asks to complete a statement about combinations.

Step 1: Understand the statement

"The number of combinations of $n$ distinct objects taken $n$ at a time" means we are finding the number of ways to choose all $n$ objects from a set of $n$ objects. Since the order of selection does not matter in combinations, we are essentially asking "How many groups of size $n$ can be formed from a set of $n$ objects?".

Step 2: Symbolic Representation

The symbolic representation for "the number of combinations of $n$ distinct objects taken $n$ at a time" is ${}_nC_n$. So, option (B) is a correct way to complete the statement symbolically.

Step 3: Calculate the value

Let's calculate the value of ${}_nC_n$ using the combination formula:

${}_nC_r = \frac{n!}{r!(n-r)!}$

Substitute $r=n$:

${}_nC_n = \frac{n!}{n!(n-n)!} = \frac{n!}{n!0!}$

Since $0! = 1$, the expression becomes:

${}_nC_n = \frac{n!}{n! \times 1} = \frac{\cancel{n!}}{\cancel{n!}} = 1$

This means there is only one way to choose all $n$ objects from a set of $n$ objects (you must take everything). So, the value is 1. This means option (C) is also a correct way to complete the statement.

Step 4: Analyze the options

• (A) ${}_nC_0$: This represents choosing 0 objects, which also equals 1. While the value is the same, it is not the correct symbolic representation for the question asked.
• (B) ${}_nC_n$: This is the correct symbolic representation of the phrase.
• (C) 1: This is the correct numerical value of ${}_nC_n$.
• (D) Both (B) and (C): Since both ${}_nC_n$ and its value, 1, are correct ways to complete the statement, this option encompasses both correct answers.

The statement can be completed either with the symbolic notation or its value. Both are valid. Therefore, the best option is the one that includes both.

Hence, the correct option is (D) Both (B) and (C).

Given:

• The number of permutations of $n$ objects taken $r$ at a time, ${}_nP_r = 720$.
• The number of combinations of $n$ objects taken $r$ at a time, ${}_nC_r = 120$.

To Find:

The value of $r!$.

Solution:

We need to use the relationship between permutations and combinations. The formula for combinations can be expressed in terms of permutations:

This relationship exists because the number of combinations (unordered selections) is found by taking the number of permutations (ordered arrangements) and dividing by the number of ways to order the $r$ selected items, which is $r!$.

We are given the values of ${}_nP_r$ and ${}_nC_r$. We can substitute these values into the formula (i) to solve for $r!$.

Substituting the given values:

$120 = \frac{720}{r!}$

To find $r!$, we can rearrange the equation by multiplying both sides by $r!$ and then dividing by 120:

$120 \times r! = 720$

$r! = \frac{720}{120}$

Now, we perform the division:

$r! = \frac{\cancel{720}^6}{\cancel{120}_1} = 6$

So, the value of $r!$ is 6.

Optional: Finding the value of r and n

We found that $r! = 6$. We know that $3! = 3 \times 2 \times 1 = 6$. So, $r=3$.

We also know ${}_nP_r = 720$, which means ${}_nP_3 = 720$.

$\frac{n!}{(n-3)!} = n(n-1)(n-2) = 720$

We need to find three consecutive integers whose product is 720. We can estimate that $10 \times 9 \times 8 = 720$.

So, $n=10$, $n-1=9$, and $n-2=8$. This gives $n=10$.

The question only asks for the value of $r!$.

Comparing our result for $r!$ with the given options:

(A) 5

(B) 6

(C) 120

(D) 720

The calculated value matches option (B).

Hence, the correct option is (B) 6.

Given:

The equation involving combinations is:

To Find:

The value of $x$.

Solution:

We will use a key property of combinations. The property states that if ${}_nC_a = {}_nC_b$, then there are two possibilities:

1. $a = b$
2. $a + b = n$

In the given equation (i), we have $n = 10$, $a = x$, and $b = x+2$. Let's examine both cases.

Case 1: $a = b$

If we apply this case, we get:

$x = x + 2$

Subtracting $x$ from both sides of the equation gives:

$0 = 2$

This is a contradiction, which means this case is not possible. So, we must consider the second case.

Case 2: $a + b = n$

If we apply this case, we substitute the values of $a$, $b$, and $n$:

$x + (x + 2) = 10$

Now, we solve this linear equation for $x$:

$2x + 2 = 10$

Subtract 2 from both sides:

$2x = 10 - 2$

$2x = 8$

Divide by 2:

$x = \frac{8}{2}$

$x = 4$

This is a valid solution. Let's check: if $x=4$, the equation becomes ${}_{10}C_4 = {}_{10}C_6$. This is true because $4+6=10$.

Therefore, the value of $x$ is 4.

Comparing this with the given options:

(A) 4

(B) 5

(C) 6

(D) 8

The calculated value matches option (A).

Hence, the correct option is (A) 4.

Solution:

We need to find the number of diagonals that can be drawn in a hexagon. A hexagon is a polygon with 6 vertices and 6 sides.

A diagonal is a line segment that connects two non-adjacent vertices of a polygon. We can solve this problem using the principles of combinations.

Step 1: Find the total number of line segments possible.

A line segment is formed by choosing any 2 vertices from the 6 available vertices. The order of choosing the vertices does not matter (the line from vertex A to B is the same as from B to A). Therefore, we use combinations.

The total number of line segments is the number of ways to choose 2 vertices from 6, which is ${}_6C_2$.

Using the combination formula ${}_nC_r = \frac{n!}{r!(n-r)!}$:

${}_6C_2 = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!}$

${}_6C_2 = \frac{6 \times 5 \times 4!}{2 \times 1 \times 4!} = \frac{6 \times 5}{2} = \frac{30}{2} = 15$

So, a total of 15 line segments can be drawn between the vertices of a hexagon.

Step 2: Subtract the number of sides.

The 15 line segments we calculated include both the diagonals and the sides of the hexagon. The sides of a polygon connect adjacent vertices, so they are not diagonals.

A hexagon has 6 sides.

To find the number of diagonals, we subtract the number of sides from the total number of line segments.

Number of diagonals = (Total line segments) - (Number of sides)

Number of diagonals = $15 - 6 = 9$

Alternate Solution (Using the Direct Formula):

There is a direct formula to calculate the number of diagonals in a polygon with $n$ sides:

Number of diagonals = $\frac{n(n-3)}{2}$

For a hexagon, the number of sides (and vertices) is $n=6$.

Substituting $n=6$ into the formula:

Number of diagonals = $\frac{6(6-3)}{2}$

Number of diagonals = $\frac{6(3)}{2} = \frac{18}{2} = 9$

Both methods confirm that a hexagon has 9 diagonals.

Comparing this result with the given options:

(A) 6

(B) 9

(C) 12

(D) 15

The calculated value matches option (B).

Hence, the correct option is (B) 9.

Given:

• There are 10 distinct points in a plane.
• No three of these points are collinear (meaning no three points lie on the same straight line).

To Find:

The total number of straight lines that can be drawn by joining any two of these points.

Solution:

A straight line is uniquely determined by any two distinct points. To form a line, we need to choose 2 points from the 10 available points.

The order in which we choose the two points does not matter. The line joining point A and point B is the same as the line joining point B and point A. Therefore, this is a combination problem.

We need to find the number of ways to choose 2 points from a set of 10. This is represented by the combination notation ${}_{10}C_2$.

The condition "no three of which are collinear" is important because it ensures that every pair of points we choose will form a unique line. If three or more points were collinear, choosing different pairs from that collinear set would result in the same line, which would complicate the calculation.

Calculation:

We use the combination formula:

${}_nC_r = \frac{n!}{r!(n-r)!}$

In this problem, $n = 10$ and $r = 2$.

${}_{10}C_2 = \frac{10!}{2!(10-2)!} = \frac{10!}{2!8!}$

Now, we expand the numerator to simplify:

${}_{10}C_2 = \frac{10 \times 9 \times 8!}{2! \times 8!}$

We can cancel the $8!$ term:

${}_{10}C_2 = \frac{10 \times 9}{2!}$

Since $2! = 2 \times 1 = 2$, we have:

${}_{10}C_2 = \frac{10 \times 9}{2} = \frac{90}{2} = 45$

Therefore, 45 distinct straight lines can be drawn.

Comparing this result with the given options:

(A) 10

(B) 20

(C) 45

(D) 90

The calculated value matches option (C).

Hence, the correct option is (C) 45.

Given:

• The set of available digits is {1, 2, 3, 4, 5, 6, 7}. This is a set of 7 distinct digits.
• We need to form 4-digit numbers.
• Repetition of digits is not allowed.

To Find:

The correct expression for the number of 4-digit numbers that can be formed.

Solution:

When forming a number, the order of the digits matters. For example, the number 1234 is different from the number 4321, even though they use the same digits. Because the order is important, this is a permutation problem.

We need to find the number of ways to arrange 4 distinct digits chosen from a set of 7 distinct digits. This is a classic permutation of $n$ objects taken $r$ at a time.

Here:

• $n = 7$ (the total number of available distinct digits)
• $r = 4$ (the number of digits in the number to be formed)

The number of such arrangements is given by the permutation formula ${}_nP_r$.

Substituting the values of $n$ and $r$, the expression is ${}_7P_4$.

Let's analyze the options:

• (A) ${}_7P_4$: This correctly represents the number of permutations (ordered arrangements) of 4 digits chosen from 7. This is the correct answer.
• (B) ${}_7C_4$: This represents the number of ways to choose 4 digits, but it does not account for the order. This would be incorrect for forming numbers.
• (C) $7^4$: This would be the correct answer if repetition of digits were allowed. Since repetition is not allowed, this is incorrect.
• (D) 7!: This would be the number of ways to form a 7-digit number using all 7 digits. This is incorrect.

Calculation for verification (not required by the question):

${}_7P_4 = \frac{7!}{(7-4)!} = \frac{7!}{3!} = 7 \times 6 \times 5 \times 4 = 840$.

The question asks for the correct expression, which is ${}_7P_4$.

Hence, the correct option is (A) ${}_7P_4$.

Given:

• A group of 5 boys.
• A group of 3 girls.

Condition:

They must be arranged in a row such that no two girls are together.

To Find:

The total number of such arrangements.

Solution:

This is a classic problem of arrangement with constraints. The "gap method" is the most effective way to solve it.

Step 1: Arrange the group without the restriction.

The condition is on the girls ("no two girls are together"). So, we first arrange the boys, who have no restrictions relative to each other.

The 5 distinct boys can be arranged in a row in $5!$ ways.

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$ ways.

Step 2: Create gaps for the restricted group.

Let's represent the arranged boys by 'B' and the possible positions (gaps) for the girls by underscores (_):

_ B _ B _ B _ B _ B _

By placing the boys first, we create spaces where the girls can be placed. Placing a girl in any of these gaps ensures that no two girls will be adjacent. Counting the gaps, we have 5 gaps between the boys and 2 gaps at the ends, for a total of $5 + 1 = 6$ possible gaps.

Step 3: Arrange the restricted group in the gaps.

Now, we need to place the 3 girls into these 6 available gaps. Since the girls are distinct, the order in which we place them matters. For example, placing Girl 1 in the first gap and Girl 2 in the second is different from placing Girl 2 in the first and Girl 1 in the second.

Therefore, we need to find the number of ways to arrange (permute) 3 girls in 6 available spots. This is a permutation problem, represented by ${}_6P_3$.

The number of ways to place the girls is ${}_6P_3$.

Step 4: Combine the results using the Multiplication Principle.

The total number of arrangements is the product of the number of ways to arrange the boys and the number of ways to place the girls.

Total arrangements = (Ways to arrange boys) $\times$ (Ways to place girls)

Total arrangements = $5! \times {}_6P_3$

Analysis of Options:

• (A) $5! \times {}_6P_3$: This perfectly matches our derived expression.
• (B) $5! \times 3!$: This would be the number of ways if the boys and girls were treated as two separate blocks to be arranged. Incorrect.
• (C) ${}_8P_8 - (6! \times 3!)$: This suggests a "total minus unwanted" approach, where $(6! \times 3!)$ would be the number of ways the 3 girls are together, which is also incorrect. The number of arrangements where girls are together is $6! \times 3!$, but subtracting this from the total does not account for cases where only two girls are together.
• (D) $5! \times {}_6C_3$: This would be correct if the girls were identical (indistinguishable), as combinations do not account for order. Since the girls are distinct, this is incorrect.

The correct expression for the number of arrangements is $5! \times {}_6P_3$.

Hence, the correct option is (A) $5! \times {}_6P_3$.

Solution:

We need to find the number of different arrangements (words) that can be formed using the letters of the word "ASSASSINATION". This is a problem of permutations with repetitions, as several letters appear more than once.

Step 1: Analyze the letters in the word "ASSASSINATION"

First, let's count the total number of letters and the frequency of each distinct letter.

The word is A-S-S-A-S-S-I-N-A-T-I-O-N.

• Total number of letters (n): 13
• Frequency of 'A': 3 times
• Frequency of 'S': 4 times
• Frequency of 'I': 2 times
• Frequency of 'N': 2 times
• Frequency of 'T': 1 time
• Frequency of 'O': 1 time

Step 2: Use the formula for permutations with repetitions

The formula for the number of distinct permutations of $n$ objects, where there are $p_1$ identical items of one kind, $p_2$ of a second kind, ..., and $p_k$ of a k-th kind is:

Number of arrangements = $\frac{n!}{p_1! \times p_2! \times \dots \times p_k!}$

Step 3: Apply the formula to the given word

Using the counts from Step 1, we can set up the expression:

$n = 13$

The denominators will be the factorials of the frequencies of the repeated letters: 'A' (3!), 'S' (4!), 'I' (2!), and 'N' (2!).

Number of arrangements = $\frac{13!}{3! \times 4! \times 2! \times 2!}$

Analysis of Options:

Let's compare our derived expression, $\frac{13!}{3!4!2!2!}$, with the given options.

• (A) $\frac{13!}{3!4!2!}$
• (B) $\frac{13!}{3!4!}$
• (C) $\frac{13!}{3!2!}$
• (D) $\frac{13!}{4!2!}$

It appears there is a typographical error in all the options provided, as none of them match the correct expression $\frac{13!}{3!4!2!2!}$. All options are missing one of the $2!$ terms in the denominator (for the repetition of the letter 'N').

However, if we are forced to choose the "closest" answer, option (A) is the most similar, differing only by one factor of $2!$ in the denominator. This suggests that the question intended to ask for this word but had an error in its options.

Assuming a typo in the options and that it should have included both $2!$ terms, none of the answers are technically correct. If the question intended to miss one of the repeated letters, say 'N', then the answer would be (A).

Given the context of multiple-choice questions, it is highly probable that option (A) contains a typo and is the intended answer.

Hence, based on the assumption of a typo in the options, the intended answer is most likely (A) $\frac{13!}{3!4!2!}$, despite it being technically incorrect.

Given:

Total number of players in the cricket team = 11.

Two positions to be chosen: Captain and Vice-Captain.

To Find:

The number of ways to choose a captain and a vice-captain from the 11 players.

Solution:

We need to select 2 players from 11 players for two distinct positions: Captain and Vice-Captain.

The order in which the players are chosen matters here. For example, choosing Player A as Captain and Player B as Vice-Captain is different from choosing Player B as Captain and Player A as Vice-Captain.

Since the order matters, this is a problem of permutation.

The number of permutations of selecting $r$ objects from a set of $n$ distinct objects is given by the formula:

In this problem, $n = 11$ (total number of players) and $r = 2$ (number of positions to be filled).

Substitute these values into the permutation formula:

${}_{11}P_2 = \frac{11!}{(11-2)!}$

${}_{11}P_2 = \frac{11!}{9!}$

${}_{11}P_2 = \frac{11 \times 10 \times 9!}{9!}$

Cancel out $9!$ from the numerator and the denominator:

${}_{11}P_2 = 11 \times 10$

${}_{11}P_2 = 110$

So, there are 110 ways to choose a captain and a vice-captain from 11 players.

Now let's look at the given options:

(A) ${}_{11}C_2$: This represents the number of combinations, where order does not matter. ${}_{11}C_2 = \frac{11!}{2!9!} = \frac{11 \times 10}{2} = 55$. This is incorrect.

(B) ${}_{11}P_2$: This represents the number of permutations, where order matters. ${}_{11}P_2 = 11 \times 10 = 110$. This is correct.

(C) 11: This is incorrect.

(D) $11 \times 10 \times 9!$: This is equal to $11!$, which is the total number of ways to arrange all 11 players. This is incorrect.

The correct option is (B) ${}_{11}P_2$.

Given:

A group consists of 5 men and 4 women.

A committee of 3 is to be formed.

The committee must consist of exactly 2 men and 1 woman.

To Find:

The number of ways the committee can be formed with exactly 2 men and 1 woman.

Solution:

We need to select a committee of 3 members with a specific composition: 2 men and 1 woman.

Since the order in which the members are selected for a committee does not matter, this is a problem of combination.

We need to perform two independent selections:

1. Select 2 men from the 5 men.

2. Select 1 woman from the 4 women.

The number of ways to select $r$ items from a set of $n$ distinct items where order does not matter is given by the combination formula:

Number of ways to choose 2 men from 5 men:

${}_5C_2 = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4 \times 3!}{2 \times 1 \times 3!}$

${}_5C_2 = \frac{5 \times 4}{2} = \frac{20}{2} = 10$

Number of ways to choose 1 woman from 4 women:

${}_4C_1 = \frac{4!}{1!(4-1)!} = \frac{4!}{1!3!} = \frac{4 \times 3!}{1 \times 3!}$

${}_4C_1 = \frac{4}{1} = 4$

To find the total number of ways to form the committee with exactly 2 men and 1 woman, we multiply the number of ways to select the men by the number of ways to select the women (using the Multiplication Principle).

Total number of ways = (Number of ways to choose 2 men) $\times$ (Number of ways to choose 1 woman)

Total number of ways = ${}_5C_2 \times {}_4C_1$

Comparing this result with the given options:

(A) ${}_5C_2 + {}_4C_1 = 10 + 4 = 14$. Incorrect.

(B) ${}_5P_2 \times {}_4P_1 = (\frac{5!}{(5-2)!}) \times (\frac{4!}{(4-1)!}) = (\frac{5!}{3!}) \times (\frac{4!}{3!}) = (5 \times 4) \times 4 = 20 \times 4 = 80$. Incorrect, as order doesn't matter in committee formation.

(C) ${}_5C_2 \times {}_4C_1 = 10 \times 4 = 40$. Correct.

(D) ${}_9C_3 = \frac{9!}{3!(9-3)!} = \frac{9!}{3!6!} = \frac{9 \times 8 \times 7 \times 6!}{3 \times 2 \times 1 \times 6!} = \frac{9 \times 8 \times 7}{6} = 3 \times 4 \times 7 = 84$. This is the total number of ways to choose any 3 people from 9, without any gender constraints. Incorrect.

The correct answer is (C) ${}_5C_2 \times {}_4C_1$.

Given:

Total number of persons = 10.

Size of the committee to be formed = 5.

A particular person must always be included in the committee.

To Find:

The number of different committees of 5 that can be formed under the given condition.

Solution:

We need to form a committee of 5 members from a group of 10 persons.

The condition is that a specific person must always be included in the committee.

Since this particular person is already selected for the committee, we can consider them as one of the 5 members.

Now, we need to select the remaining members for the committee.

Number of committee members already selected = 1 (the particular person)

Number of members still needed for the committee = $5 - 1 = 4$.

Since the particular person is already included, they are removed from the pool of available persons for the remaining selection.

Number of persons remaining from whom we need to select the remaining members = $10 - 1 = 9$.

So, we need to choose 4 members from the remaining 9 persons.

The number of ways to do this is given by the combination formula ${}_nC_r$, where $n$ is the total number of items to choose from, and $r$ is the number of items to choose.

Let's calculate the value of ${}_9C_4$:

${}_9C_4 = \frac{9!}{4!(9-4)!} = \frac{9!}{4!5!} = \frac{9 \times 8 \times 7 \times 6 \times 5!}{4 \times 3 \times 2 \times 1 \times 5!}$

Cancel out $5!$ from numerator and denominator:

${}_9C_4 = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}$

${}_9C_4 = \frac{\cancel{9}^{3} \times \cancel{8}^{2} \times 7 \times \cancel{6}^{1}}{\cancel{4}_{1} \times \cancel{3}_{1} \times \cancel{2}_{1} \times 1}$

${}_9C_4 = 3 \times 2 \times 7 \times 1 = 42 \times 1 = 126$

So, there are 126 ways to form the committee with the particular person always included.

Comparing this result with the given options:

(A) ${}_{10}C_5$: This is the number of ways to choose any 5 persons from 10 without any restriction. Incorrect.

(B) ${}_{10}C_4$: This is incorrect.

(C) ${}_9C_5$: This would be the number of ways to choose 5 persons from the remaining 9 if the particular person was *not* to be included. Incorrect.

(D) ${}_9C_4$: This represents choosing 4 additional members from the remaining 9 persons after the particular person is included. Correct.

The correct answer is (D) ${}_9C_4$.

Given:

The equation ${}_nC_8 = {}_nC_2$.

To Find:

The value of ${}_nC_2$.

Solution:

We are given the equation ${}_nC_8 = {}_nC_2$.

We know a property of combinations that states if ${}_nC_a = {}_nC_b$, then either $a = b$ or $a + b = n$.

In this case, we have $a = 8$ and $b = 2$.

Case 1: $a = b$

$8 = 2$, which is false.

Case 2: $a + b = n$

$8 + 2 = n$

Now that we have found the value of $n$, we can find the value of ${}_nC_2$, which is ${}_{10}C_2$.

Using the combination formula ${}_nC_r = \frac{n!}{r!(n-r)!}$, we calculate ${}_{10}C_2$:

${}_{10}C_2 = \frac{10!}{2!(10-2)!}$

${}_{10}C_2 = \frac{10!}{2!8!}$

${}_{10}C_2 = \frac{10 \times 9 \times 8!}{2 \times 1 \times 8!}$

Cancel out $8!$ from the numerator and the denominator:

${}_{10}C_2 = \frac{10 \times 9}{2}$

${}_{10}C_2 = \frac{90}{2}$

Thus, the value of ${}_nC_2$ is 45.

Comparing this with the given options:

(A) 45

(B) 55

(C) 66

(D) 99

The correct answer is (A) 45.

Given:

Four types of counting problems and four counting techniques to match.

Problems:

(i) Arranging books on a shelf

(ii) Selecting a team from a group

(iii) Forming numbers with distinct digits

(iv) Choosing items from a menu

Techniques:

(a) Combination

(b) Fundamental Principle of Counting

(c) Permutation

(d) Permutation

To Match:

Match each problem type with the appropriate counting technique.

Solution:

Let's analyze each problem type:

(i) Arranging books on a shelf: When arranging items in a specific order, the order matters. This is the definition of a Permutation.

Match: (i) - (c) or (d) Permutation

(ii) Selecting a team from a group: When selecting a group of people or items where the order of selection does not matter (a team of A, B, C is the same as a team of B, A, C), this is a Combination.

Match: (ii) - (a) Combination

(iii) Forming numbers with distinct digits: When forming numbers using distinct digits, the position of each digit matters (e.g., 123 is different from 321). This involves selecting and arranging digits, which is a Permutation problem. Alternatively, it can be solved using the Fundamental Principle of Counting by considering the number of choices for each digit position sequentially.

Match: (iii) - (c) or (d) Permutation (or potentially (b) Fundamental Principle of Counting depending on interpretation, but Permutation is a direct application when order matters with distinct items)

(iv) Choosing items from a menu: A common type of menu problem involves selecting one item from each of several categories (e.g., one appetizer, one main course, one dessert). The total number of possibilities is found by multiplying the number of options in each category. This is an application of the Fundamental Principle of Counting (specifically, the multiplication rule).

Match: (iv) - (b) Fundamental Principle of Counting

Now let's look at the options provided and find the one that matches our analysis, keeping in mind there are two labels for Permutation.

We need:

• (i) matched with Permutation ((c) or (d))
• (ii) matched with Combination ((a))
• (iii) matched with Permutation ((c) or (d))
• (iv) matched with Fundamental Principle of Counting ((b))

Let's check option (A):

• (i)-(c): Arranging books - Permutation. Matches.
• (ii)-(a): Selecting a team - Combination. Matches.
• (iii)-(d): Forming numbers with distinct digits - Permutation. Matches (using the other Permutation label).
• (iv)-(b): Choosing items from a menu - Fundamental Principle of Counting. Matches.

Option (A) successfully matches all problem types with the appropriate techniques.

Let's quickly check why other options are incorrect:

• (B) (ii)-(b) (Selecting a team - FPC) is incorrect.
• (C) (i)-(a) (Arranging books - Combination) is incorrect.
• (D) (iii)-(b) (Forming numbers with distinct digits - FPC) could be argued as applicable, but (iv)-(d) (Choosing from a menu - Permutation) is generally incorrect for typical menu problems. Given the clear matches in (A), it is the intended answer.

The correct matching is (i)-(c), (ii)-(a), (iii)-(d), (iv)-(b).

The correct answer is (A) (i)-(c), (ii)-(a), (iii)-(d), (iv)-(b).

Given:

The equation ${}_nP_5 = 20 \times {}_nP_3$.

To Find:

The value of $n$.

Solution:

We are given the equation ${}_nP_5 = 20 \times {}_nP_3$.

The formula for the number of permutations of $n$ distinct items taken $r$ at a time is ${}_nP_r = \frac{n!}{(n-r)!}$.

Using this formula, we can write the terms in the given equation:

For the permutations to be defined, we must have $n \geq 5$ and $n \geq 3$. Combining these conditions, $n$ must be an integer and $n \geq 5$.

Substitute the expressions for ${}_nP_5$ and ${}_nP_3$ into the given equation:

$\frac{n!}{(n-5)!} = 20 \times \frac{n!}{(n-3)!}$

Since $n \geq 5$, $n!$ is non-zero. We can divide both sides by $n!$:

$\frac{1}{(n-5)!} = \frac{20}{(n-3)!}$

Rearrange the equation:

$(n-3)! = 20 \times (n-5)!$

Expand the factorial $(n-3)!$ until we reach $(n-5)!$:

Substitute this expansion back into the equation:

$(n-3)(n-4)(n-5)! = 20 \times (n-5)!$

Since $n \geq 5$, $(n-5)!$ is defined and non-zero. We can divide both sides by $(n-5)!$:

$(n-3)(n-4) = 20$

Expand the left side of the equation:

$n^2 - 4n - 3n + 12 = 20$}

$n^2 - 7n + 12 = 20$

Move the constant term to the left side to form a quadratic equation:

$n^2 - 7n + 12 - 20 = 0$

$n^2 - 7n - 8 = 0$

Factor the quadratic equation. We need two numbers that multiply to -8 and add up to -7. These numbers are -8 and 1.

This equation gives two possible solutions for $n$:

$n - 8 = 0 \implies n = 8$

$n + 1 = 0 \implies n = -1$

As established earlier, $n$ must be an integer and $n \geq 5$ for the permutations to be valid.

The solution $n = 8$ satisfies the condition $n \geq 5$.

The solution $n = -1$ does not satisfy the condition $n \geq 5$.

Therefore, the only valid value for $n$ is 8.

Comparing this result with the given options:

(A) 8

(B) 9

(C) 10

(D) 12

The value $n=8$ matches option (A).

The correct answer is (A) 8.

Given:

The word is "DELHI".

The word has 5 distinct letters: D, E, L, H, I.

Letters are used at most once to form a "word".

To Find:

The total number of distinct "words" (arrangements) that can be formed using the letters of "DELHI" where each letter appears 0 or 1 times.

Solution:

The phrase "letters are used at most once" means that in any formed word, each letter from "DELHI" can appear zero times or one time. This implies we are forming words of different lengths using subsets of the letters, where the order of letters matters (since we are forming "words").

This is a problem involving Permutations because the order of the letters in the formed word is important.

We can form words of length 1, length 2, length 3, length 4, or length 5 using the 5 distinct letters.

1. Number of words of length 1:

We choose 1 letter from 5 and arrange it. This is ${}_5P_1$.

2. Number of words of length 2:

We choose 2 letters from 5 and arrange them. This is ${}_5P_2$.

3. Number of words of length 3:

We choose 3 letters from 5 and arrange them. This is ${}_5P_3$.

4. Number of words of length 4:

We choose 4 letters from 5 and arrange them. This is ${}_5P_4$.

5. Number of words of length 5:

We choose 5 letters from 5 and arrange them. This is ${}_5P_5$ (or $5!$).

The total number of words that can be formed is the sum of the number of words of each possible length (1 to 5), because these are mutually exclusive cases.

Total number of words = (Number of words of length 1) + (Number of words of length 2) + (Number of words of length 3) + (Number of words of length 4) + (Number of words of length 5)

Total words = $5 + 20 + 60 + 120 + 120 = 325$.

Comparing this result with the given options:

(A) $5!$: This is ${}_5P_5$, which is only the number of 5-letter words. Incorrect.

(B) ${}_5P_1 + {}_5P_2 + {}_5P_3 + {}_5P_4 + {}_5P_5$: This matches our derived expression for the total number of words. Correct.

(C) $5^5$: This would be the number of 5-letter words if repetition of letters were allowed. Incorrect.

(D) ${}_5C_1 + {}_5C_2 + {}_5C_3 + {}_5C_4 + {}_5C_5$: This is the total number of non-empty subsets of the letters (sum of combinations), which doesn't account for the order of letters within the "word". Incorrect.

The correct answer is (B) ${}_5P_1 + {}_5P_2 + {}_5P_3 + {}_5P_4 + {}_5P_5$.

Given:

The word is "ENGINEERING".

To Find:

The number of different arrangements (permutations) that can be made with the letters of the word "ENGINEERING".

Solution:

The word "ENGINEERING" has a total of 11 letters.

To find the number of different arrangements, we first count the frequency of each distinct letter in the word:

• The letter 'E' appears 3 times.
• The letter 'N' appears 3 times.
• The letter 'G' appears 2 times.
• The letter 'I' appears 2 times.
• The letter 'R' appears 1 time.

Total number of letters, $n = 11$.

Frequencies of repeated letters:

$n_E = 3$

$n_N = 3$

$n_G = 2$

$n_I = 2$

$n_R = 1$

The formula for the number of permutations of $n$ objects where there are $n_1$ identical objects of type 1, $n_2$ identical objects of type 2, ..., $n_k$ identical objects of type k is:

Using this formula for the word "ENGINEERING":

Number of arrangements = $\frac{11!}{3! \times 3! \times 2! \times 2! \times 1!}$

Since $1! = 1$, the formula simplifies to:

Comparing this result with the given options:

(A) $\frac{11!}{3!3!2!2!}$

(B) $\frac{11!}{3!3!2!}$

(C) $\frac{11!}{3!2!2!}$

(D) $\frac{11!}{3!3!}$

Our calculated formula matches option (A).

The correct answer is (A) $\frac{11!}{3!3!2!2!}$.

Given:

Number of distinct Maths books = 6.

Number of distinct Physics books = 5.

A committee of 4 books is to be selected.

The selection must consist of exactly 2 books from Maths and 2 books from Physics.

To Find:

The number of ways to select 4 books with the specified composition.

Solution:

We need to select a total of 4 books, with the condition that 2 must be Maths books and 2 must be Physics books.

Since the order in which the books are selected does not matter (a collection of books forms the committee), this is a problem involving combinations.

We need to perform two independent selections:

1. Select 2 Maths books from the 6 available Maths books.

2. Select 2 Physics books from the 5 available Physics books.

The number of ways to select $r$ items from a set of $n$ distinct items where order does not matter is given by the combination formula:

Number of ways to choose 2 Maths books from 6 Maths books:

Calculating ${}_6C_2$:

${}_6C_2 = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} = \frac{6 \times 5 \times \cancel{4!}}{2 \times 1 \times \cancel{4!}} = \frac{30}{2} = 15$

Number of ways to choose 2 Physics books from 5 Physics books:

Calculating ${}_5C_2$:

${}_5C_2 = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4 \times \cancel{3!}}{2 \times 1 \times \cancel{3!}} = \frac{20}{2} = 10$

Since the selection of Maths books and the selection of Physics books are independent events, we use the Multiplication Principle to find the total number of ways to select the committee.

Total number of ways = $15 \times 10 = 150$

Comparing our derived expression with the given options:

(A) ${}_6C_2 + {}_5C_2$: This would be the number of ways if we selected either 2 Maths books OR 2 Physics books, which is not the condition. Incorrect.

(B) ${}_6P_2 \times {}_5P_2$: This would be used if the order of selection mattered within the group of Maths books and within the group of Physics books, which is not the case for a committee/selection of books. Incorrect.

(C) ${}_6C_2 \times {}_5C_2$: This matches our calculation and represents the number of ways to choose 2 Maths books AND 2 Physics books. Correct.

(D) ${}_{11}C_4$: This is the total number of ways to select any 4 books from the 11 available books without any constraints on the subject. Incorrect.

The correct answer is (C) ${}_6C_2 \times {}_5C_2$.

Given:

The property of combinations: ${}_nC_r = {}_{n}C_{n-r}$.

To Interpret:

The meaning or implication of this property.

Solution:

The term ${}_nC_r$ represents the number of ways to choose a subset of $r$ elements from a set of $n$ distinct elements, where the order of selection does not matter.

The term ${}_{n}C_{n-r}$ represents the number of ways to choose a subset of $n-r$ elements from the same set of $n$ distinct elements.

The property ${}_nC_r = {}_{n}C_{n-r}$ means that the number of ways to choose $r$ items from a set of $n$ is equal to the number of ways to choose $n-r$ items from the same set.

Consider a set of $n$ items. If we choose $r$ items to be in a subset, the remaining $n-r$ items are automatically *not* in that subset. Conversely, if we choose $n-r$ items to be *not* in a subset, the remaining $r$ items are automatically *in* the subset.

Therefore, choosing $r$ items is equivalent to choosing the $n-r$ items that will be left behind. Every distinct subset of size $r$ corresponds to a unique complement subset of size $n-r$, and vice versa.

The property ${}_nC_r = {}_{n}C_{n-r}$ mathematically expresses this equivalence.

Let's evaluate the given options based on this understanding:

(A) The number of ways to choose r items is the same as the number of ways to choose the items not selected. This accurately describes the relationship explained above. Choosing $r$ items is selecting them, and the $n-r$ items not selected are the items left out.

(B) The order of selection matters. This property is related to combinations (${}_nC_r$), where order does not matter. Permutations (${}_nP_r$) are used when order matters.

(C) This is a property of permutations. This is a property of combinations, not permutations. The equivalent property for permutations, ${}_nP_r = {}_{n}P_{n-r}$, is generally not true.

(D) The number of permutations equals the number of combinations. In general, ${}_nP_r \geq {}_nC_r$ for $r \geq 0$, and equality holds only for $r=0, 1$ or $r=n$. This statement is not the meaning of the given property.

Therefore, the property ${}_nC_r = {}_{n}C_{n-r}$ indicates that the number of ways to choose $r$ items is the same as the number of ways to choose the items not selected (which number $n-r$).

The correct answer is (A) The number of ways to choose r items is the same as the number of ways to choose the items not selected.

Given:

The value ${}_8C_3$ and four options.

To Find:

Which of the given options is NOT equal to ${}_8C_3$.

Solution:

First, let's calculate the value of ${}_8C_3$.

The formula for combinations is ${}_nC_r = \frac{n!}{r!(n-r)!}$.

${}_8C_3 = \frac{8!}{3!5!}$

${}_8C_3 = \frac{8 \times 7 \times 6 \times 5!}{3 \times 2 \times 1 \times 5!}$

Cancel out $5!$ from the numerator and the denominator:

${}_8C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1}$

${}_8C_3 = \frac{336}{6}$

Now, let's evaluate each option:

Option (A) ${}_8C_5$:

Using the property ${}_nC_r = {}_{n}C_{n-r}$, we have ${}_8C_5 = {}_{8}C_{8-5} = {}_8C_3$.

So, ${}_8C_5$ is equal to ${}_8C_3$.

Alternatively, calculating ${}_8C_5 = \frac{8!}{5!(8-5)!} = \frac{8!}{5!3!} = \frac{8 \times 7 \times 6 \times 5!}{5! \times 3 \times 2 \times 1} = \frac{336}{6} = 56$. This is equal to ${}_8C_3$.

Option (B) $\frac{8 \times 7 \times 6}{3 \times 2 \times 1}$:

This is the simplified form of the calculation for ${}_8C_3$ derived above.

$\frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56$. This is equal to ${}_8C_3$.

Option (C) 56:

This is the calculated value of ${}_8C_3$. This is equal to ${}_8C_3$.

Option (D) ${}_8P_3$:

This is a permutation. The formula for permutations is ${}_nP_r = \frac{n!}{(n-r)!}$.

${}_8P_3 = \frac{8!}{5!}$

${}_8P_3 = \frac{8 \times 7 \times 6 \times 5!}{5!}$

Cancel out $5!$ from the numerator and the denominator:

${}_8P_3 = 8 \times 7 \times 6$

Comparing the value of ${}_8P_3$ (336) with the value of ${}_8C_3$ (56), we see that $336 \neq 56$.

Therefore, ${}_8P_3$ is NOT equal to ${}_8C_3$.

Based on the evaluation of each option, the value that is NOT equal to ${}_8C_3$ is ${}_8P_3$.

The correct answer is (D) ${}_8P_3$.

Given:

The ratio of combinations: ${}_{2n}C_3 : {}_nC_3 = 11 : 1$.

To Find:

The value of $n$.

Solution:

The formula for the number of combinations of $n$ distinct items taken $r$ at a time is ${}_nC_r = \frac{n!}{r!(n-r)!}$.

For the combinations to be defined, we must have $2n \geq 3$ and $n \geq 3$. Since $n$ must be an integer, this implies $n \geq 3$. Also $n$ and $2n$ must be integers.

We can write the given ratio as an equation:

Now, apply the combination formula to the terms ${}_{2n}C_3$ and ${}_nC_3$:

${}_{2n}C_3 = \frac{(2n)!}{3!(2n-3)!}$

${}_nC_3 = \frac{n!}{3!(n-3)!}$

Substitute these into equation (i):

$\frac{\frac{(2n)!}{3!(2n-3)!}}{\frac{n!}{3!(n-3)!}} = 11$

Simplify the expression:

$\frac{(2n)!}{3!(2n-3)!} \times \frac{3!(n-3)!}{n!} = 11$

Cancel out $3!$ from the numerator and denominator:

$\frac{(2n)!}{(2n-3)!} \times \frac{(n-3)!}{n!} = 11$

Expand the factorials $(2n)!$ and $n!$ until we reach $(2n-3)!$ and $(n-3)!$ respectively:

Substitute these expansions into the simplified equation:

$\frac{(2n)(2n-1)(2n-2)(2n-3)!}{(2n-3)!} \times \frac{(n-3)!}{n(n-1)(n-2)(n-3)!} = 11$

Assuming $n \geq 3$, $(2n-3)!$ and $(n-3)!$ are non-zero, so we can cancel them:

$(2n)(2n-1)(2n-2) \times \frac{1}{n(n-1)(n-2)} = 11$

Factor out a 2 from $(2n-2)$: $2n(2n-1)2(n-1) \times \frac{1}{n(n-1)(n-2)} = 11$

$4n(2n-1)(n-1) \times \frac{1}{n(n-1)(n-2)} = 11$

Since $n \geq 3$, $n \neq 0$ and $n-1 \neq 0$. We can cancel $n$ and $(n-1)$ from the numerator and denominator:

Now, solve for $n$ by cross-multiplying:

$4(2n-1) = 11(n-2)$

Distribute on both sides:

$8n - 4 = 11n - 22$

Gather terms with $n$ on one side and constants on the other:

$22 - 4 = 11n - 8n$

$18 = 3n$

Divide by 3:

$n = 6$

Check if $n=6$ satisfies the condition $n \geq 3$: Yes, $6 \geq 3$. The combination terms are valid.

Comparing this result with the given options:

(A) 4

(B) 5

(C) 6

(D) 8

Our calculated value $n=6$ matches option (C).

The correct answer is (C) 6.

Given:

Available digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (Total 10 distinct digits).

We need to form 5-digit numbers.

Repetition of digits is not allowed.

To Find:

The number of distinct 5-digit numbers that can be formed.

Solution:

A 5-digit number has 5 places: ten thousands, thousands, hundreds, tens, and units place.

Let the 5-digit number be represented by the places $\_$ $\_$ $\_$ $\_$ $\_$.

Since the number must be a 5-digit number, the digit in the ten thousands place (the first place) cannot be 0.

Also, repetition of digits is not allowed.

We can solve this using the Fundamental Principle of Counting or by considering permutations and excluding invalid cases.

Method 1: Using the Fundamental Principle of Counting

Consider the choices for each position:

1. Ten Thousands Place (1st digit): The digit cannot be 0. So there are 9 choices (1, 2, 3, 4, 5, 6, 7, 8, 9).

2. Thousands Place (2nd digit): One digit has been used in the first place. We have 9 remaining digits (including 0, which is now allowed). So there are 9 choices.

3. Hundreds Place (3rd digit): Two digits have been used in the first two places. We have 8 remaining digits. So there are 8 choices.

4. Tens Place (4th digit): Three digits have been used. We have 7 remaining digits. So there are 7 choices.

5. Units Place (5th digit): Four digits have been used. We have 6 remaining digits. So there are 6 choices.

By the Fundamental Principle of Counting (Multiplication Rule), the total number of 5-digit numbers is the product of the number of choices for each position:

Notice that the product $9 \times 8 \times 7 \times 6$ is the number of permutations of 9 distinct items taken 4 at a time, which is ${}_9P_4$.

$9 \times 8 \times 7 \times 6 = {}_9P_4 = \frac{9!}{(9-4)!} = \frac{9!}{5!}$

So, the total number of 5-digit numbers is $9 \times {}_9P_4$.

Method 2: Using Total Permutations minus Invalid Permutations

Consider all possible arrangements of 5 distinct digits chosen from the 10 available digits (0-9). The number of such arrangements is the number of permutations of 10 items taken 5 at a time, ${}_{10}P_5$.

This includes sequences that start with 0 (e.g., 01234). These are not valid 5-digit numbers.

We need to subtract the number of invalid sequences, which are the 5-digit sequences starting with 0.

If the first digit is 0, we need to arrange the remaining 4 digits using the other 9 available digits (1, 2, 3, 4, 5, 6, 7, 8, 9). The number of ways to arrange 4 distinct digits from 9 available digits is the number of permutations of 9 items taken 4 at a time, ${}_9P_4$.

The number of valid 5-digit numbers is the total number of sequences minus the number of invalid sequences:

As shown in Method 1, $10 \times 9 \times 8 \times 7 \times 6 = 10 \times (9 \times 8 \times 7 \times 6) = 10 \times {}_9P_4$.

So, ${}_{10}P_5 - {}_9P_4 = 10 \times {}_9P_4 - {}_9P_4 = (10 - 1) \times {}_9P_4 = 9 \times {}_9P_4$.

Both methods yield the same result, $9 \times {}_9P_4$, which is equal to ${}_{10}P_5 - {}_9P_4$.

$9 \times {}_9P_4 = 9 \times (9 \times 8 \times 7 \times 6) = 9 \times 3024 = 27216$

Comparing our results with the given options:

(A) ${}_{10}P_5$: Incorrect ($30240$).

(B) ${}_{10}C_5$: Incorrect (Combinations do not account for order).

(C) ${}_{10}P_5 - {}_{9}P_4$: Matches our calculation ($27216$).

(D) $9 \times {}_9P_4$: Matches our calculation ($27216$).

Both options (C) and (D) represent the correct answer. However, option (D) directly reflects the logic of Method 1 (9 choices for the first digit, and then arranging 4 out of the remaining 9 digits for the rest). Option (C) reflects the logic of Method 2 (total permutations minus invalid ones). Both are valid ways to express the result.

Assuming the question expects one of the provided symbolic forms, both (C) and (D) are mathematically equivalent and correct. In multiple-choice questions like this, sometimes there might be a preferred form based on how the topic was taught or the source of the question. However, based strictly on mathematical equivalence, both (C) and (D) are correct.

Let's choose option (D) as it aligns directly with the step-by-step construction using the Fundamental Principle of Counting, which is a fundamental approach to such problems.

The correct answer is (D) $9 \times {}_9P_4$.

Given:

The word "OPTICAL".

The condition is that the vowels must always come together in any arrangement.

To Find:

The number of different arrangements of the letters of the word "OPTICAL" such that all vowels are together.

Solution:

The word "OPTICAL" has 7 distinct letters: O, P, T, I, C, A, L.

Identify the vowels and consonants in the word:

• Vowels: O, I, A (3 distinct vowels)
• Consonants: P, T, C, L (4 distinct consonants)

The condition is that the vowels must always come together. To handle this, we treat the group of vowels (O, I, A) as a single unit or block.

Now, we have the following items to arrange:

• The block of vowels (OIA)
• The 4 consonants (P, T, C, L)

So, we are essentially arranging 5 distinct items: the vowel block and the 4 individual consonants. The number of ways to arrange these 5 distinct items is given by the factorial of the number of items.

Calculate $5!$:

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

Next, the vowels within the vowel block (OIA) can arrange themselves in any order. Since there are 3 distinct vowels (O, I, A) within the block, they can be arranged in $3!$ ways.

Calculate $3!$:

$3! = 3 \times 2 \times 1 = 6$

To find the total number of arrangements where the vowels are together, we multiply the number of ways to arrange the 5 items (vowel block + consonants) by the number of ways to arrange the vowels within their block. This is because for each arrangement of the 5 items, there are $3!$ ways to arrange the vowels inside their block.

Total number of arrangements = $120 \times 6 = 720$

Comparing our derived expression with the given options:

(A) $5! \times 3!$: Matches our calculation. Correct.

(B) $7!$: This is the total number of arrangements of 7 distinct letters without any restrictions. Incorrect.

(C) $5! + 3!$: Addition is used when dealing with mutually exclusive cases, not sequential events like arranging groups and then items within groups. Incorrect.

(D) $7!/3!$: This would be the number of arrangements if there were 3 identical letters in the word. Incorrect.

The correct answer is (A) $5! \times 3!$.

Given:

Total number of students = 10.

Team size to be selected = 4.

Condition: Two particular students must always be excluded from the team.

To Find:

The number of ways to select a team of 4 students under the given condition.

Solution:

We need to select a team of 4 students from a group of 10.

The condition states that two particular students must always be excluded from the team. This means these two students cannot be part of the selection pool.

First, we determine the number of students available for selection after excluding the two particular students.

Now, we need to select a team of 4 students from these 8 available students.

Since the order in which students are selected for a team does not matter, this is a problem of combination.

The number of ways to select $r$ items from a set of $n$ distinct items where order does not matter is given by the combination formula ${}_nC_r = \frac{n!}{r!(n-r)!}$.

In this case, $n = 8$ (available students) and $r = 4$ (students to be selected for the team).

Let's calculate the value of ${}_8C_4$:

${}_8C_4 = \frac{8!}{4!(8-4)!}

${}_8C_4 = \frac{8!}{4!4!}

${}_8C_4 = \frac{8 \times 7 \times 6 \times 5 \times 4!}{4 \times 3 \times 2 \times 1 \times 4!}

Cancel out the $4!$ terms:

${}_8C_4 = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}

${}_8C_4 = \frac{\cancel{8}^{1} \times 7 \times \cancel{6}^{1} \times 5}{\cancel{4}_{1} \times \cancel{3}_{1} \times \cancel{2}_{1} \times 1}$

So, there are 70 ways to select a team of 4 students from the group of 10, such that two particular students are always excluded.

Comparing our calculation with the given options:

(A) ${}_{10}C_4$: This is the number of ways to select 4 students from 10 without any restrictions. Incorrect.

(B) ${}_{10}C_4 - {}_2C_2 \times {}_8C_2$: This expression seems to subtract cases where the two excluded students are included, which cannot happen under the given condition. Incorrect formulation.

(C) ${}_8C_4$: This represents the number of ways to choose 4 students from the 8 students who are available for selection (after excluding the two). This matches our logic and calculation. Correct.

(D) ${}_{10}C_4 - {}_2C_0 \times {}_8C_4$: This subtracts the number of ways to choose 0 from the excluded students and 4 from the remaining 8, from the total number of ways to choose 4 from 10. While ${}_2C_0 \times {}_8C_4 = 1 \times {}_8C_4 = {}_8C_4$, the form ${}_{10}C_4 - {}_8C_4$ is generally not the correct way to handle exclusion. The direct method of reducing the pool of available items is preferred and leads directly to ${}_8C_4$.

The correct answer is (C) ${}_8C_4$.

Given:

The relationship between permutation and combination: ${}_nP_r = k \times {}_nC_r$.

To Find:

The value of $k$.

Solution:

We know the formulas for permutations and combinations are:

The given equation is:

Substitute the formulas from (i) and (ii) into equation (iii):

$\frac{n!}{(n-r)!} = k \times \frac{n!}{r!(n-r)!}$

To solve for $k$, we can divide both sides of the equation by ${}_nC_r$, assuming ${}_nC_r \neq 0$ (which is true for $0 \leq r \leq n$):

$k = \frac{{}_nP_r}{{}_nC_r}$

Substitute the factorial expressions:

$k = \frac{\frac{n!}{(n-r)!}}{\frac{n!}{r!(n-r)!}}$

When dividing by a fraction, we multiply by its reciprocal:

$k = \frac{n!}{(n-r)!} \times \frac{r!(n-r)!}{n!}$

Assuming $n \geq r$, $n!$, $r!$, and $(n-r)!$ are defined. We can cancel out common terms $n!$ and $(n-r)!$ from the numerator and denominator:

This leaves us with:

So, the value of $k$ is $r!$.

Comparing this result with the given options:

(A) $r!$

(B) $(n-r)!$

(C) $n!$

(D) $1/r!$

Our derived value $k = r!$ matches option (A).

The correct answer is (A) $r!$.

Given:

Number of red balls = 6.

Number of blue balls = 4.

Total number of balls = $6 + 4 = 10$.

Number of balls to be drawn = 3.

Condition: At least one red ball must be included in the selection.

To Find:

The number of ways to draw 3 balls such that at least one red ball is included.

Solution (Method 1: Summing up favorable cases):

"At least one red ball" means the selection of 3 balls can have:

Case 1: Exactly 1 red ball and 2 blue balls.

Case 2: Exactly 2 red balls and 1 blue ball.

Case 3: Exactly 3 red balls and 0 blue balls.

These cases are mutually exclusive, so we can find the number of ways for each case and add them up.

The number of ways to choose $r$ items from $n$ distinct items is given by ${}_nC_r = \frac{n!}{r!(n-r)!}$.

Case 1: 1 Red and 2 Blue

Number of ways to choose 1 red ball from 6 = ${}_6C_1$

Number of ways to choose 2 blue balls from 4 = ${}_4C_2$

${}_6C_1 = \frac{6!}{1!5!} = 6$

${}_4C_2 = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6$

Ways for Case 1 = $6 \times 6 = 36$

Case 2: 2 Red and 1 Blue

Number of ways to choose 2 red balls from 6 = ${}_6C_2$

Number of ways to choose 1 blue ball from 4 = ${}_4C_1$

${}_6C_2 = \frac{6!}{2!4!} = \frac{6 \times 5}{2 \times 1} = 15$

${}_4C_1 = \frac{4!}{1!3!} = 4$

Ways for Case 2 = $15 \times 4 = 60$

Case 3: 3 Red and 0 Blue

Number of ways to choose 3 red balls from 6 = ${}_6C_3$

Number of ways to choose 0 blue balls from 4 = ${}_4C_0$

${}_6C_3 = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$

${}_4C_0 = \frac{4!}{0!4!} = 1$

Ways for Case 3 = $20 \times 1 = 20$

Total number of ways = Ways for Case 1 + Ways for Case 2 + Ways for Case 3

Total ways = $36 + 60 + 20 = 116$

This matches option (B).

Alternate Solution (Method 2: Using the complement):

The total number of ways to draw any 3 balls from the 10 balls (without any restriction) is given by ${}_{10}C_3$.

${}_{10}C_3 = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120$

The opposite of having "at least one red ball" is having "no red balls". If there are no red balls, it means all 3 balls drawn must be blue.

The number of ways to draw 3 blue balls from the 4 blue balls is given by ${}_4C_3$.

${}_4C_3 = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = \frac{4 \times 3 \times 2 \times 1}{3 \times 2 \times 1 \times 1} = 4$

The number of ways to draw at least one red ball is the total number of ways to draw 3 balls minus the number of ways to draw 3 blue balls (i.e., no red balls).

Ways (at least one red) = $120 - 4 = 116$

This matches option (A).

Both methods lead to the same result (116) and correspond to the expressions given in options (A) and (B).

Therefore, the correct answer is the option that states both (A) and (B) are correct.

The correct answer is (C) Both (A) and (B).

Given:

A polygon having 12 sides.

A polygon with 12 sides has 12 vertices.

To Find:

The number of different triangles that can be formed by joining the vertices of the polygon.

Solution:

To form a triangle, we need to select any 3 distinct vertices from the available vertices of the polygon.

The number of vertices in a polygon with 12 sides is 12.

We are selecting 3 vertices from a set of 12 vertices.

The order in which we select the vertices does not matter; selecting vertex A, then B, then C forms the same triangle as selecting B, then C, then A.

Since the order of selection does not matter, this is a problem of combination.

The number of ways to choose $r$ items from a set of $n$ distinct items where order does not matter is given by the combination formula:

In this case, $n = 12$ (total number of vertices) and $r = 3$ (number of vertices required to form a triangle).

Now, we calculate the value of ${}_{12}C_3$:

${}_{12}C_3 = \frac{12!}{3!(12-3)!}

${}_{12}C_3 = \frac{12!}{3!9!}

${}_{12}C_3 = \frac{12 \times 11 \times 10 \times 9!}{3 \times 2 \times 1 \times 9!}$

Cancel out $9!$ from the numerator and the denominator:

${}_{12}C_3 = \frac{12 \times 11 \times 10}{3 \times 2 \times 1}$

${}_{12}C_3 = \frac{1320}{6}

So, 220 different triangles can be formed by joining the vertices of a polygon having 12 sides.

Comparing our derived expression and calculated value with the given options:

(A) ${}_{12}P_3$: This is the number of permutations, $12 \times 11 \times 10 = 1320$. Incorrect.

(B) ${}_{12}C_3$: This matches our derived expression and calculation ($220$). Correct.

(C) 12: This is the number of vertices/sides. Incorrect.

(D) ${}_{12}C_3 - 12$: This expression does not represent the number of triangles formed from polygon vertices. Incorrect.

The correct answer is (B) ${}_{12}C_3$.

Given:

The equation: ${}_nC_r + {}_nC_{r-1} = {}_{n+1}C_k$

To Find:

The value of $k$.

Solution:

We are given the equation ${}_nC_r + {}_nC_{r-1} = {}_{n+1}C_k$.

This equation involves the sum of two combinations with the same upper index ($n$) and consecutive lower indices ($r$ and $r-1$).

There is a well-known identity in combinatorics called Pascal's Identity, which states that:

This identity is fundamental in the construction of Pascal's triangle, where each entry is the sum of the two entries directly above it.

Comparing the given equation ${}_nC_r + {}_nC_{r-1} = {}_{n+1}C_k$ with Pascal's Identity ${}_nC_r + {}_nC_{r-1} = {}_{n+1}C_r$, we can equate the right-hand sides:

For combinations with the same upper index to be equal, the lower indices must either be equal or sum up to the upper index.

So, from ${}_{n+1}C_k = {}_{n+1}C_r$, we have two possibilities:

Possibility 1: $k = r$

Possibility 2: $k + r = n + 1$, which implies $k = n + 1 - r$.

However, the identity ${}_nC_r + {}_nC_{r-1}$ specifically results in the element in the $(n+1)$-th row at position $r$ (or $r+1$ if you start counting from $0$). The direct result of the sum ${}_nC_r + {}_nC_{r-1}$ is ${}_{n+1}C_r$. Thus, in the context of this identity, the value of $k$ is precisely $r$. While ${}_{n+1}C_r = {}_{n+1}C_{n+1-r}$ is also true by the symmetry property of combinations, the identity ${}_nC_r + {}_nC_{r-1}$ equals ${}_{n+1}C_r$.

Therefore, based on Pascal's Identity, $k$ is equal to $r$.

Comparing our result with the given options:

(A) $r-1$

(B) $r$

(C) $n$

(D) $n+1$

Our derived value $k=r$ matches option (B).

The correct answer is (B) $r$.

Given:

Four mathematical expressions involving factorials, permutations, and combinations.

To Find:

Which of the given options represents a NOT valid calculation.

Solution:

Let's analyze each option based on the definitions of factorials, permutations, and combinations:

(A) $7!$:

The factorial of a non-negative integer $n$, denoted by $n!$, is the product of all positive integers less than or equal to $n$. $n! = n \times (n-1) \times ... \times 2 \times 1$. The factorial is defined for $n \geq 0$, with $0! = 1$ by definition.

Here, $n=7$, which is a non-negative integer. So, $7!$ is a valid calculation:

(B) ${}_5P_7$:

The number of permutations of $n$ distinct items taken $r$ at a time, denoted by ${}_nP_r$, is the number of ways to select $r$ items from a set of $n$ distinct items and arrange them in order. The formula is ${}_nP_r = \frac{n!}{(n-r)!}$. This is defined only when $n \geq r \geq 0$.

Here, $n=5$ and $r=7$. Since $r > n$ (7 > 5), it is not possible to select and arrange 7 distinct items from a set of only 5 distinct items. Applying the formula would involve $\frac{5!}{(5-7)!} = \frac{5!}{(-2)!}$, and the factorial of a negative integer is undefined in this context.

(C) ${}_9C_3$:

The number of combinations of $n$ distinct items taken $r$ at a time, denoted by ${}_nC_r$, is the number of ways to select $r$ items from a set of $n$ distinct items where the order of selection does not matter. The formula is ${}_nC_r = \frac{n!}{r!(n-r)!}$. This is defined only when $n \geq r \geq 0$.

Here, $n=9$ and $r=3$. Since $9 \geq 3 \geq 0$, ${}_9C_3$ is a valid calculation:

(D) $0!$:

As mentioned in option (A), the factorial of 0 is defined as $0! = 1$.

Based on the analysis, the only expression that represents a NOT valid calculation is ${}_5P_7$ because the number of items to be arranged (7) is greater than the number of available items (5).

The correct answer is (B) ${}_5P_7$.

Given:

Available digits: 0, 1, 2, 3, 4 (a set of 5 distinct digits).

We need to form numbers greater than 1000 and less than 4000.

Repetition of digits is not allowed.

To Find:

The number of distinct integers that can be formed using the given digits, which are greater than 1000 and less than 4000, without repetition.

Solution:

The numbers must be greater than $1000$ and less than $4000$. This implies that the numbers must be 4-digit numbers.

A 4-digit number can be represented by four positions: thousands, hundreds, tens, and units place.

Let the 4-digit number be $\textsf{abcd}$, where a, b, c, and d are the digits in the thousands, hundreds, tens, and units place, respectively.

The condition that the number is greater than 1000 and less than 4000 restricts the possible values for the thousands digit (a).

The thousands digit (a) cannot be 0 (as it would be a 3-digit number).

The thousands digit (a) cannot be 4 or greater (as the number would be 4000 or greater).

The available digits are {0, 1, 2, 3, 4}.

Therefore, the thousands digit (a) must be one of the digits from {1, 2, 3}.

Since repetition is not allowed, the digits used for the remaining three positions (hundreds, tens, units) must be chosen from the remaining digits in the set.

After choosing the first digit, there are $5 - 1 = 4$ digits remaining from the original set of 5 digits.

For the hundreds digit (b): Any of the 4 remaining digits can be used.

After choosing the first two digits, there are $5 - 2 = 3$ digits remaining.

For the tens digit (c): Any of the 3 remaining digits can be used.

After choosing the first three digits, there are $5 - 3 = 2$ digits remaining.

For the units digit (d): Any of the 2 remaining digits can be used.

By the Fundamental Principle of Counting (Multiplication Rule), the total number of different 4-digit numbers that can be formed is the product of the number of choices for each position.

Calculate the product:

$3 \times 4 = 12$

$12 \times 3 = 36$

$36 \times 2 = 72$

So, there are 72 numbers greater than 1000 and less than 4000 that can be formed using the digits 0, 1, 2, 3, 4 if repetition is not allowed.

Comparing this result with the given options:

(A) 144

(B) 108

(C) 72

(D) 60

Our calculated value 72 matches option (C).

The correct answer is (C) 72.

Given:

Question paper has two parts: Part A and Part B.

Number of questions in Part A = 10.

Number of questions in Part B = 10.

Total number of questions to attempt = 8.

Constraint: Student must choose at least 4 questions from each part.

To Find:

The number of ways the student can select the questions satisfying the conditions.

Solution:

Let $n_A$ be the number of questions selected from Part A, and $n_B$ be the number of questions selected from Part B.

According to the problem statement, the total number of questions attempted is 8:

The constraint is that the student must choose at least 4 questions from each part. This translates to:

We need to find the possible integer values for $n_A$ and $n_B$ that satisfy equation (i) and both inequality constraints.

Let's list the possible pairs $(n_A, n_B)$ that sum to 8, keeping in mind that $0 \leq n_A \leq 10$ and $0 \leq n_B \leq 10$ since there are 10 questions in each part:

• (0, 8) - $0 \not\geq 4$ (violates $n_A \ge 4$)
• (1, 7) - $1 \not\geq 4$ (violates $n_A \ge 4$)
• (2, 6) - $2 \not\geq 4$ (violates $n_A \ge 4$)
• (3, 5) - $3 \not\geq 4$ (violates $n_A \ge 4$)
• (4, 4) - $4 \geq 4$ (satisfies $n_A \ge 4$) and $4 \geq 4$ (satisfies $n_B \ge 4$). This case is valid.
• (5, 3) - $3 \not\geq 4$ (violates $n_B \ge 4$)
• (6, 2) - $2 \not\geq 4$ (violates $n_B \ge 4$)
• (7, 1) - $1 \not\geq 4$ (violates $n_B \ge 4$)
• (8, 0) - $0 \not\geq 4$ (violates $n_B \ge 4$)

The only combination of questions from Part A and Part B that satisfies all conditions is selecting exactly 4 questions from Part A and exactly 4 questions from Part B.

Now, we calculate the number of ways to select questions for this valid case:

Number of ways to select 4 questions from 10 questions in Part A = ${}_{10}C_4$

Number of ways to select 4 questions from 10 questions in Part B = ${}_{10}C_4$

Since the selection from Part A and the selection from Part B are independent events, the total number of ways for this case is the product of the individual ways (by the Multiplication Principle).

Let's calculate the value:

${}_{10}C_4 = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 10 \times 3 \times 7 = 210$

Total number of ways = $210 \times 210 = 44100$

Comparing our result with the given options:

(A) ${}_{10}C_4 \times {}_{10}C_4 + {}_{10}C_5 \times {}_{10}C_3 + {}_{10}C_3 \times {}_{10}C_5$: This includes cases like (5 from A, 3 from B) and (3 from A, 5 from B), which violate the "at least 4 from each part" constraint. Incorrect.

(B) ${}_{10}C_4 \times {}_{10}C_4 + {}_{10}C_5 \times {}_{10}C_3$: This includes the case (5 from A, 3 from B), which violates the "at least 4 from each part" constraint. Incorrect.

(C) ${}_{10}C_8$: This represents choosing 8 questions from a pool of 10, ignoring the existence of two parts and the constraint. Incorrect (and the total pool is 20, not 10).

(D) ${}_{10}C_4 \times {}_{10}C_4$: This matches our calculation for the only valid case. Correct.

The correct answer is (D) ${}_{10}C_4 \times {}_{10}C_4$.

Given:

The word "LEADING".

The condition is that the vowels must occupy only the odd positions.

To Find:

The number of different arrangements of the letters of the word "LEADING" such that the vowels are always in odd positions.

Solution:

The word "LEADING" has 7 distinct letters: L, E, A, D, I, N, G.

Identify the vowels and consonants:

• Vowels: E, A, I (3 distinct vowels)
• Consonants: L, D, N, G (4 distinct consonants)

There are 7 positions in the word, which can be labelled 1st, 2nd, 3rd, 4th, 5th, 6th, 7th.

Identify the odd and even positions:

• Odd positions: 1st, 3rd, 5th, 7th (4 positions)
• Even positions: 2nd, 4th, 6th (3 positions)

The condition is that the 3 vowels must occupy only the odd positions. There are 4 available odd positions for the 3 vowels.

We can solve this problem by considering the arrangements in steps:

Step 1: Place the vowels in the odd positions.

We need to choose 3 of the 4 odd positions and arrange the 3 distinct vowels in these chosen 3 positions. The number of ways to do this is the number of permutations of 4 items taken 3 at a time, ${}_4P_3$.

Alternatively, this can be broken down as:

- Choose 3 odd positions out of 4: ${}_4C_3$ ways.

- Arrange the 3 distinct vowels in the 3 chosen positions: $3!$ ways.

Using the multiplication principle, the number of ways to place the vowels is ${}_4C_3 \times 3!$.

Note that ${}_4C_3 \times 3! = \frac{4!}{3!1!} \times 3! = \frac{4!}{1!} = 4!$. Also, ${}_4P_3 = \frac{4!}{(4-3)!} = \frac{4!}{1!} = 4!$. So, both expressions are equal to $4!$.

Step 2: Place the consonants in the remaining positions.

There are 7 total positions. 3 positions have been occupied by vowels (chosen from the 4 odd positions). So, the number of remaining positions is $7 - 3 = 4$.

These 4 remaining positions consist of the 3 even positions and the 1 odd position that was not selected for a vowel.

We have 4 distinct consonants (L, D, N, G) to arrange in these 4 remaining positions. The number of ways to arrange 4 distinct items in 4 distinct positions is $4!$.

Step 3: Combine the steps.

Since the placement of vowels and the placement of consonants are independent events, we multiply the number of ways for each step to get the total number of arrangements.

Total number of arrangements = (Ways to place vowels) $\times$ (Ways to place consonants)

Or, using the expanded form for placing vowels:

Calculate the value:

${}_4P_3 = 4 \times 3 \times 2 = 24$

$4! = 4 \times 3 \times 2 \times 1 = 24$

Total arrangements = $24 \times 24 = 576$.

Let's check the options:

(A) ${}_4P_3 \times 4!$: This matches one of our derived formulas.

(B) ${}_4C_3 \times 3! \times 4!$: This matches the other derived formula, and is equivalent to option (A).

(C) $3! \times 4!$: This would imply placing 3 items in 3 slots and 4 items in 4 slots, which is not the scenario here (vowels must go into specific *types* of slots, and there are more odd slots than vowels). Incorrect.

(D) $7!$: This is the total number of arrangements without any restrictions. Incorrect.

Both (A) and (B) are mathematically correct and represent the number of ways. However, option (B) better represents the standard step-by-step process of choosing positions first and then arranging the items, which might be the intended formulation.

The correct answer is (A) ${}_4P_3 \times 4!$ (which is equivalent to option B).

Given:

${}_nC_r = 35$

${}_nP_r = 210$

To Find:

The values of $n$ and $r$.

Solution:

We are given the values of a combination and a permutation involving the same $n$ and $r$.

We know the relationship between permutations and combinations:

Substitute the given values ${}_nP_r = 210$ and ${}_nC_r = 35$ into this relationship:

Now, we can solve for $r!$ by dividing both sides by 35:

$r! = 6$

We need to find the integer $r$ such that its factorial is 6.

Let's calculate factorials of small integers:

$1! = 1$

$2! = 2 \times 1 = 2$}

$3! = 3 \times 2 \times 1 = 6$

So, we find that $r! = 6$ when $r = 3$.

Now that we have the value of $r$, we can use either of the original equations to find $n$. Let's use the combination equation ${}_nC_r = 35$ with $r=3$:

Using the formula ${}_nC_r = \frac{n!}{r!(n-r)!}$:

$\frac{n!}{3!(n-3)!} = 35$

Substitute the value of $3!$ (which is 6):

$\frac{n!}{6(n-3)!} = 35$}

$n! = 35 \times 6 \times (n-3)!$

$n! = 210 \times (n-3)!$

Expand $n!$ until we reach $(n-3)!$:

Assuming $n \geq 3$, $(n-3)!$ is non-zero, so we can divide both sides by $(n-3)!$:

We need to find an integer $n$ such that the product of $n$ and the two consecutive integers $(n-1)$ and $(n-2)$ is 210. We know $n$ must be $\geq r$, so $n \geq 3$.

Let's test integer values for $n$ starting from 3:

If $n=3$: $3(3-1)(3-2) = 3 \times 2 \times 1 = 6 \neq 210$

If $n=4$: $4(4-1)(4-2) = 4 \times 3 \times 2 = 24 \neq 210$

If $n=5$: $5(5-1)(5-2) = 5 \times 4 \times 3 = 60 \neq 210$

If $n=6$: $6(6-1)(6-2) = 6 \times 5 \times 4 = 120 \neq 210$

If $n=7$: $7(7-1)(7-2) = 7 \times 6 \times 5 = 210$

So, the value of $n$ is 7.

The values of $n$ and $r$ are 7 and 3, respectively.

Let's verify these values using the given permutation equation ${}_nP_r = 210$:

${}_7P_3 = \frac{7!}{(7-3)!} = \frac{7!}{4!} = \frac{7 \times 6 \times 5 \times 4!}{4!} = 7 \times 6 \times 5 = 210$.

This confirms our values for $n$ and $r$ are correct.

Comparing our result with the given options:

(A) $n=7, r=3$

(B) $n=6, r=3$

(C) $n=5, r=3$

(D) $n=7, r=2$

Our calculated values $n=7$ and $r=3$ match option (A).

The correct answer is (A) $n=7, r=3$.

Given:

Total number of points = 10.

Condition: 3 of these points are collinear (lie on the same straight line).

To Find:

The number of distinct straight lines that can be drawn by joining any two of these points.

Solution:

A straight line is uniquely determined by selecting any two distinct points.

If all 10 points were non-collinear, the number of lines that could be drawn would be the number of ways to choose 2 points from 10, which is given by the combination ${}_{10}C_2$.

${}_{10}C_2 = \frac{10!}{2!(10-2)!} = \frac{10!}{2!8!} = \frac{10 \times 9}{2 \times 1} = 45$

However, we are given that 3 of the 10 points are collinear. Let these 3 points be P, Q, and R.

If these 3 points were non-collinear, choosing any 2 of them would form a distinct line. The number of lines formed by these 3 points if they were non-collinear would be ${}_3C_2$.

${}_3C_2 = \frac{3!}{2!(3-2)!} = \frac{3!}{2!1!} = \frac{3 \times 2 \times 1}{2 \times 1 \times 1} = 3$

These 3 lines would be PQ, QR, and PR.

But since these 3 points are collinear, choosing any two of them (P and Q, Q and R, or P and R) results in the same single straight line.

The combinations ${}_3C_2$ count the pairs of points among the collinear ones. These ${}_3C_2 = 3$ pairs would form 3 distinct lines if the points were not collinear. However, because they are collinear, they form only 1 line.

So, from the total count ${}_{10}C_2$ (assuming all points are non-collinear), we have overcounted the lines formed by the 3 collinear points. We have counted ${}_3C_2 = 3$ lines from these 3 points, whereas there is actually only 1 line.

To correct the overcounting, we subtract the number of lines that would be formed by the 3 collinear points if they were non-collinear (${}_3C_2$) and add back the single line that is actually formed by these 3 collinear points (which is 1 line).

Number of lines = (Total lines assuming non-collinear) - (Lines formed by collinear points if non-collinear) + (Actual line formed by collinear points)

Substitute the calculated values:

Number of lines = $45 - 3 + 1$}

Number of lines = $42 + 1 = 43$

Comparing our derived expression and calculated value with the given options:

(A) ${}_{10}C_2$: This is the total number of lines if all points were non-collinear. Incorrect.

(B) ${}_{10}C_2 - {}_3C_2$: This subtracts the overcounted lines but doesn't add back the single line formed by the collinear points. Incorrect ($45 - 3 = 42$).

(C) ${}_{10}C_2 - {}_3C_2 + 1$: This correctly subtracts the overcounted lines and adds back the single line formed by the collinear points. Correct ($45 - 3 + 1 = 43$).

(D) ${}_{10}P_2$: This is the number of permutations, which counts ordered pairs of points, and does not apply to the formation of lines where order does not matter. Incorrect.

The correct answer is (C) ${}_{10}C_2 - {}_3C_2 + 1$.

Given:

Number of different prizes = 5.

Number of students = 3.

Condition: Each student can get any number of prizes.

To Find:

The total number of ways the prizes can be distributed among the students.

Solution:

We have 5 distinct prizes to distribute among 3 distinct students.

For each prize, we need to decide which student receives it.

Consider the first prize. It can be given to any of the 3 students.

Consider the second prize. Since each student can get any number of prizes, the second prize can also be given to any of the 3 students, regardless of who received the first prize.

Similarly, for each of the 5 prizes, there are 3 independent choices for the recipient.

Number of choices for the 3rd prize = 3

Number of choices for the 4th prize = 3

Number of choices for the 5th prize = 3

By the Fundamental Principle of Counting (Multiplication Rule), the total number of ways to distribute the 5 prizes is the product of the number of choices for each prize.

Total number of ways = $3^5$

$3^5 = 3 \times 3 \times 3 \times 3 \times 3 = 9 \times 9 \times 3 = 81 \times 3 = 243$

So, there are $3^5$ ways to distribute the 5 different prizes among 3 students such that each student can get any number of prizes.

Comparing our derived expression with the given options:

(A) $5^3$: This would be the number of ways to distribute 3 different prizes among 5 students where each student can get any number of prizes. Incorrect.

(B) $3^5$: This matches our derived expression. Correct.

(C) ${}_5P_3$: This is the number of ways to arrange 3 items out of 5, or distribute 3 distinct items into 5 distinct ordered slots without repetition. Incorrect.

(D) ${}_5C_3$: This is the number of ways to choose 3 items out of 5, where order doesn't matter. Incorrect.

The correct answer is (B) $3^5$.

Given:

Four different scenarios.

To Identify:

Which scenario involves a combination.

Solution:

A combination is a selection of items from a set where the order of selection does not matter. A permutation is an arrangement of items from a set where the order of selection or arrangement does matter.

Let's analyze each scenario:

(A) Arranging books on a shelf: The order in which the books are placed on the shelf creates a different arrangement. For example, placing book A then book B is different from placing book B then book A. Since the order matters, this involves a Permutation.

(B) Forming a password using digits and letters: A password is a sequence of characters. The order of the characters is crucial. For example, "abc1" is a different password from "1abc". Since the order matters, this involves a Permutation (or the Fundamental Principle of Counting for ordered selections with or without repetition).

(C) Selecting a team of 5 players from 11: A team is a group of players. The order in which the players are selected does not change the composition of the team. For example, selecting player A, then B, then C, then D, then E results in the same team as selecting player E, then D, then C, then B, then A. Since the order of selection does not matter for the team composition, this involves a Combination.

(D) Assigning roles (like President, Vice-President) to members of a committee: Assigning specific roles to individuals involves arrangement or ordering. For example, assigning A as President and B as Vice-President is different from assigning B as President and A as Vice-President. Since specific positions/roles are assigned, the order of assignment matters, involving a Permutation.

Based on the analysis, the scenario that involves a combination is selecting a team of players from a group.

The correct answer is (C) Selecting a team of 5 players from 11.

Given:

The equation ${}_nC_{10} = {}_nC_{12}$.

To Find:

The value of ${}_{n}C_n$.

Solution:

We are given the equation ${}_nC_{10} = {}_nC_{12}$.

We use the property of combinations which states that if ${}_nC_a = {}_nC_b$, then either $a = b$ or $a + b = n$.

In our case, we have $a = 10$ and $b = 12$.

Case 1: $a = b$

$10 = 12$, which is false.

Case 2: $a + b = n$

$10 + 12 = n$

For the combinations ${}_nC_{10}$ and ${}_nC_{12}$ to be defined, $n$ must be an integer such that $n \geq 10$ and $n \geq 12$. Our value $n=22$ satisfies these conditions ($22 \geq 10$ and $22 \geq 12$).

Now we need to find the value of ${}_{n}C_n$. Substitute $n = 22$ into this expression:

Using the combination formula ${}_nC_r = \frac{n!}{r!(n-r)!}$, with $n=22$ and $r=22$:

${}_{22}C_{22} = \frac{22!}{22!(22-22)!}

${}_{22}C_{22} = \frac{22!}{22!0!}

By definition, $0! = 1$.

${}_{22}C_{22} = \frac{22!}{22! \times 1}

${}_{22}C_{22} = \frac{22!}{22!} = 1$

Alternatively, ${}_nC_n$ represents the number of ways to choose $n$ items from a set of $n$ distinct items. There is only one way to do this, which is to select all the items.

Therefore, the value of ${}_{n}C_n$ is 1.

Comparing this result with the given options:

(A) 0

(B) 1

(C) 22

(D) Undefined

Our calculated value 1 matches option (B).

The correct answer is (B) 1.

Given:

The word "COMMITTEE".

To Find:

The number of different arrangements (words) that can be formed using the letters of the word "COMMITTEE".

Solution:

The word "COMMITTEE" has a total of 9 letters.

To find the number of different arrangements, we first count the frequency of each distinct letter in the word:

• The letter 'C' appears 1 time.
• The letter 'O' appears 1 time.
• The letter 'M' appears 2 times.
• The letter 'I' appears 1 time.
• The letter 'T' appears 2 times.
• The letter 'E' appears 2 times.

Total number of letters, $n = 9$.

The letters that are repeated are M, T, and E, each appearing 2 times.

Frequencies of repeated letters:

$n_M = 2$

$n_T = 2$

$n_E = 2$

The formula for the number of distinct permutations of $n$ objects where there are $n_1$ identical objects of type 1, $n_2$ identical objects of type 2, ..., $n_k$ identical objects of type k is:

Using this formula for the word "COMMITTEE":

Number of arrangements = $\frac{9!}{2! \times 2! \times 2!}$

We only include the factorials for the letters that appear more than once.

Comparing this result with the given options:

(A) $\frac{9!}{2!2!}$

(B) $\frac{9!}{2!2!2!}$

(C) $\frac{9!}{2!}$

(D) $9!$

Our calculated formula matches option (B).

The correct answer is (B) $\frac{9!}{2!2!2!}$.

Given:

Number of distinct rings = 5.

Number of fingers = 4.

Condition: Each ring can be worn on any finger.

To Find:

The number of ways the 5 distinct rings can be worn on the 4 fingers.

Solution:

We have 5 distinct rings, and for each ring, we need to decide which of the 4 fingers it will be placed on.

Let's consider the distribution of the rings one by one.

Consider the first ring. It can be placed on any of the 4 fingers. So there are 4 choices for the first ring.

Consider the second ring. It is distinct from the first ring and can also be placed on any of the 4 fingers (including the finger where the first ring was placed, as the problem implies no restriction on the number of rings per finger). So there are 4 choices for the second ring.

Similarly, for each of the 5 distinct rings, there are 4 independent choices of finger.

Number of choices for Ring 3 = 4

Number of choices for Ring 4 = 4

Number of choices for Ring 5 = 4

By the Fundamental Principle of Counting (Multiplication Rule), the total number of ways to distribute the 5 rings is the product of the number of choices for each ring.

Total number of ways = $4^5$

Calculate the value:

$4^5 = 4 \times 4 \times 4 \times 4 \times 4 = 16 \times 16 \times 4 = 256 \times 4 = 1024$

So, there are $4^5$ ways to wear 5 distinct rings on 4 fingers.

Comparing our derived expression with the given options:

(A) $4^5$: This matches our derived expression. Correct.

(B) $5^4$: This would be the number of ways to distribute 4 distinct items among 5 distinct categories where each item can go into any category (e.g., distributing 4 different prizes among 5 students). Incorrect.

(C) ${}_5P_4$: This is the number of permutations of 5 items taken 4 at a time, $\frac{5!}{(5-4)!} = 5! = 120$. This would represent arranging 4 out of 5 distinct items, or assigning 4 distinct items to 5 distinct ordered slots without repetition. Incorrect.

(D) ${}_5C_4$: This is the number of combinations of 5 items taken 4 at a time, $\frac{5!}{4!(5-4)!} = 5$. This represents choosing a group of 4 items from 5, where order doesn't matter. Incorrect.

The correct answer is (A) $4^5$.

Given:

The set of numbers $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.

We need to select a pair of numbers from this set such that their sum is odd.

The pair consists of two distinct numbers, and the order of selection does not matter (since $\{a, b\}$ is the same pair as $\{b, a\}$).

To Find:

The number of pairs of distinct numbers from the set $S$ whose sum is odd.

Solution:

The set $S$ contains 10 distinct numbers.

Let's classify the numbers in the set as odd or even:

• Odd numbers: $\{1, 3, 5, 7, 9\}$. There are 5 odd numbers.
• Even numbers: $\{2, 4, 6, 8, 10\}$. There are 5 even numbers.

We are looking for pairs of numbers whose sum is odd.

The sum of two integers is odd if and only if one integer is odd and the other integer is even.

Let the two numbers in the pair be $x$ and $y$. The sum $x+y$ is odd if (x is odd and y is even) or (x is even and y is odd). Since the pair $\{x, y\}$ is the same as $\{y, x\}$, these two conditions are covered by selecting one odd number and one even number.

So, to form a pair whose sum is odd, we need to select 1 number from the set of odd numbers and 1 number from the set of even numbers.

Number of ways to select 1 odd number from the 5 odd numbers = ${}_5C_1$.

${}_5C_1 = \frac{5!}{1!(5-1)!} = \frac{5!}{1!4!} = 5$

Number of ways to select 1 even number from the 5 even numbers = ${}_5C_1$.

${}_5C_1 = \frac{5!}{1!(5-1)!} = \frac{5!}{1!4!} = 5$

Since the selection of an odd number and the selection of an even number are independent events, we use the Multiplication Principle to find the total number of ways to select one of each.

Total number of pairs = $5 \times 5 = 25$

So, there are 25 pairs of numbers from the set $\{1, 2, ..., 10\}$ whose sum is odd.

Let's consider other cases of selecting two numbers:

Case A: Selecting two odd numbers. Sum is even (Odd + Odd = Even).

Number of ways to select 2 odd numbers from 5 = ${}_5C_2 = \frac{5!}{2!3!} = 10$.

Case B: Selecting two even numbers. Sum is even (Even + Even = Even).

Number of ways to select 2 even numbers from 5 = ${}_5C_2 = \frac{5!}{2!3!} = 10$.

The total number of ways to select any 2 numbers from the 10 numbers is ${}_{10}C_2 = \frac{10!}{2!8!} = 45$.

The total number of pairs with an even sum is the sum of Case A and Case B: ${}_5C_2 + {}_5C_2 = 10 + 10 = 20$.

The total number of pairs must be the sum of pairs with odd sum and pairs with even sum: $25 + 20 = 45$, which matches ${}_{10}C_2$. This confirms our logic.

Comparing our derived expression with the given options:

(A) ${}_{10}C_2$: This is the total number of pairs, not necessarily with an odd sum. Incorrect.

(B) ${}_5C_1 \times {}_5C_1$: This matches our derived expression for selecting one odd and one even number. Correct.

(C) ${}_5C_2 + {}_5C_2$: This is the number of pairs with an even sum (selecting two odd OR two even). Incorrect.

(D) ${}_{10}C_2 - ({}_5C_2 + {}_5C_2)$: This represents the total pairs minus the pairs with an even sum, which equals the pairs with an odd sum. This is also a correct way to express the answer ($45 - (10 + 10) = 45 - 20 = 25$).

Both (B) and (D) are mathematically correct and represent the answer. However, option (B) directly reflects the structure of pairs with an odd sum (one odd, one even), which is often the primary approach for this type of problem.

The correct answer is (B) ${}_5C_1 \times {}_5C_1$.

Given:

The equation ${}_nC_4 = {}_nC_6$.

To Find:

The value of ${}_{12}C_n$.

Solution:

We are given the equation ${}_nC_4 = {}_nC_6$.

We know a property of combinations that states if ${}_nC_a = {}_nC_b$, then either $a = b$ or $a + b = n$.

In this case, we have $a = 4$ and $b = 6$.

Case 1: $a = b$

$4 = 6$, which is false.

Case 2: $a + b = n$

$4 + 6 = n$

For the combinations ${}_nC_4$ and ${}_nC_6$ to be defined, $n$ must be an integer such that $n \geq 4$ and $n \geq 6$. Our value $n=10$ satisfies these conditions ($10 \geq 4$ and $10 \geq 6$), so the value of $n$ is valid.

Now we need to find the value of ${}_{12}C_n$. Substitute $n = 10$ into this expression:

Using the combination formula ${}_nC_r = \frac{n!}{r!(n-r)!}$, we can calculate ${}_{12}C_{10}$:

${}_{12}C_{10} = \frac{12!}{10!(12-10)!}$

${}_{12}C_{10} = \frac{12!}{10!2!}$

${}_{12}C_{10} = \frac{12 \times 11 \times 10!}{10! \times 2 \times 1}$

Cancel out $10!$ from the numerator and the denominator:

${}_{12}C_{10} = \frac{12 \times 11}{2}$}

${}_{12}C_{10} = \frac{132}{2}$}

The value of ${}_{12}C_n$ is 66.

Comparing this result with the given options:

(A) ${}_{12}C_{10}$: This is the symbolic form that evaluates to 66.

(B) ${}_{12}C_{12}$: This evaluates to 1.

(C) 1: This is the value of ${}_{12}C_{12}$.

(D) ${}_{12}C_2$: This also evaluates to $\frac{12!}{2!10!} = \frac{12 \times 11}{2} = 66$.

Both options (A) and (D) are equal to the calculated value of ${}_{12}C_n$. However, option (A) directly uses the found value of $n$.

The correct answer is (A) ${}_{12}C_{10}$.

Given:

Available digits: {1, 2, 3, 4, 5} (a set of 5 distinct digits).

We need to form 3-digit numbers.

Repetition of digits is allowed.

To Find:

The number of distinct 3-digit numbers that can be formed using the given digits, with repetition allowed.

Solution:

A 3-digit number has three places: hundreds, tens, and units place.

Let the 3-digit number be represented by the places $\_$ $\_$ $\_$.

The available digits are {1, 2, 3, 4, 5}. There are 5 distinct digits available.

Since repetition is allowed, the choice of a digit for one position does not affect the choices for the other positions.

1. Hundreds Place (1st digit): The first digit can be any of the 5 available digits.

2. Tens Place (2nd digit): Since repetition is allowed, the second digit can also be any of the 5 available digits.

3. Units Place (3rd digit): Similarly, the third digit can be any of the 5 available digits.

By the Fundamental Principle of Counting (Multiplication Rule), the total number of different 3-digit numbers that can be formed is the product of the number of choices for each position.

Total number of 3-digit numbers = $5^3$

Calculate the value:

$5^3 = 5 \times 5 \times 5 = 25 \times 5 = 125$

So, there are $5^3$ different 3-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if repetition is allowed.

Comparing our derived expression with the given options:

(A) ${}_5P_3$: This is the number of permutations without repetition ($5 \times 4 \times 3 = 60$). Incorrect.

(B) ${}_5C_3$: This is the number of combinations without repetition ($\frac{5 \times 4}{2} = 10$). Incorrect.

(C) $5^3$: This matches our derived expression. Correct.

(D) $3^5$: This would be forming 5-digit numbers using 3 distinct digits with repetition, or distributing 5 distinct items into 3 categories. Incorrect.

The correct answer is (C) $5^3$.

Given:

The word "BANANA".

To Find:

The number of different arrangements (words) that can be formed using the letters of the word "BANANA".

Solution:

The word "BANANA" has a total of 6 letters.

To find the number of different arrangements, we first count the frequency of each distinct letter in the word:

• The letter 'B' appears 1 time.
• The letter 'A' appears 3 times.
• The letter 'N' appears 2 times.

Total number of letters, $n = 6$.

The letters that are repeated are A and N, appearing 3 and 2 times respectively.

Frequencies of repeated letters:

$n_A = 3$

$n_N = 2$

$n_B = 1$

The formula for the number of distinct permutations of $n$ objects where there are $n_1$ identical objects of type 1, $n_2$ identical objects of type 2, ..., $n_k$ identical objects of type k is:

Using this formula for the word "BANANA":

Number of arrangements = $\frac{6!}{3! \times 2! \times 1!}$

Since $1! = 1$, the formula simplifies to:

Let's calculate the value:

$\frac{6!}{3!2!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1) \times (2 \times 1)}$

$\frac{720}{6 \times 2} = \frac{720}{12} = 60$

Comparing our derived formula and calculated value with the given options:

(A) $6!$: This would be the number of arrangements if all letters were distinct. Incorrect ($720$).

(B) $\frac{6!}{3!2!}$: This matches our derived formula. Correct ($60$).

(C) $\frac{6!}{3!}$: This would only account for one set of repeated letters (A's). Incorrect.

(D) $\frac{6!}{2!}$: This would only account for one set of repeated letters (N's). Incorrect.

The correct answer is (B) $\frac{6!}{3!2!}$.

Given:

Number of teachers = 5.

Number of students = 10.

Total number of people = $5 + 10 = 15$.

A committee of 6 is to be formed.

Condition: The committee must include exactly 2 teachers.

To Find:

The number of ways the committee of 6 can be formed with exactly 2 teachers.

Solution:

We need to form a committee of 6 members with a specific composition: exactly 2 teachers and the rest from the students.

Since the committee size is 6 and we need exactly 2 teachers, the number of students in the committee must be $6 - 2 = 4$.

We need to perform two independent selections:

1. Select 2 teachers from the 5 available teachers.

2. Select 4 students from the 10 available students.

Since the order of selection does not matter for forming a committee, we use the combination formula ${}_nC_r = \frac{n!}{r!(n-r)!}$.

Number of ways to choose 2 teachers from 5 teachers:

Calculating ${}_5C_2$:

${}_5C_2 = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10$

Number of ways to choose 4 students from 10 students:

Calculating ${}_{10}C_4$:

${}_{10}C_4 = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 10 \times 3 \times 7 = 210$

To find the total number of ways to form the committee with exactly 2 teachers (and hence 4 students), we multiply the number of ways to select the teachers by the number of ways to select the students (using the Multiplication Principle).

Total number of ways = $10 \times 210 = 2100$

Comparing our derived expression with the given options:

(A) ${}_5C_2 \times {}_{10}C_4$: This matches our calculation. Correct.

(B) ${}_5P_2 \times {}_{10}P_4$: This would involve permutations, where order matters, which is not the case for forming a committee. Incorrect.

(C) ${}_5C_2 + {}_{10}C_4$: Addition is used for mutually exclusive cases, not for combining selections to form a single group. Incorrect.

(D) ${}_{15}C_6 - {}_{10}C_6$: ${}_{15}C_6$ is the total number of ways to choose any 6 people from 15. ${}_{10}C_6$ is the number of ways to choose 6 students from the 10 students (i.e., 0 teachers). Subtracting this from the total would give the number of committees with *at least* 1 teacher, not exactly 2. Incorrect.

The correct answer is (A) ${}_5C_2 \times {}_{10}C_4$.

To evaluate $8! - 6!$, we first calculate the values of $8!$ and $6!$ separately.

The factorial of a non-negative integer $n$, denoted by $n!$, is the product of all positive integers less than or equal to $n$.

First, calculate $8!$:

$8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$

$8! = 40320$

Next, calculate $6!$:

$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$

$6! = 720$

Now, we subtract $6!$ from $8!$:

$8! - 6! = 40320 - 720$

Performing the subtraction:

$40320 - 720 = 39600$

Thus, $8! - 6! = 39600$.

The final answer is $\mathbf{39600}$.

We need to compute the value of the expression $\frac{12!}{9! \cdot 3!}$.

Recall that the factorial of a non-negative integer $n$, denoted by $n!$, is the product of all positive integers less than or equal to $n$. Thus, $n! = n \times (n-1) \times \dots \times 2 \times 1$.

We can expand the larger factorial in the numerator in terms of the largest factorial in the denominator.

$12! = 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$

We can write this as:

$12! = 12 \times 11 \times 10 \times (9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1)$

Which is:

$12! = 12 \times 11 \times 10 \times 9!$

Now, we also calculate the value of $3!$:

$3! = 3 \times 2 \times 1$

$3! = 6$

Substitute these into the given expression:

$\frac{12!}{9! \cdot 3!} = \frac{12 \times 11 \times 10 \times 9!}{9! \times (3 \times 2 \times 1)}$

We can cancel the $9!$ term from the numerator and the denominator:

$\frac{12 \times 11 \times 10 \times \cancel{9!}}{\cancel{9!} \times 6} = \frac{12 \times 11 \times 10}{6}$

Now, we simplify the remaining expression:

$\frac{12 \times 11 \times 10}{6} = \frac{1320}{6}$

Alternatively, we can cancel before multiplying fully:

$\frac{\cancel{12}^{2} \times 11 \times 10}{\cancel{6}_{1}} = 2 \times 11 \times 10$

$2 \times 11 \times 10 = 22 \times 10 = 220$

Thus, $\frac{12!}{9! \cdot 3!} = 220$.

The final answer is $\mathbf{220}$.

We are given the equation:

$\frac{1}{8!} + \frac{1}{9!} = \frac{x}{10!}$

To solve for $x$, we can simplify the left side of the equation or multiply the entire equation by $10!$ to eliminate the denominators. Multiplying by $10!$ is often easier.

Multiply both sides of the equation by $10!$:

$10! \left( \frac{1}{8!} + \frac{1}{9!} \right) = 10! \left( \frac{x}{10!} \right)$

Distribute $10!$ on the left side:

$\frac{10!}{8!} + \frac{10!}{9!} = x$

Now, we need to simplify the factorial terms. Recall that $n! = n \times (n-1)!$.

$\frac{10!}{8!} = \frac{10 \times 9 \times 8!}{8!}$

Cancel out $8!$ from the numerator and denominator:

$\frac{10 \times 9 \times \cancel{8!}}{\cancel{8!}} = 10 \times 9 = 90$

$\frac{10!}{9!} = \frac{10 \times 9!}{9!}$

Cancel out $9!$ from the numerator and denominator:

$\frac{10 \times \cancel{9!}}{\cancel{9!}} = 10$

Substitute these simplified values back into the equation $\frac{10!}{8!} + \frac{10!}{9!} = x$:

$90 + 10 = x$

$100 = x$

Thus, the value of $x$ is $100$.

Alternate Solution (Finding a common denominator on the left side):

Start with the given equation:

$\frac{1}{8!} + \frac{1}{9!} = \frac{x}{10!}$

We can write $9! = 9 \times 8!$ and $10! = 10 \times 9! = 10 \times 9 \times 8!$.

Find a common denominator for the left side, which is $9!$.

$\frac{1}{8!} = \frac{1 \times 9}{8! \times 9} = \frac{9}{9!}$

So the equation becomes:

$\frac{9}{9!} + \frac{1}{9!} = \frac{x}{10!}$

Combine the terms on the left side:

$\frac{9+1}{9!} = \frac{x}{10!}$

$\frac{10}{9!} = \frac{x}{10!}$

Now, we have $\frac{10}{9!} = \frac{x}{10 \times 9!}$.

To solve for $x$, multiply both sides by $10 \times 9!$:

$(10 \times 9!) \times \frac{10}{9!} = (10 \times 9!) \times \frac{x}{10 \times 9!}$

Cancel terms on both sides:

$(10 \times \cancel{9!}) \times \frac{10}{\cancel{9!}} = \cancel{(10 \times 9!)} \times \frac{x}{\cancel{(10 \times 9!)}}$

$10 \times 10 = x$

$100 = x$

Both methods yield the same result. The value of $x$ is $100$.

The final answer is $\mathbf{100}$.

We need to find the number of 3-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 without repetition.

This is a problem of arranging 3 distinct digits out of a set of 5 distinct digits. The order of the digits matters (e.g., 123 is different from 321).

This can be solved using permutations or the fundamental principle of counting.

Method 1: Using the Fundamental Principle of Counting

We need to fill three places representing the hundreds, tens, and units digits.

For the hundreds digit, there are 5 possible choices (any of the digits 1, 2, 3, 4, 5).

Since repetition is not allowed, for the tens digit, there are only 4 remaining choices.

For the units digit, there are only 3 remaining choices.

By the fundamental principle of counting, the total number of 3-digit numbers is the product of the number of choices for each position.

Number of 3-digit numbers = (Choices for hundreds) $\times$ (Choices for tens) $\times$ (Choices for units)

Number of 3-digit numbers = $5 \times 4 \times 3$

$5 \times 4 \times 3 = 20 \times 3 = 60$

Method 2: Using Permutations

This problem is equivalent to finding the number of permutations of 5 distinct items taken 3 at a time, denoted as $P(5, 3)$ or $_5P_3$.

The formula for permutations is $P(n, r) = \frac{n!}{(n-r)!}$, where $n$ is the total number of items, and $r$ is the number of items to choose.

Here, $n=5$ (the total number of digits) and $r=3$ (the number of digits in the 3-digit number).

$P(5, 3) = \frac{5!}{(5-3)!} = \frac{5!}{2!}$

Calculate the factorials:

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

$2! = 2 \times 1 = 2$

Substitute the values into the formula:

$P(5, 3) = \frac{120}{2} = 60$

Both methods show that there are 60 different 3-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 without repetition.

The final answer is $\mathbf{60}$.

Given:

Number of routes from City A to City B = 4

Number of routes from City B to City C = 3

To Find:

The total number of ways a person can travel from City A to City C via City B.

Solution:

This problem can be solved using the Fundamental Principle of Counting (also known as the multiplication principle).

The principle states that if an event can occur in $m$ ways, and after it occurs, another event can occur in $n$ ways, then the total number of ways both events can occur in sequence is $m \times n$.

In this case, the journey from City A to City C via City B consists of two consecutive events:

1. Travelling from City A to City B.

2. Travelling from City B to City C.

The number of ways to perform the first event (A to B) is 4.

The number of ways to perform the second event (B to C) is 3.

According to the Fundamental Principle of Counting, the total number of ways to travel from City A to City C via City B is the product of the number of ways for each leg of the journey.

Total number of ways = (Number of routes from A to B) $\times$ (Number of routes from B to C)

Total number of ways = $4 \times 3$

Total number of ways = $12$

Therefore, there are 12 ways a person can travel from City A to City C via City B.

The final answer is $\mathbf{12}$.

We have 5 different flags available.

A signal is generated by arranging at least two flags in order on a vertical staff.

"At least two flags" means the signal can be formed using 2 flags, 3 flags, 4 flags, or 5 flags.

Since the flags are different and the order matters (as they are arranged on a staff), this is a problem involving permutations.

We need to calculate the number of permutations for each possible number of flags in the signal and sum them up.

The number of permutations of $n$ distinct items taken $r$ at a time is given by the formula $P(n, r) = \frac{n!}{(n-r)!}$.

In this case, the total number of different flags is $n=5$. The number of flags used in a signal is $r$.

Case 1: Signal with 2 flags ($r=2$)

The number of ways to arrange 2 flags out of 5 is $P(5, 2)$.

$P(5, 2) = \frac{5!}{(5-2)!} = \frac{5!}{3!}$

$P(5, 2) = \frac{5 \times 4 \times 3!}{3!} = 5 \times 4 = 20$

Case 2: Signal with 3 flags ($r=3$)

The number of ways to arrange 3 flags out of 5 is $P(5, 3)$.

$P(5, 3) = \frac{5!}{(5-3)!} = \frac{5!}{2!}$

$P(5, 3) = \frac{5 \times 4 \times 3 \times 2!}{2!} = 5 \times 4 \times 3 = 60$

Case 3: Signal with 4 flags ($r=4$)

The number of ways to arrange 4 flags out of 5 is $P(5, 4)$.

$P(5, 4) = \frac{5!}{(5-4)!} = \frac{5!}{1!}$

$P(5, 4) = \frac{5 \times 4 \times 3 \times 2 \times 1}{1} = 120$

Case 4: Signal with 5 flags ($r=5$)

The number of ways to arrange 5 flags out of 5 is $P(5, 5)$.

$P(5, 5) = \frac{5!}{(5-5)!} = \frac{5!}{0!}$

Since $0! = 1$, we have:

$P(5, 5) = \frac{120}{1} = 120$

The total number of different signals is the sum of the number of signals possible in each case (since these cases are mutually exclusive).

Total signals = (Signals with 2 flags) + (Signals with 3 flags) + (Signals with 4 flags) + (Signals with 5 flags)

Total signals = $P(5, 2) + P(5, 3) + P(5, 4) + P(5, 5)$

Total signals = $20 + 60 + 120 + 120$

Total signals = $80 + 240 = 320$

Therefore, 320 different signals can be generated.

The final answer is $\mathbf{320}$.

The given word is 'JODHPUR'.

Let's count the number of letters in the word 'JODHPUR'.

The letters are J, O, D, H, P, U, R.

There are 7 letters in the word 'JODHPUR'.

Now, let's check if any letter is repeated. By inspecting the letters, we can see that all 7 letters are distinct.

The problem is to find the number of permutations of 7 distinct objects.

The number of permutations of $n$ distinct objects is given by $n!$ (n factorial).

In this case, the number of distinct letters is $n=7$.

So, the number of permutations of the letters of the word 'JODHPUR' is $7!$.

Calculate $7!$:

$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$

$7! = 42 \times 20 \times 6$

$7! = 840 \times 6$

$7! = 5040$

Therefore, the number of permutations of the letters of the word 'JODHPUR' is 5040.

The final answer is $\mathbf{5040}$.

The given word is 'JAIPUR'.

Let's identify the letters and classify them as vowels and consonants.

The letters in the word 'JAIPUR' are J, A, I, P, U, R.

The vowels are A, I, U. There are 3 vowels.

The consonants are J, P, R. There are 3 consonants.

The total number of letters in the word is $3 + 3 = 6$.

We are asked to find the number of arrangements where the vowels are always together.

To handle this constraint, we treat the set of vowels {A, I, U} as a single block or unit.

Now, the items to be arranged are this vowel block and the three consonants (J, P, R).

So, we are arranging the following items: (AIU), J, P, R.

There are $1 + 3 = 4$ items to arrange.

The number of ways to arrange these 4 distinct items is $4!$.

$4! = 4 \times 3 \times 2 \times 1 = 24$

Within the vowel block (AIU), the vowels can be arranged among themselves. There are 3 distinct vowels (A, I, U).

The number of ways to arrange these 3 distinct vowels is $3!$.

$3! = 3 \times 2 \times 1 = 6$

To find the total number of arrangements where the vowels are always together, we multiply the number of ways to arrange the items (vowel block and consonants) by the number of ways to arrange the vowels within their block.

Total arrangements = (Arrangements of items) $\times$ (Arrangements within vowel block)

Total arrangements = $4! \times 3!$

Total arrangements = $24 \times 6$

$24 \times 6 = 144$

Therefore, there are 144 arrangements of the letters of the word 'JAIPUR' such that the vowels are always together.

The final answer is $\mathbf{144}$.

We are given the equation $P(n, 2) = 30$.

The formula for the number of permutations of $n$ distinct objects taken $r$ at a time is $P(n, r) = \frac{n!}{(n-r)!}$.

In this problem, we have $r=2$. So, the formula becomes:

$P(n, 2) = \frac{n!}{(n-2)!}$

We are given that $P(n, 2) = 30$, so we can write the equation:

$\frac{n!}{(n-2)!} = 30$

Now, we expand the factorial in the numerator. We know that $n! = n \times (n-1) \times (n-2)!$ for $n \ge 2$.

Substitute this into the equation:

$\frac{n \times (n-1) \times (n-2)!}{(n-2)!} = 30$

Assuming $n \ge 2$, we can cancel out the $(n-2)!$ term from the numerator and the denominator:

$n \times (n-1) = 30$

Expand the left side of the equation:

$n^2 - n = 30$

Rearrange the equation into a quadratic form by moving all terms to one side:

$n^2 - n - 30 = 0$

Now, we need to solve this quadratic equation for $n$. We can factor the quadratic expression. We look for two numbers that multiply to -30 and add up to -1. These numbers are -6 and 5.

So, we can factor the equation as:

$(n - 6)(n + 5) = 0$

This gives two possible solutions for $n$:

$n - 6 = 0 \implies n = 6$

$n + 5 = 0 \implies n = -5$

In the context of permutations $P(n, r)$, $n$ must be a non-negative integer and $n \ge r$. In this case, $r=2$.

The value $n = -5$ is not a valid solution because $n$ must be non-negative.

The value $n = 6$ is a valid solution because it is a non-negative integer and $6 \ge 2$.

Therefore, the value of $n$ is 6.

The final answer is $\mathbf{6}$.

Given:

Total number of students in the group, $n = 10$.

Number of students to be selected, $r = 3$.

To Find:

The number of ways to select 3 students from 10 students.

Solution:

Since the order in which the students are selected does not matter (selecting student A, then B, then C is the same as selecting B, then C, then A), this is a problem involving combinations.

The number of combinations of $n$ distinct objects taken $r$ at a time is given by the formula:

$C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}$

Substitute the given values $n=10$ and $r=3$ into the formula:

$C(10, 3) = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!}$

Expand the factorials:

$10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$

$3! = 3 \times 2 \times 1 = 6$

$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$

Substitute the expanded factorials (or partially expand $10!$ to cancel $7!$) into the expression for $C(10, 3)$:

$C(10, 3) = \frac{10 \times 9 \times 8 \times 7!}{3! \times 7!}$

Cancel out the $7!$ term from the numerator and the denominator:

$C(10, 3) = \frac{10 \times 9 \times 8}{3!}$

$C(10, 3) = \frac{10 \times 9 \times 8}{3 \times 2 \times 1}$

$C(10, 3) = \frac{10 \times 9 \times 8}{6}$

Perform the multiplication in the numerator:

$10 \times 9 \times 8 = 90 \times 8 = 720$

So, $C(10, 3) = \frac{720}{6}$

Now, perform the division:

$\frac{720}{6} = 120$

Thus, the number of ways to select 3 students from a group of 10 students is 120.

The final answer is $\mathbf{120}$.

We are given the equation $\binom{n}{2} = 45$.

The symbol $\binom{n}{r}$ represents the number of combinations of $n$ distinct objects taken $r$ at a time, and it is defined by the formula:

$\binom{n}{r} = \frac{n!}{r!(n-r)!}$

In this problem, we have $r=2$. So, the formula becomes:

$\binom{n}{2} = \frac{n!}{2!(n-2)!}$

We are given that $\binom{n}{2} = 45$, so we can write the equation:

$\frac{n!}{2!(n-2)!} = 45$

Now, we expand the factorial in the numerator. We know that $n! = n \times (n-1) \times (n-2)!$ for $n \ge 2$. Also, $2! = 2 \times 1 = 2$.

Substitute these into the equation:

$\frac{n \times (n-1) \times (n-2)!}{2 \times (n-2)!} = 45$

Assuming $n \ge 2$, we can cancel out the $(n-2)!$ term from the numerator and the denominator:

$\frac{n \times (n-1)}{2} = 45$

Multiply both sides by 2 to eliminate the denominator:

$n \times (n-1) = 45 \times 2$

$n(n-1) = 90$

Expand the left side of the equation:

$n^2 - n = 90$

Rearrange the equation into a quadratic form by moving all terms to one side:

$n^2 - n - 90 = 0$

Now, we need to solve this quadratic equation for $n$. We can factor the quadratic expression. We look for two numbers that multiply to -90 and add up to -1. These numbers are -10 and 9.

So, we can factor the equation as:

$(n - 10)(n + 9) = 0$

This gives two possible solutions for $n$:

$n - 10 = 0 \implies n = 10$

$n + 9 = 0 \implies n = -9$

In the context of combinations $\binom{n}{r}$, $n$ must be a non-negative integer and $n \ge r$. In this case, $r=2$.

The value $n = -9$ is not a valid solution because $n$ must be non-negative.

The value $n = 10$ is a valid solution because it is a non-negative integer and $10 \ge 2$.

Therefore, the value of $n$ is 10.

The final answer is $\mathbf{10}$.

Given:

Total number of men available = 5

Total number of women available = 6

Number of men to be selected for the committee = 2

Number of women to be selected for the committee = 3

To Find:

The total number of distinct committees consisting of 2 men and 3 women that can be formed.

Solution:

Forming a committee involves selecting a group of individuals, and the order in which they are selected does not matter. Therefore, this problem requires the use of combinations.

We need to perform two independent selections:

1. Select 2 men from the available 5 men.

2. Select 3 women from the available 6 women.

The number of ways to select $r$ objects from a set of $n$ distinct objects, without regard to order, is given by the combination formula:

$C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}$

First, calculate the number of ways to select 2 men from 5 men ($n=5, r=2$).

Number of ways to select men = $\binom{5}{2}$

$\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!}$

$\binom{5}{2} = \frac{5 \times 4 \times 3!}{ (2 \times 1) \times 3!} = \frac{5 \times 4}{2} = \frac{20}{2} = 10$

There are 10 ways to select 2 men from 5 men.

Next, calculate the number of ways to select 3 women from 6 women ($n=6, r=3$).

Number of ways to select women = $\binom{6}{3}$

$\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!}$

$\binom{6}{3} = \frac{6 \times 5 \times 4 \times 3!}{ (3 \times 2 \times 1) \times 3!} = \frac{6 \times 5 \times 4}{6} = \frac{120}{6} = 20$

There are 20 ways to select 3 women from 6 women.

To find the total number of committees with 2 men AND 3 women, we multiply the number of ways to select the men by the number of ways to select the women. This is based on the Fundamental Principle of Counting, as the selection of men and the selection of women are independent events that occur together to form the committee.

Total number of committees = (Number of ways to select men) $\times$ (Number of ways to select women)

Total number of committees = $\binom{5}{2} \times \binom{6}{3}$

Total number of committees = $10 \times 20$

$10 \times 20 = 200$

Therefore, 200 different committees consisting of 2 men and 3 women can be formed from 5 men and 6 women.

The final answer is $\mathbf{200}$.

Given:

The factorial of a non-negative integer $n$, denoted by $n!$, is defined as the product of all positive integers less than or equal to $n$.

For $n \ge 1$, $n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1$.

By definition, $0! = 1$.

To Prove:

$n! = n \cdot (n-1)!$ for $n \ge 1$.

Proof:

We start with the definition of $n!$ for $n \ge 1$:

$n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1$

Consider the terms that follow $n$ in the product $(n-1) \times (n-2) \times \dots \times 2 \times 1$.

By the definition of factorial, this product is exactly $(n-1)!$.

$(n-1) \times (n-2) \times \dots \times 2 \times 1 = (n-1)!$

Substitute this back into the expression for $n!$:

$n! = n \times \underbrace{((n-1) \times (n-2) \times \dots \times 2 \times 1)}_{\text{(n-1)!}}$

Thus, we have:

$n! = n \times (n-1)!$

This identity holds for $n \ge 2$ based on the definition $n! = n \times (n-1) \times \dots \times 1$.

Let's check the case for $n=1$.

Left side: $1! = 1$ (by definition).

Right side: $1 \cdot (1-1)! = 1 \cdot 0!$

Since $0! = 1$, the right side is $1 \cdot 1 = 1$.

So, $1! = 1 \cdot 0!$ is true.

The identity $n! = n \cdot (n-1)!$ is true for all $n \ge 1$.

This completes the proof.

Given:

The set of available digits is $\{0, 1, 2, 3, 4, 5, 6\}$.

The requirements are to form odd numbers less than 1000 without repetition of digits.

To Find:

The total number of such odd numbers.

Solution:

Odd numbers less than 1000 can be 1-digit, 2-digit, or 3-digit numbers.

The given digits are 0, 1, 2, 3, 4, 5, 6. Total number of digits is 7.

For a number to be odd, its last digit must be odd. The odd digits in the given set are 1, 3, and 5.

We will consider each case separately based on the number of digits.

Case 1: 1-digit odd numbers

A 1-digit number formed from the given digits is odd if it is one of 1, 3, or 5.

The possible 1-digit odd numbers are 1, 3, 5.

Number of 1-digit odd numbers = 3.

Case 2: 2-digit odd numbers

A 2-digit number is of the form AB, where A is the tens digit and B is the units digit.

The units digit (B) must be an odd digit from the given set, so B $\in \{1, 3, 5\}$. There are 3 choices for the units digit.

The tens digit (A) cannot be 0, and it cannot be the digit used for the units place (since repetition is not allowed).

Total digits available = 7.

One digit is used for the units place.

Remaining digits = $7 - 1 = 6$.

The tens digit cannot be 0. Since the units digit is one of $\{1, 3, 5\}$ (which are not 0), 0 is always among the 6 remaining digits.

So, the number of choices for the tens digit is (Remaining digits) - (Digit 0) = $6 - 1 = 5$.

By the Fundamental Principle of Counting, the number of 2-digit odd numbers = (Choices for Units digit) $\times$ (Choices for Tens digit)

Number of 2-digit odd numbers = $3 \times 5 = 15$.

Case 3: 3-digit odd numbers

A 3-digit number is of the form ABC, where A is the hundreds digit, B is the tens digit, and C is the units digit.

The units digit (C) must be an odd digit from the given set, so C $\in \{1, 3, 5\}$. There are 3 choices for the units digit.

The hundreds digit (A) cannot be 0, and it cannot be the digit used for the units place.

Total digits available = 7.

One digit is used for the units place.

Remaining digits = $7 - 1 = 6$.

The hundreds digit cannot be 0. Since the units digit is one of $\{1, 3, 5\}$ (which are not 0), 0 is always among the 6 remaining digits.

So, the number of choices for the hundreds digit is (Remaining digits) - (Digit 0) = $6 - 1 = 5$.

Let's say one digit was chosen for the units place and one for the hundreds place. Two distinct digits have been used.

Digits remaining = $7 - 2 = 5$.

The tens digit (B) can be any of the remaining 5 digits (it can be 0).

Number of choices for the tens digit = 5.

By the Fundamental Principle of Counting, the number of 3-digit odd numbers = (Choices for Units digit) $\times$ (Choices for Hundreds digit) $\times$ (Choices for Tens digit)

Number of 3-digit odd numbers = $3 \times 5 \times 5 = 75$.

The total number of odd numbers less than 1000 formed without repetition is the sum of the numbers from the three cases:

Total odd numbers = (1-digit odd numbers) + (2-digit odd numbers) + (3-digit odd numbers)

Total odd numbers = $3 + 15 + 75 = 93$.

The final answer is $\mathbf{93}$.

The given word is 'BHOPAL'.

The letters in the word 'BHOPAL' are B, H, O, P, A, L.

There are 6 letters in the word 'BHOPAL'.

All the letters are distinct.

Part 1: Total number of arrangements

The total number of arrangements of the 6 distinct letters is $6!$.

$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$

So, there are 720 arrangements of the letters of the word 'BHOPAL'.

Part 2: Number of arrangements that begin with B and end with L

We need to find the number of arrangements where the first letter is B and the last letter is L.

Consider the 6 positions for the letters in an arrangement:

_ _ _ _ _ _

The first position is fixed with B.

The last position is fixed with L.

B _ _ _ _ L

The letters B and L are used for the first and last positions, respectively. We are left with the remaining letters to arrange in the middle positions.

The remaining letters are H, O, P, A.

There are 4 remaining letters, and there are 4 middle positions to fill.

The letters H, O, P, A are distinct.

The number of ways to arrange these 4 distinct letters in the 4 middle positions is $4!$.

$4! = 4 \times 3 \times 2 \times 1 = 24$

Since the first position is fixed as B (1 choice) and the last position is fixed as L (1 choice), and the remaining 4 letters can be arranged in $4!$ ways in the middle, the total number of arrangements that begin with B and end with L is:

Number of arrangements = (Choices for 1st position) $\times$ (Arrangements of middle letters) $\times$ (Choices for last position)

Number of arrangements = $1 \times 4! \times 1 = 4! = 24$

So, there are 24 arrangements that begin with B and end with L.

The total number of arrangements is $\mathbf{720}$.

The number of arrangements that begin with B and end with L is $\mathbf{24}$.

Given:

$P(n, r) = 336$

$\binom{n}{r} = 56$

To Find:

The values of $n$ and $r$.

Solution:

We know the formulas for permutations and combinations:

$P(n, r) = \frac{n!}{(n-r)!}$

$\binom{n}{r} = \frac{n!}{r!(n-r)!}$

There is a relationship between permutations and combinations:

$\binom{n}{r} = \frac{1}{r!} \times \frac{n!}{(n-r)!}$

Substituting the formula for $P(n, r)$ into this relationship, we get:

$\binom{n}{r} = \frac{P(n, r)}{r!}$

We are given the values of $P(n, r)$ and $\binom{n}{r}$. Substitute these values into the relationship:

$56 = \frac{336}{r!}$

Now, we can solve for $r!$:

$r! = \frac{336}{56}$

$r! = 6$

We need to find the integer $r$ whose factorial is 6. Let's calculate the first few factorials:

$1! = 1$

$2! = 2 \times 1 = 2$

$3! = 3 \times 2 \times 1 = 6$

$4! = 4 \times 3 \times 2 \times 1 = 24$

From this, we see that $r! = 6$ implies $r = 3$.

Now that we have the value of $r$, we can use the equation for permutations $P(n, r) = 336$ to find $n$. Substitute $r=3$:

$P(n, 3) = 336$

Using the formula for $P(n, r)$: $P(n, 3) = \frac{n!}{(n-3)!}$.

So, $\frac{n!}{(n-3)!} = 336$

Expand $n!$ as $n \times (n-1) \times (n-2) \times (n-3)!$ for $n \ge 3$:

$\frac{n \times (n-1) \times (n-2) \times (n-3)!}{(n-3)!} = 336$

Cancel out $(n-3)!$ from the numerator and the denominator:

$n(n-1)(n-2) = 336$

We need to find three consecutive integers whose product is 336. We can try integer values for $n$, starting from values slightly larger than $r=3$.

If $n=4$: $4 \times 3 \times 2 = 24$

If $n=5$: $5 \times 4 \times 3 = 60$

If $n=6$: $6 \times 5 \times 4 = 120$

If $n=7$: $7 \times 6 \times 5 = 210$

If $n=8$: $8 \times 7 \times 6 = 56 \times 6 = 336$

So, the value of $n$ is 8.

We can verify these values using the combination formula:

$\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6 \times 5!}{(3 \times 2 \times 1) \times 5!} = \frac{8 \times 7 \times 6}{6} = 8 \times 7 = 56$.

This matches the given value of $\binom{n}{r}$.

The values of $n$ and $r$ are $n=8$ and $r=3$.

The final answer is $\mathbf{n=8, r=3}$.

Given:

Total number of different books, $n = 10$.

Number of books to be selected, $r = 4$.

To Find:

The number of different selections of 4 books from 10 different books.

Solution:

Since we are asked for the number of selections, the order in which the books are chosen does not matter. This is a problem of finding the number of combinations.

The number of combinations of $n$ distinct objects taken $r$ at a time is given by the formula:

$C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}$

Substitute the given values $n=10$ and $r=4$ into the formula:

$C(10, 4) = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!}$

Expand the factorials. We can partially expand $10!$ to cancel out $6!$:

$10! = 10 \times 9 \times 8 \times 7 \times 6!$

$4! = 4 \times 3 \times 2 \times 1 = 24$

$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$

Substitute these into the expression for $C(10, 4)$:

$C(10, 4) = \frac{10 \times 9 \times 8 \times 7 \times 6!}{4! \times 6!}$

Cancel out the $6!$ term from the numerator and the denominator:

$C(10, 4) = \frac{10 \times 9 \times 8 \times 7}{4!}$

$C(10, 4) = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}$

$C(10, 4) = \frac{10 \times 9 \times 8 \times 7}{24}$

Perform the multiplication in the numerator:

$10 \times 9 \times 8 \times 7 = 90 \times 56 = 5040$

So, $C(10, 4) = \frac{5040}{24}$

Perform the division:

$\frac{5040}{24} = 210$

Alternatively, we can simplify by cancelling terms directly:

$C(10, 4) = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = \frac{10}{\cancel{2}} \times \frac{\cancel{9}}{\cancel{3}} \times \frac{\cancel{8}}{\cancel{4}} \times 7$

$C(10, 4) = 5 \times 3 \times 2 \times 7 = 15 \times 14 = 210$

Thus, the number of different selections of 4 books from 10 different books is 210.

The final answer is $\mathbf{210}$.

Given:

Total number of men available = 7.

Total number of women available = 4.

Size of the committee to be formed = 5.

Requirement for the committee: exactly 3 men.

To Find:

The number of ways to form a committee of 5 with exactly 3 men from the given group.

Solution:

A committee is a selection of people where the order of selection does not matter. Therefore, this problem involves combinations.

The committee must have exactly 3 men. Since the total committee size is 5, the remaining members must be women.

Number of women in the committee = Total committee size - Number of men

Number of women in the committee = $5 - 3 = 2$.

So, the committee must consist of exactly 3 men and exactly 2 women.

We need to perform two independent selections:

1. Select 3 men from the available 7 men.

2. Select 2 women from the available 4 women.

The number of ways to select 3 men from 7 men is given by $\binom{7}{3}$.

$\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!}$

$\binom{7}{3} = \frac{7 \times 6 \times 5 \times 4!}{(3 \times 2 \times 1) \times 4!} = \frac{7 \times 6 \times 5}{6} = 7 \times 5 = 35$

There are 35 ways to select 3 men from 7 men.

The number of ways to select 2 women from 4 women is given by $\binom{4}{2}$.

$\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!}$

$\binom{4}{2} = \frac{4 \times 3 \times 2!}{ (2 \times 1) \times 2!} = \frac{4 \times 3}{2} = \frac{12}{2} = 6$

There are 6 ways to select 2 women from 4 women.

To find the total number of committees with exactly 3 men and exactly 2 women, we multiply the number of ways to select the men by the number of ways to select the women. This is based on the Fundamental Principle of Counting, as the selection of men and the selection of women are independent events that occur together to form the committee.

Total number of committees = (Number of ways to select men) $\times$ (Number of ways to select women)

Total number of committees = $\binom{7}{3} \times \binom{4}{2}$

Total number of committees = $35 \times 6$

$35 \times 6 = 210$

Therefore, there are 210 ways to form a committee of 5 consisting of exactly 3 men from a group of 7 men and 4 women.

The final answer is $\mathbf{210}$.

We need to evaluate the expression $\frac{P(10, 3)}{P(8, 2)}$.

We use the formula for permutations: $P(n, r) = \frac{n!}{(n-r)!}$.

First, let's calculate $P(10, 3)$. Here $n=10$ and $r=3$.

$P(10, 3) = \frac{10!}{(10-3)!} = \frac{10!}{7!}$

Expand $10!$ until $7!$:

$10! = 10 \times 9 \times 8 \times 7!$

Substitute this back into the expression for $P(10, 3)$:

$P(10, 3) = \frac{10 \times 9 \times 8 \times 7!}{7!}$

Cancel out the $7!$ terms:

$P(10, 3) = 10 \times 9 \times 8 = 90 \times 8 = 720$

So, $P(10, 3) = 720$.

Next, let's calculate $P(8, 2)$. Here $n=8$ and $r=2$.

$P(8, 2) = \frac{8!}{(8-2)!} = \frac{8!}{6!}$

Expand $8!$ until $6!$:

$8! = 8 \times 7 \times 6!$

Substitute this back into the expression for $P(8, 2)$:

$P(8, 2) = \frac{8 \times 7 \times 6!}{6!}$

Cancel out the $6!$ terms:

$P(8, 2) = 8 \times 7 = 56$

So, $P(8, 2) = 56$.

Now, we evaluate the given expression $\frac{P(10, 3)}{P(8, 2)}$ by substituting the calculated values:

$\frac{P(10, 3)}{P(8, 2)} = \frac{720}{56}$

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor. Both numbers are divisible by 8.

$720 \div 8 = 90$

$56 \div 8 = 7$

So, $\frac{720}{56} = \frac{90}{7}$

The fraction $\frac{90}{7}$ cannot be simplified further.

Therefore, the value of $\frac{P(10, 3)}{P(8, 2)}$ is $\frac{90}{7}$.

The final answer is $\mathbf{\frac{90}{7}}$.

The given word is 'DELHI'.

The letters in the word 'DELHI' are D, E, L, H, I.

There are 5 letters in the word 'DELHI'.

All the letters are distinct.

We need to find the number of arrangements such that the letter 'L' is always at the first position.

Consider the 5 positions for the letters in an arrangement:

_ _ _ _ _

The constraint is that the first position must be occupied by the letter 'L'.

L _ _ _ _

The letter 'L' is fixed in the first position. There is only 1 choice for the first position (which is L).

We are left with the remaining 4 letters to arrange in the remaining 4 positions.

The remaining letters are D, E, H, I.

These 4 letters are distinct, and we need to arrange them in the remaining 4 positions.

The number of ways to arrange 4 distinct objects in 4 positions is given by $4!$ (4 factorial).

$4! = 4 \times 3 \times 2 \times 1 = 24$

Since the first position is fixed (1 way) and the remaining 4 letters can be arranged in the other 4 positions in $4!$ ways, the total number of arrangements that begin with 'L' is the product of the number of choices for each step.

Number of arrangements = (Choices for 1st position) $\times$ (Arrangements of remaining letters)

Number of arrangements = $1 \times 4!$

Number of arrangements = $1 \times 24 = 24$

Therefore, there are 24 arrangements of the letters of the word 'DELHI' such that L is always at the first position.

The final answer is $\mathbf{24}$.

Given:

The set of available digits is $\{0, 1, 2, 3, 4, 5\}$.

The requirements are to form 4-digit numbers with repetition allowed.

To Find:

The total number of such 4-digit numbers.

Solution:

A 4-digit number has four positions: thousands, hundreds, tens, and units.

_ _ _ _

We need to determine the number of choices for each position.

For the thousands digit (the leftmost digit), it cannot be 0 (otherwise, it would be a 3-digit number or less).

The available digits are 0, 1, 2, 3, 4, 5. There are 6 digits in total.

The possible digits for the thousands place are 1, 2, 3, 4, 5. There are 5 choices.

For the hundreds digit, since repetition is allowed, any of the 6 digits can be used.

The possible digits for the hundreds place are 0, 1, 2, 3, 4, 5. There are 6 choices.

For the tens digit, repetition is allowed, so any of the 6 digits can be used.

The possible digits for the tens place are 0, 1, 2, 3, 4, 5. There are 6 choices.

For the units digit, repetition is allowed, so any of the 6 digits can be used.

The possible digits for the units place are 0, 1, 2, 3, 4, 5. There are 6 choices.

By the Fundamental Principle of Counting, the total number of 4-digit numbers that can be formed is the product of the number of choices for each position.

Total number of 4-digit numbers = (Choices for thousands) $\times$ (Choices for hundreds) $\times$ (Choices for tens) $\times$ (Choices for units)

Total number of 4-digit numbers = $5 \times 6 \times 6 \times 6$

$5 \times 6 \times 6 \times 6 = 5 \times 216$

$5 \times 216 = 1080$

Therefore, 1080 different 4-digit numbers can be formed using the digits 0, 1, 2, 3, 4, 5 with repetition allowed.

The final answer is $\mathbf{1080}$.

Given:

Total number of points, $n = 10$.

Condition: No three of the points are collinear.

To Find:

The number of straight lines that can be drawn from these 10 points.

Solution:

A straight line is determined by selecting any two distinct points.

Since the order of selecting the two points does not matter (the line formed by point A and point B is the same as the line formed by point B and point A), this is a problem involving combinations.

We need to select 2 points from the available 10 points.

The number of combinations of $n$ distinct objects taken $r$ at a time is given by the formula:

$C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}$

In this problem, we have $n=10$ (total number of points) and $r=2$ (number of points needed to form a line).

The number of straight lines is given by $C(10, 2)$.

$C(10, 2) = \frac{10!}{2!(10-2)!} = \frac{10!}{2!8!}$

Expand the factorials. We can partially expand $10!$ to cancel out $8!$:

$10! = 10 \times 9 \times 8!$

$2! = 2 \times 1 = 2$

$8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$

Substitute these into the expression for $C(10, 2)$:

$C(10, 2) = \frac{10 \times 9 \times 8!}{2! \times 8!}$

Cancel out the $8!$ term from the numerator and the denominator:

$C(10, 2) = \frac{10 \times 9}{2!}$

$C(10, 2) = \frac{10 \times 9}{2 \times 1}$

$C(10, 2) = \frac{90}{2}$

Perform the division:

$\frac{90}{2} = 45$

The condition that "no three of which are collinear" is important because if three or more points were collinear, they would lie on the same single straight line, and selecting different pairs of points from that collinear set would result in the same line, reducing the total number of unique lines. In this problem, the condition ensures that every pair of points defines a unique straight line.

Therefore, the number of straight lines that can be drawn from 10 points, no three of which are collinear, is 45.

The final answer is $\mathbf{45}$.

We are given the equation $\binom{n}{8} = \binom{n}{6}$.

We use the property of combinations which states that if $\binom{n}{a} = \binom{n}{b}$, then either $a = b$ or $a + b = n$.

In the given equation $\binom{n}{8} = \binom{n}{6}$, we have $a = 8$ and $b = 6$.

According to the property, there are two possibilities:

Case 1: $a = b$

$8 = 6$

This is clearly false, so this case does not give a solution.

Case 2: $a + b = n$}

$8 + 6 = n$

$14 = n$

So, $n = 14$.

We should check if this value of $n$ is valid in the context of combinations. For $\binom{n}{r}$ to be defined, $n$ must be a non-negative integer, and $n \ge r$.

In this case, $n=14$, which is a non-negative integer. We also have $r=8$ and $r=6$.

Check if $n \ge 8$: $14 \ge 8$ (True).

Check if $n \ge 6$: $14 \ge 6$ (True).

Since $n=14$ satisfies these conditions, it is a valid solution.

We can also solve this by expanding the combination formulas, although it is more involved:

$\frac{n!}{8!(n-8)!} = \frac{n!}{6!(n-6)!}$

Since $n! \ne 0$ (assuming $n \ge 8$), we can divide both sides by $n!$:

$\frac{1}{8!(n-8)!} = \frac{1}{6!(n-6)!}$

Cross-multiply:

$6!(n-6)! = 8!(n-8)!$

Expand the larger factorials in terms of the smaller ones:

$6! = 6 \times 5 \times \dots \times 1$

$8! = 8 \times 7 \times 6!$

$(n-6)! = (n-6) \times (n-7) \times (n-8)!$

Substitute these into the equation:

$6! \times (n-6) \times (n-7) \times (n-8)! = (8 \times 7 \times 6!) \times (n-8)!$

Assuming $(n-8)! \ne 0$ (which requires $n \ge 8$) and $6! \ne 0$, we can divide both sides by $6! (n-8)!$:

$(n-6)(n-7) = 8 \times 7$

$(n-6)(n-7) = 56$

Expand the left side:

$n^2 - 7n - 6n + 42 = 56$

$n^2 - 13n + 42 = 56$

Move all terms to one side to form a quadratic equation:

$n^2 - 13n + 42 - 56 = 0$

$n^2 - 13n - 14 = 0$

Factor the quadratic equation. We look for two numbers that multiply to -14 and add up to -13. These numbers are -14 and 1.

$(n - 14)(n + 1) = 0$

This gives two possible solutions for $n$:

$n - 14 = 0 \implies n = 14$

$n + 1 = 0 \implies n = -1$

Again, in the context of combinations, $n$ must be a non-negative integer and $n \ge r$. Since $r=8$, $n$ must be at least 8. The value $n=-1$ is not valid.

The value $n=14$ is valid as $14 \ge 8$ and $14 \ge 6$.

Both methods confirm that the value of $n$ is 14.

The final answer is $\mathbf{n=14}$.

Given:

Total number of players available = 15.

Size of the cricket team to be chosen = 11.

The team must include a captain and a vice-captain.

The process involves selection of the team followed by assignment of roles.

To Find:

The total number of ways to choose a cricket team of 11 from 15 players, and then assign a captain and a vice-captain from the selected team.

Solution:

This problem can be broken down into two sequential steps:

Step 1: Select the 11 players for the team from the 15 available players.

Step 2: From the 11 selected players, choose one as the captain and another as the vice-captain.

Step 1: Selecting the team of 11 players

The order in which the 11 players are selected does not matter, so this is a combination problem.

Number of ways to select 11 players from 15 is given by the combination formula $C(n, r) = \binom{n}{r}$, where $n=15$ and $r=11$.

Number of ways to select the team = $\binom{15}{11}$

Using the property $\binom{n}{r} = \binom{n}{n-r}$, we have $\binom{15}{11} = \binom{15}{15-11} = \binom{15}{4}$.

$\binom{15}{4} = \frac{15!}{4!(15-4)!} = \frac{15!}{4!11!}$

$\binom{15}{4} = \frac{15 \times 14 \times 13 \times 12 \times 11!}{ (4 \times 3 \times 2 \times 1) \times 11! }$

$\binom{15}{4} = \frac{15 \times 14 \times 13 \times 12}{24}$

Simplify the expression:

$\binom{15}{4} = \frac{15 \times 14 \times 13 \times \cancel{12}^1}{\cancel{24}_2} = \frac{15 \times 14 \times 13}{2} = 15 \times \cancel{14}^7 \times 13 = 15 \times 7 \times 13$

$15 \times 7 = 105$

$105 \times 13 = 1365$

So, there are 1365 ways to select the team of 11 players.

Step 2: Assigning Captain and Vice-Captain from the selected 11 players

From the 11 selected players, we need to choose one as captain and one as vice-captain. The order matters here (choosing player A as captain and B as vice-captain is different from choosing player B as captain and A as vice-captain). This is a permutation problem.

Number of ways to choose a captain and vice-captain from 11 players is given by the permutation formula $P(n, r) = \frac{n!}{(n-r)!}$, where $n=11$ and $r=2$.

Number of ways to assign roles = $P(11, 2)$

$P(11, 2) = \frac{11!}{(11-2)!} = \frac{11!}{9!}$

$P(11, 2) = \frac{11 \times 10 \times 9!}{9!}$

$P(11, 2) = 11 \times 10 = 110$

So, there are 110 ways to assign the roles of captain and vice-captain from the 11 selected players.

Total number of ways

By the Fundamental Principle of Counting, the total number of ways to complete both steps is the product of the number of ways for each step.

Total number of ways = (Number of ways to select the team) $\times$ (Number of ways to assign roles)

Total number of ways = $\binom{15}{11} \times P(11, 2)$

Total number of ways = $1365 \times 110$

$1365 \times 110 = 150150$

$\begin{array}{cc}& & 1 & 3 & 6 & 5 \\ \times & & & 1 & 1 & 0 \\ \hline & & 0 & 0 & 0 & 0 \\ & 1 & 3 & 6 & 5 & \times \\ 1 & 3 & 6 & 5 & \times & \times \\ \hline 1 & 5 & 0 & 1 & 5 & 0 \\ \hline \end{array}$

Therefore, there are 150150 ways to choose a cricket team of 11 from 15 players, including a captain and a vice-captain, by first selecting the team and then assigning the roles.

Alternate Solution (Assignment of roles first):

This problem can also be viewed as first choosing the captain and vice-captain from the 15 players, and then selecting the remaining 9 players for the team from the remaining players.

Step 1: Choose a captain and a vice-captain from 15 players.

This is a permutation problem, as the roles are distinct. We are selecting 2 players from 15 and arranging them in the roles of captain and vice-captain.

Number of ways to choose Captain and Vice-Captain = $P(15, 2)$

$P(15, 2) = \frac{15!}{(15-2)!} = \frac{15!}{13!} = \frac{15 \times 14 \times 13!}{13!} = 15 \times 14 = 210$

So, there are 210 ways to choose the captain and vice-captain from the 15 players.

Step 2: Select the remaining 9 players for the team.

Two players have already been chosen as captain and vice-captain. We need to select $11 - 2 = 9$ more players for the team from the remaining $15 - 2 = 13$ players.

The order of selecting these 9 players does not matter, so this is a combination problem.

Number of ways to select the remaining 9 players = $\binom{13}{9}$

$\binom{13}{9} = \frac{13!}{9!(13-9)!} = \frac{13!}{9!4!}$

$\binom{13}{9} = \frac{13 \times 12 \times 11 \times 10 \times 9!}{9! \times (4 \times 3 \times 2 \times 1)} = \frac{13 \times 12 \times 11 \times 10}{24}$

Simplify the expression:

$\binom{13}{9} = \frac{13 \times \cancel{12}^1 \times 11 \times \cancel{10}^5}{\cancel{24}_2} = \frac{13 \times 11 \times 5}{1} = 143 \times 5 = 715$

So, there are 715 ways to select the remaining 9 players.

Total number of ways = (Number of ways to choose Captain and Vice-Captain) $\times$ (Number of ways to select remaining players)

Total number of ways = $P(15, 2) \times \binom{13}{9}$

Total number of ways = $210 \times 715$

$210 \times 715 = 150150$

Both methods yield the same result.

The final answer is $\mathbf{150150}$.

We need to expand $(n+2)!$ in terms of $n!$.

Recall the definition of the factorial of a non-negative integer $k$, denoted by $k!$, which is the product of all positive integers less than or equal to $k$.

For $k \ge 1$, $k! = k \times (k-1) \times (k-2) \times \dots \times 2 \times 1$.

Also, we use the recursive definition $k! = k \times (k-1)!$ for $k \ge 1$.

We start with $(n+2)!$. Using the recursive definition with $k = n+2$ (assuming $n+2 \ge 1$, i.e., $n \ge -1$), we have:

$(n+2)! = (n+2) \times ((n+2)-1)!$

$(n+2)! = (n+2) \times (n+1)!$

Now, we expand $(n+1)!$ using the same recursive definition with $k = n+1$ (assuming $n+1 \ge 1$, i.e., $n \ge 0$):

$(n+1)! = (n+1) \times ((n+1)-1)!$

$(n+1)! = (n+1) \times n!$

Substitute the expression for $(n+1)!$ back into the equation for $(n+2)!$:

$(n+2)! = (n+2) \times \underbrace{(n+1)!}_{\text{equals} (n+1) \times n!}$

$(n+2)! = (n+2) \times (n+1) \times n!$

This expansion is valid for non-negative integer values of $n$. For $n=0$, it becomes $2! = (0+2)(0+1)0! \implies 2 = 2 \times 1 \times 1$, which is true. For $n > 0$, $(n+2)!$ and $n!$ are defined as per the product definition, and the expansion holds.

Thus, the expansion of $(n+2)!$ in terms of $n!$ is $(n+2)(n+1)n!$.

The final answer is $\mathbf{(n+2)(n+1)n!}$.

The given word is 'ASSAM'.

Let's list the letters in the word and count their frequencies:

• A: appears 2 times
• S: appears 2 times
• M: appears 1 time

The total number of letters in the word 'ASSAM' is $2 + 2 + 1 = 5$.

This is a problem of finding the number of permutations of a set of objects where some objects are identical.

The formula for the number of distinct permutations of $n$ objects, where $n_1$ are of one type, $n_2$ are of a second type, ..., $n_k$ are of a $k$-th type, and $n_1 + n_2 + \dots + n_k = n$, is given by:

$\frac{n!}{n_1! n_2! \dots n_k!}$

In this case, we have:

• Total number of letters, $n = 5$.
• Number of 'A's, $n_1 = 2$.
• Number of 'S's, $n_2 = 2$.
• Number of 'M's, $n_3 = 1$.

The number of distinct words that can be formed is:

$\frac{5!}{2! 2! 1!}$

Calculate the factorials:

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

$2! = 2 \times 1 = 2$}

$1! = 1$

Substitute the factorial values into the formula:

$\frac{120}{(2) \times (2) \times (1)}$

$\frac{120}{4}$

$120 \div 4 = 30$

Therefore, 30 distinct words can be formed using the letters of the word 'ASSAM'.

The final answer is $\mathbf{30}$.

The number of permutations of $n$ distinct things taken $r$ at a time, denoted by $P(n, r)$ or $_nP_r$, is the number of ways to arrange $r$ objects chosen from a set of $n$ distinct objects, where the order of arrangement matters.

The formula for the number of permutations of $n$ distinct things taken $r$ at a time is:

$P(n, r) = \frac{n!}{(n-r)!}$

This formula is valid for non-negative integers $n$ and $r$ such that $0 \le r \le n$.

Let's explain how this formula is derived using the Fundamental Principle of Counting.

We want to select and arrange $r$ objects from $n$ distinct objects. Imagine $r$ positions that need to be filled.

• For the first position, there are $n$ choices (any of the $n$ distinct objects).
• After filling the first position, there are $n-1$ objects remaining. So, for the second position, there are $n-1$ choices.
• For the third position, there are $n-2$ remaining objects, giving $n-2$ choices.
• This continues until the $r$-th position.

For the $r$-th position, there are $n - (r-1) = n - r + 1$ objects remaining, giving $n - r + 1$ choices.

By the Fundamental Principle of Counting, the total number of ways to fill the $r$ positions is the product of the number of choices for each position:

$P(n, r) = n \times (n-1) \times (n-2) \times \dots \times (n-r+1)$

This product can be expressed using factorials. We can write it as:

$P(n, r) = \frac{n \times (n-1) \times \dots \times (n-r+1) \times (n-r) \times (n-r-1) \times \dots \times 1}{(n-r) \times (n-r-1) \times \dots \times 1}$

The numerator is the product of integers from $n$ down to 1, which is $n!$.

The denominator is the product of integers from $(n-r)$ down to 1, which is $(n-r)!$.

So, $P(n, r) = \frac{n!}{(n-r)!}$.

The formula for the number of permutations of $n$ distinct things taken $r$ at a time is $\mathbf{P(n, r) = \frac{n!}{(n-r)!}}$.

The number of combinations of $n$ distinct things taken $r$ at a time, denoted by $\binom{n}{r}$ or $C(n, r)$ or $_nC_r$, is the number of ways to select a group of $r$ objects from a set of $n$ distinct objects, where the order of selection does not matter.

The formula for the number of combinations of $n$ distinct things taken $r$ at a time is:

$\binom{n}{r} = \frac{n!}{r!(n-r)!}$

This formula is valid for non-negative integers $n$ and $r$ such that $0 \le r \le n$.

Let's explain how this formula is related to permutations.

Consider the number of permutations of $n$ distinct things taken $r$ at a time, $P(n, r)$. This involves two steps:

1. Selecting $r$ objects from $n$ (this is a combination, which we denote by $\binom{n}{r}$).

2. Arranging the selected $r$ objects (there are $r!$ ways to arrange $r$ distinct objects).

By the Fundamental Principle of Counting, the total number of permutations is the product of the number of ways to perform each step:

$P(n, r) = \binom{n}{r} \times r!$

We know the formula for permutations: $P(n, r) = \frac{n!}{(n-r)!}$.

Substitute this into the relationship above:

$\frac{n!}{(n-r)!} = \binom{n}{r} \times r!$

To find the formula for combinations, we divide both sides by $r!$:

$\binom{n}{r} = \frac{1}{r!} \times \frac{n!}{(n-r)!}$

$\binom{n}{r} = \frac{n!}{r!(n-r)!}$

This confirms the formula for combinations.

The formula for the number of combinations of $n$ distinct things taken $r$ at a time is $\mathbf{\binom{n}{r} = \frac{n!}{r!(n-r)!}}$.

We are given the property of combinations: $\binom{n}{r} = \binom{n}{n-r}$.

This property states that the number of ways to choose $r$ objects from a set of $n$ distinct objects is the same as the number of ways to choose $n-r$ objects from the set of $n$ distinct objects (since choosing $r$ objects to include is equivalent to choosing $n-r$ objects to exclude).

We need to calculate $\binom{15}{12}$.

Using the given property with $n=15$ and $r=12$, we have:

$\binom{15}{12} = \binom{15}{15-12}$

$\binom{15}{12} = \binom{15}{3}$

Calculating $\binom{15}{3}$ is simpler than calculating $\binom{15}{12}$ because the factorial in the denominator ($3!$) is smaller than $12!$, leading to fewer terms to expand in the numerator.

We use the formula for combinations: $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.

For $\binom{15}{3}$, we have $n=15$ and $r=3$.

$\binom{15}{3} = \frac{15!}{3!(15-3)!} = \frac{15!}{3!12!}$

Expand the factorials. We can partially expand $15!$ to cancel out $12!$:

$15! = 15 \times 14 \times 13 \times 12!$

$3! = 3 \times 2 \times 1 = 6$

$12! = 12 \times 11 \times \dots \times 1$

Substitute these into the expression for $\binom{15}{3}$:

$\binom{15}{3} = \frac{15 \times 14 \times 13 \times 12!}{3! \times 12!}$

Cancel out the $12!$ term from the numerator and the denominator:

$\binom{15}{3} = \frac{15 \times 14 \times 13}{3!}$

$\binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1}$

$\binom{15}{3} = \frac{15 \times 14 \times 13}{6}$

Perform the multiplication in the numerator:

$15 \times 14 \times 13 = 210 \times 13 = 2730$

So, $\binom{15}{3} = \frac{2730}{6}$

Perform the division:

$\frac{2730}{6} = 455$

Alternatively, simplify by cancelling terms directly:

$\binom{15}{3} = \frac{\cancel{15}^5 \times \cancel{14}^7 \times 13}{\cancel{3}_1 \times \cancel{2}_1 \times 1} = 5 \times 7 \times 13$

$5 \times 7 \times 13 = 35 \times 13 = 455$

Thus, using the property $\binom{n}{r} = \binom{n}{n-r}$, we find that $\binom{15}{12} = \binom{15}{3} = 455$.

The final answer is $\mathbf{455}$.

Given:

Total number of points on the circle, $n = 21$.

To Find:

The number of chords that can be drawn through these 21 points.

Solution:

A chord of a circle is a line segment that connects two distinct points on the circle.

To draw a chord, we need to select 2 points from the given 21 points on the circle.

Since the order in which the two points are selected does not matter (selecting point A and then point B results in the same chord as selecting point B and then point A), this is a problem involving combinations.

We need to find the number of ways to choose 2 points from a set of 21 distinct points.

The number of combinations of $n$ distinct objects taken $r$ at a time is given by the formula:

$C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}$

In this problem, we have $n=21$ (total number of points) and $r=2$ (number of points needed to form a chord).

The number of chords is given by $C(21, 2)$.

$C(21, 2) = \frac{21!}{2!(21-2)!} = \frac{21!}{2!19!}$

Expand the factorials. We can partially expand $21!$ to cancel out $19!$:

$21! = 21 \times 20 \times 19!$

$2! = 2 \times 1 = 2$

$19! = 19 \times 18 \times \dots \times 1$

Substitute these into the expression for $C(21, 2)$:

$C(21, 2) = \frac{21 \times 20 \times 19!}{2! \times 19!}$

Cancel out the $19!$ term from the numerator and the denominator:

$C(21, 2) = \frac{21 \times 20}{2!}$

$C(21, 2) = \frac{21 \times 20}{2 \times 1}$

$C(21, 2) = \frac{420}{2}$

Perform the division:

$\frac{420}{2} = 210$

Therefore, 210 chords can be drawn through 21 points on a circle.

The final answer is $\mathbf{210}$.

Given:

Number of red balls = 5

Number of blue balls = 4

Number of black balls = 3

Total number of balls = $5 + 4 + 3 = 12$.

Size of the selection to be made = 3 balls.

Requirement for the selection: exactly 1 red ball.

To Find:

The number of ways to select 3 balls such that exactly 1 red ball is included.

Solution:

We need to select a committee of 3 balls from the total collection. The order of selection does not matter, so this is a combination problem.

The selection of 3 balls must contain exactly 1 red ball. This means the remaining $3 - 1 = 2$ balls must be selected from the non-red balls.

The non-red balls consist of the blue balls and the black balls. The total number of non-red balls is $4 \text{ (blue)} + 3 \text{ (black)} = 7$.

We need to perform two independent selections:

1. Select exactly 1 red ball from the 5 available red balls.

2. Select the remaining 2 balls from the 7 available non-red balls.

The number of ways to select 1 red ball from 5 red balls is given by $\binom{5}{1}$.

$\binom{5}{1} = \frac{5!}{1!(5-1)!} = \frac{5!}{1!4!} = \frac{5 \times 4!}{1 \times 4!} = 5$

There are 5 ways to select 1 red ball.

The number of ways to select 2 non-red balls from the 7 available non-red balls is given by $\binom{7}{2}$.

$\binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7!}{2!5!}$

$\binom{7}{2} = \frac{7 \times 6 \times 5!}{ (2 \times 1) \times 5!} = \frac{7 \times 6}{2} = \frac{42}{2} = 21$

There are 21 ways to select 2 non-red balls.

To find the total number of selections with exactly 1 red ball, we multiply the number of ways to select the red ball by the number of ways to select the non-red balls. This is based on the Fundamental Principle of Counting, as the selection of the red ball and the selection of the non-red balls are independent events that occur together to form the desired group of 3 balls.

Total number of selections = (Ways to select 1 red ball) $\times$ (Ways to select 2 non-red balls)

Total number of selections = $\binom{5}{1} \times \binom{7}{2}$

Total number of selections = $5 \times 21$

$5 \times 21 = 105$

Therefore, there are 105 ways to make a selection of 3 balls such that there is exactly 1 red ball.

The final answer is $\mathbf{105}$.

We are given the equation $n! = 5040$.

We need to find the non-negative integer $n$ whose factorial is equal to 5040.

We can calculate the factorials of small non-negative integers until we reach 5040.

Calculate the factorials step by step:

$0! = 1$}

$1! = 1$}

$2! = 2 \times 1 = 2$}

$3! = 3 \times 2 \times 1 = 6$}

$4! = 4 \times 3 \times 2 \times 1 = 24$}

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$}

$6! = 6 \times 5! = 6 \times 120 = 720$}

$7! = 7 \times 6! = 7 \times 720 = 5040$}

We found that $7! = 5040$.

Since $n! = 5040$, and we found that $7! = 5040$, by comparing the two, we can conclude that $n=7$.

The value of $n$ is 7.

The final answer is $\mathbf{n=7}$.

Given:

We need to find the number of integers between 100 and 1000 that have distinct digits.

To Find:

The count of such numbers.

Solution:

Numbers between 100 and 1000 are 3-digit numbers.

A 3-digit number is of the form ABC, where A is the hundreds digit, B is the tens digit, and C is the units digit. The digits must be distinct, and the hundreds digit cannot be 0.

We have 10 digits available: $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$.

We will fill the positions for the hundreds, tens, and units digits considering the restrictions.

_ _ _ (Hundreds, Tens, Units)

For the hundreds digit (A), it cannot be 0. So there are 9 choices (any digit from 1 to 9).

For the tens digit (B), it must be different from the hundreds digit. Since the hundreds digit is one of the non-zero digits, 0 is still available. So we can choose any of the remaining 9 digits.

Number of choices for the tens digit = 9.

For the units digit (C), it must be different from both the hundreds and the tens digits. Two distinct digits have already been used.

Total digits available = 10.

Number of digits used = 2.

Number of remaining digits = $10 - 2 = 8$.

Number of choices for the units digit = 8.

By the Fundamental Principle of Counting, the total number of 3-digit numbers with distinct digits is the product of the number of choices for each position.

Total number of 3-digit numbers with distinct digits = (Choices for Hundreds) $\times$ (Choices for Tens) $\times$ (Choices for Units)

Total number = $9 \times 9 \times 8$

$9 \times 9 = 81$

$81 \times 8 = 648$

Since the numbers must be between 100 and 1000, they are exactly the 3-digit numbers.

Therefore, there are 648 numbers between 100 and 1000 that have distinct digits.

The final answer is $\mathbf{648}$.

The given word is 'PUNJAB'.

Let's identify the letters and classify them as vowels and consonants.

The letters in the word 'PUNJAB' are P, U, N, J, A, B.

The vowels are U, A. There are 2 vowels.

The consonants are P, N, J, B. There are 4 consonants.

The total number of letters in the word is $2 + 4 = 6$.

All the letters are distinct.

We want to find the number of arrangements such that no two vowels are together.

The strategy for this type of problem is to first arrange the consonants, and then place the vowels in the spaces created between the consonants (and at the ends).

Step 1: Arrange the consonants

There are 4 distinct consonants (P, N, J, B).

The number of ways to arrange these 4 consonants is $4!$.

$4! = 4 \times 3 \times 2 \times 1 = 24$

Step 2: Create spaces for the vowels

When we arrange the 4 consonants, there are 5 possible positions where the vowels can be placed so that no two vowels are adjacent. These positions are indicated by underscores below, where C represents a consonant:

_ C _ C _ C _ C _

There are $4 + 1 = 5$ spaces.

Step 3: Place the vowels in the spaces

We have 2 distinct vowels (U, A) that need to be placed in these 5 available spaces.

We need to select 2 of the 5 spaces and arrange the 2 vowels in those selected spaces. The order of the vowels matters.

This is a permutation problem: arranging 2 distinct vowels in 5 distinct positions.

The number of ways to place the 2 vowels in the 5 spaces is $P(5, 2)$.

$P(5, 2) = \frac{5!}{(5-2)!} = \frac{5!}{3!} = \frac{5 \times 4 \times 3!}{3!} = 5 \times 4 = 20$

Step 4: Combine the arrangements

By the Fundamental Principle of Counting, the total number of arrangements where no two vowels are together is the product of the number of ways to arrange the consonants and the number of ways to place the vowels in the spaces.

Total arrangements = (Arrangements of consonants) $\times$ (Ways to place vowels)

Total arrangements = $4! \times P(5, 2)$

Total arrangements = $24 \times 20$

$24 \times 20 = 480$

Therefore, there are 480 arrangements of the letters of the word 'PUNJAB' such that no two vowels are together.

The final answer is $\mathbf{480}$.

We are given the equation $P(n, 5) = 20 \cdot P(n, 3)$.

The formula for the number of permutations of $n$ distinct objects taken $r$ at a time is $P(n, r) = \frac{n!}{(n-r)!}$.

Using the formula, we can write the expressions for $P(n, 5)$ and $P(n, 3)$:

$P(n, 5) = \frac{n!}{(n-5)!}$

$P(n, 3) = \frac{n!}{(n-3)!}$

Substitute these expressions into the given equation:

$\frac{n!}{(n-5)!} = 20 \cdot \frac{n!}{(n-3)!}$

For the permutations to be defined, $n$ must be a non-negative integer, and $n \ge r$. In this equation, we have $r=5$ and $r=3$. Therefore, we must have $n \ge 5$. This also implies $n! \ne 0$.

Divide both sides of the equation by $n!$ (since $n! \ne 0$):

$\frac{1}{(n-5)!} = 20 \cdot \frac{1}{(n-3)!}$

Rearrange the equation to isolate the factorials:

$(n-3)! = 20 \cdot (n-5)!$

Expand the larger factorial, $(n-3)!$, in terms of the smaller factorial, $(n-5)!$. We know that $k! = k \times (k-1) \times (k-2) \times \dots \times m!$ where $m < k$.

$(n-3)! = (n-3) \times ((n-3)-1)! = (n-3) \times (n-4)!$

$(n-4)! = (n-4) \times ((n-4)-1)! = (n-4) \times (n-5)!$

So, $(n-3)! = (n-3) \times (n-4) \times (n-5)!$

Substitute this expansion back into the equation $(n-3)! = 20 \cdot (n-5)!$:

$(n-3)(n-4)(n-5)! = 20 \cdot (n-5)!$

Since we established that $n \ge 5$, we have $n-5 \ge 0$. Thus $(n-5)!$ is defined and non-zero (unless $n=5$, in which case $(n-5)! = 0! = 1$). We can divide both sides by $(n-5)!$:

$(n-3)(n-4) = 20$

Expand the left side of the equation:

$n^2 - 4n - 3n + 12 = 20$}

$n^2 - 7n + 12 = 20$}

Rearrange the equation into a quadratic form by moving all terms to one side:

$n^2 - 7n + 12 - 20 = 0$}

$n^2 - 7n - 8 = 0$}

Now, we need to solve this quadratic equation for $n$. We can factor the quadratic expression. We look for two numbers that multiply to -8 and add up to -7. These numbers are -8 and 1.

So, we can factor the equation as:

$(n - 8)(n + 1) = 0$

This gives two possible solutions for $n$:

$n - 8 = 0 \implies n = 8$

$n + 1 = 0 \implies n = -1$

In the context of permutations $P(n, r)$, $n$ must be a non-negative integer and $n \ge r$. In this equation, we have $r=5$. Therefore, we must have $n \ge 5$.

The value $n = -1$ is not a valid solution because $n$ must be non-negative and $n \ge 5$.

The value $n = 8$ is a valid solution because it is a non-negative integer and $8 \ge 5$.

Therefore, the value of $n$ is 8.

The final answer is $\mathbf{n=8}$.

Given:

Total number of points, $n = 12$.

Number of collinear points, $k = 5$.

To Find:

The number of triangles that can be formed from these 12 points.

Solution:

A triangle is formed by selecting any three non-collinear points.

We can first find the total number of ways to select any 3 points from the 12 points, without considering the collinearity. This is a combination problem, as the order of selecting the points does not matter.

Total number of ways to select 3 points from 12 = $\binom{12}{3}$.

$\binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12!}{3!9!}$

$\binom{12}{3} = \frac{12 \times 11 \times 10 \times 9!}{(3 \times 2 \times 1) \times 9!} = \frac{12 \times 11 \times 10}{6} = \frac{1320}{6} = 220$

So, there are 220 ways to select any 3 points from the 12 points.

However, a triangle cannot be formed if the three selected points are collinear.

We are given that 5 of the points are collinear.

Any set of 3 points selected from these 5 collinear points will not form a triangle. We need to subtract the number of ways to select 3 points from these 5 collinear points from the total number of ways to select 3 points.

Number of ways to select 3 points from the 5 collinear points = $\binom{5}{3}$.

$\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!}$

$\binom{5}{3} = \frac{5 \times 4 \times 3!}{(3 \times 2 \times 1) \times 2!} = \frac{5 \times 4}{2} = \frac{20}{2} = 10$

So, there are 10 combinations of 3 points from the collinear set that do not form a triangle.

The number of triangles is the total number of ways to select 3 points minus the number of ways to select 3 points that are collinear.

Number of triangles = (Total ways to select 3 points from 12) - (Ways to select 3 points from the 5 collinear points)

Number of triangles = $\binom{12}{3} - \binom{5}{3}$

Number of triangles = $220 - 10$

$220 - 10 = 210$

Therefore, 210 triangles can be formed from 12 points, of which 5 are collinear.

The final answer is $\mathbf{210}$.

Given:

Total number of students = 15.

Number of boys = 10.

Number of girls = 5.

Number of students to be selected = 4.

Requirement: at least 2 girls are included in the selection.

To Find:

The number of ways to select 4 students with at least 2 girls.

Solution:

A selection of students is a combination, as the order does not matter.

"At least 2 girls" means the selection of 4 students can contain either 2 girls, 3 girls, or 4 girls.

We need to calculate the number of ways for each case and sum them up.

Let the number of girls selected be $g$ and the number of boys selected be $b$. We need $g+b = 4$ and $g \ge 2$.

Possible cases for $(g, b)$ are:

Case 1: Exactly 2 girls ($g=2$). Since $g+b=4$, $b = 4-2 = 2$ boys.

Case 2: Exactly 3 girls ($g=3$). Since $g+b=4$, $b = 4-3 = 1$ boy.

Case 3: Exactly 4 girls ($g=4$). Since $g+b=4$, $b = 4-4 = 0$ boys.

Note that the maximum number of girls we can select is 4, as there are only 5 girls available in total, and we are selecting a total of 4 students.

Case 1: Exactly 2 girls and 2 boys

Number of ways to select 2 girls from 5 = $\binom{5}{2} = \frac{5!}{2!3!} = \frac{5 \times 4}{2 \times 1} = 10$

Number of ways to select 2 boys from 10 = $\binom{10}{2} = \frac{10!}{2!8!} = \frac{10 \times 9}{2 \times 1} = 45$

Number of selections in Case 1 = $\binom{5}{2} \times \binom{10}{2} = 10 \times 45 = 450$

Case 2: Exactly 3 girls and 1 boy

Number of ways to select 3 girls from 5 = $\binom{5}{3} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10$

Number of ways to select 1 boy from 10 = $\binom{10}{1} = \frac{10!}{1!9!} = \frac{10}{1} = 10$

Number of selections in Case 2 = $\binom{5}{3} \times \binom{10}{1} = 10 \times 10 = 100$

Case 3: Exactly 4 girls and 0 boys

Number of ways to select 4 girls from 5 = $\binom{5}{4} = \frac{5!}{4!1!} = \frac{5}{1} = 5$

Number of ways to select 0 boys from 10 = $\binom{10}{0} = 1$ (There is only one way to select nothing from a set).

Number of selections in Case 3 = $\binom{5}{4} \times \binom{10}{0} = 5 \times 1 = 5$

The total number of ways to select 4 students such that at least 2 girls are included is the sum of the number of ways in each case (since these cases are mutually exclusive).

Total number of ways = (Ways for Case 1) + (Ways for Case 2) + (Ways for Case 3)

Total number of ways = $450 + 100 + 5 = 555$

Therefore, there are 555 ways to select 4 students from the class such that at least 2 girls are included.

The final answer is $\mathbf{555}$.

The symbol $n!$ (read as "n factorial") represents the product of all positive integers from 1 up to $n$ for a non-negative integer $n$.

By definition, for $n \ge 1$, $n! = n \times (n-1) \times (n-2) \times \dots \times 3 \times 2 \times 1$.

The definition of $0!$ cannot be obtained directly from the product definition, as there are no positive integers less than or equal to 0.

However, $0!$ is defined conventionally and mathematically for consistency in various formulas, such as the formula for permutations ($P(n,n) = n!/0!$) and combinations ($\binom{n}{0} = n!/(0!n!) = 1$), and in series expansions like the Taylor series.

Using the recursive definition of factorial for $n \ge 1$:

$n! = n \times (n-1)!$

We can rearrange this formula to find $(n-1)!$:

$(n-1)! = \frac{n!}{n}$

Let's use this formula for $n=1$:

$(1-1)! = \frac{1!}{1}$

$0! = \frac{1!}{1}$

Since $1! = 1$, we have:

$0! = \frac{1}{1} = 1$

This result makes the factorial definition consistent for all non-negative integers and ensures that formulas involving factorials work correctly, especially in combinatorics where $0!$ appears frequently (e.g., in $\binom{n}{0} = 1$, which represents selecting 0 items from $n$ items, and there is only 1 way to do this - select nothing).

The value of $0!$ is 1.

The final answer is $\mathbf{1}$.

The given word is 'APPLY'.

Let's list the letters in the word and count their frequencies:

• A: appears 1 time
• P: appears 2 times
• L: appears 1 time
• Y: appears 1 time

The total number of letters in the word 'APPLY' is $1 + 2 + 1 + 1 = 5$.

Part 1: Total number of distinct words

This is a problem of finding the number of permutations of a set of objects where some objects are identical.

The formula for the number of distinct permutations of $n$ objects, where $n_1$ are of one type, $n_2$ are of a second type, ..., $n_k$ are of a $k$-th type, and $n_1 + n_2 + \dots + n_k = n$, is given by:

$\frac{n!}{n_1! n_2! \dots n_k!}$

In this case, we have:

• Total number of letters, $n = 5$.
• Number of 'A's, $n_1 = 1$.
• Number of 'P's, $n_2 = 2$.
• Number of 'L's, $n_3 = 1$.
• Number of 'Y's, $n_4 = 1$.

The number of distinct words that can be formed is:

$\frac{5!}{1! 2! 1! 1!} = \frac{5!}{2!}$

Calculate the factorials:

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

$2! = 2 \times 1 = 2$}

$1! = 1$

Substitute the factorial values into the formula:

$\frac{120}{2}$

$120 \div 2 = 60$

So, there are 60 distinct words that can be formed using the letters of the word 'APPLY'.

Part 2: Number of arrangements that begin with P

We need to find the number of arrangements where the first letter is 'P'.

Consider the 5 positions for the letters in an arrangement:

_ _ _ _ _

The first position is fixed with 'P'.

P _ _ _ _

One 'P' is used for the first position. We are left with the remaining 4 letters to arrange in the remaining 4 positions.

The remaining letters are A, P, L, Y.

Among these remaining letters, the letter 'P' appears 1 time, 'A' appears 1 time, 'L' appears 1 time, and 'Y' appears 1 time. All are distinct among themselves now.

The letters are: A, L, Y and one P.

The number of ways to arrange these 4 distinct remaining letters in the remaining 4 positions is $4!$.

$4! = 4 \times 3 \times 2 \times 1 = 24$

Alternatively, if we consider the original set of letters (A, P, P, L, Y) and fix one 'P' at the start, the remaining letters to arrange are A, P, L, Y. These 4 letters are distinct. The number of permutations of these 4 distinct letters is $4!$.

Number of arrangements = (Choices for 1st position) $\times$ (Arrangements of remaining letters)

Number of arrangements = $1 \times 4! = 1 \times 24 = 24$

So, there are 24 arrangements that begin with P.

The total number of distinct words is $\mathbf{60}$.

The number of arrangements that begin with P is $\mathbf{24}$.

We are given the equation $\binom{n}{r} = \binom{n}{r-2}$ and the condition $n > 2$.

We use the property of combinations which states that if $\binom{n}{a} = \binom{n}{b}$, then either $a = b$ or $a + b = n$.

In the given equation $\binom{n}{r} = \binom{n}{r-2}$, we can identify $a = r$ and $b = r-2$.

According to the property, there are two possibilities:

Case 1: $a = b$

$r = r-2$

$r - r = -2$

$0 = -2$

This is a contradiction. Therefore, this case is impossible, and the equality $\binom{n}{r} = \binom{n}{r-2}$ cannot hold because the lower indices are equal.

Case 2: $a + b = n$

$r + (r-2) = n$

Combine the terms on the left side:

$2r - 2 = n$

This is the necessary condition for the equality $\binom{n}{r} = \binom{n}{r-2}$ to hold, given that $r \ne r-2$.

We are asked to find the value of $r$. We can solve the equation $2r - 2 = n$ for $r$:

$2r = n + 2$

$r = \frac{n+2}{2}$

For the combination notation $\binom{n}{r}$ and $\binom{n}{r-2}$ to be defined in the standard context of combinatorics, $n$ and $r$ must be non-negative integers. Additionally, the lower index cannot exceed the upper index, so we must have $r \ge 0$, $r-2 \ge 0$, $n \ge r$, and $n \ge r-2$.

The condition $r-2 \ge 0$ implies $r \ge 2$.

The derived value $r = \frac{n+2}{2}$ must be an integer and satisfy $r \ge 2$ and $r \le n$.

If $r = \frac{n+2}{2}$ is an integer, then $n+2$ must be an even number, which means $n$ must be an even integer.

The condition $n>2$ implies that the smallest possible integer value for $n$ is 3. If $n=3$, $r = (3+2)/2 = 2.5$, which is not an integer. For $r$ to be an integer, $n$ must be an even integer greater than 2, i.e., $n \in \{4, 6, 8, \dots\}$.

If $n=4$ (the smallest even integer > 2), $r = (4+2)/2 = 3$. This is an integer, $r=3 \ge 2$, and $n=4 \ge r=3$. So $\binom{4}{3} = \binom{4}{1}$, which is $4=4$. This is true.

If $n=6$, $r = (6+2)/2 = 4$. This is an integer, $r=4 \ge 2$, and $n=6 \ge r=4$. So $\binom{6}{4} = \binom{6}{2}$, which is $15=15$. This is true.

If $n=8$, $r = (8+2)/2 = 5$. This is an integer, $r=5 \ge 2$, and $n=8 \ge r=5$. So $\binom{8}{5} = \binom{8}{3}$, which is $56=56$. This is true.

The equation $\binom{n}{r} = \binom{n}{r-2}$ holds for integer $r \ge 2$ if and only if $n$ is an even integer such that $n>2$ and $r = \frac{n+2}{2}$.

The question asks for "the value of $r$". Based on the given equation and the condition $n>2$, the value of $r$ is determined by $n$ as $r = \frac{n+2}{2}$. It is not a single numerical value independent of $n$. However, if a single numerical answer is expected, it might refer to the smallest integer value $r$ can take, which occurs when $n$ is the smallest valid even integer ($n=4$), yielding $r=3$. Without further context or clarification, the relationship $r = \frac{n+2}{2}$ is the direct result.

The value of $r$ is $\mathbf{\frac{n+2}{2}}$.

Given:

Number of friends = 5.

Number of chairs in a row = 5.

To Find:

The number of ways to seat 5 friends in a row of 5 chairs.

Solution:

We need to arrange 5 distinct friends in 5 distinct positions (the chairs in the row).

This is a problem of arranging $n$ distinct objects in $n$ positions, which is a permutation of $n$ objects taken $n$ at a time.

The number of ways to arrange $n$ distinct objects is given by $n!$ (n factorial).

In this case, the number of friends (objects) is $n=5$.

The number of positions (chairs) is also 5.

The number of ways to seat the 5 friends in the 5 chairs is $5!$.

$5! = 5 \times 4 \times 3 \times 2 \times 1$

$5! = 20 \times 6 = 120$

Alternatively, using the Fundamental Principle of Counting:

We have 5 chairs to fill.

• For the first chair, there are 5 choices (any of the 5 friends).
• For the second chair, there are 4 remaining choices.
• For the third chair, there are 3 remaining choices.
• For the fourth chair, there are 2 remaining choices.
• For the fifth chair, there is 1 remaining choice.

Total number of ways = $5 \times 4 \times 3 \times 2 \times 1 = 120$

Therefore, there are 120 ways that 5 friends can be seated in a row of 5 chairs.

The final answer is $\mathbf{120}$.

Given:

Number of white balls = 6.

Number of black balls = 5.

Total number of balls = $6 + 5 = 11$.

Number of balls to be selected = 4.

Requirement: at least 3 black balls are included in the selection.

To Find:

The number of ways to select 4 balls with at least 3 black balls.

Solution:

A selection of balls is a combination, as the order does not matter.

"At least 3 black balls" means the selection of 4 balls can contain either 3 black balls or 4 black balls.

We need to calculate the number of ways for each case and sum them up.

Let the number of black balls selected be $b$ and the number of white balls selected be $w$. We need $b+w = 4$ and $b \ge 3$.

Possible cases for $(b, w)$ are:

Case 1: Exactly 3 black balls ($b=3$). Since $b+w=4$, $w = 4-3 = 1$ white ball.

Case 2: Exactly 4 black balls ($b=4$). Since $b+w=4$, $w = 4-4 = 0$ white balls.

Note that the maximum number of black balls we can select is 4, as we are selecting a total of 4 balls, and there are 5 black balls available (we cannot select 5 black balls in a group of 4).

Case 1: Exactly 3 black balls and 1 white ball

Number of ways to select 3 black balls from 5 = $\binom{5}{3} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10$

Number of ways to select 1 white ball from 6 = $\binom{6}{1} = \frac{6!}{1!5!} = \frac{6}{1} = 6$

Number of selections in Case 1 = $\binom{5}{3} \times \binom{6}{1} = 10 \times 6 = 60$

Case 2: Exactly 4 black balls and 0 white balls

Number of ways to select 4 black balls from 5 = $\binom{5}{4} = \frac{5!}{4!1!} = \frac{5}{1} = 5$

Number of ways to select 0 white balls from 6 = $\binom{6}{0} = 1$

Number of selections in Case 2 = $\binom{5}{4} \times \binom{6}{0} = 5 \times 1 = 5$

The total number of ways to select 4 balls such that at least 3 black balls are included is the sum of the number of ways in each case (since these cases are mutually exclusive).

Total number of ways = (Ways for Case 1) + (Ways for Case 2)

Total number of ways = $60 + 5 = 65$

Therefore, there are 65 ways to make a selection of 4 balls from the bag so as to include at least 3 black balls.

The final answer is $\mathbf{65}$.

Given:

$P(n, r)$ represents the number of permutations of $n$ distinct things taken $r$ at a time.

A permutation is an arrangement where the order of the selected objects matters.

The factorial of a non-negative integer $k$, $k!$, is the product of all positive integers up to $k$, with $0! = 1$.

To Prove:

$P(n, r) = \frac{n!}{(n-r)!}$ for $0 \le r \le n$, where $n$ and $r$ are non-negative integers.

Proof:

Consider the process of selecting and arranging $r$ distinct objects from a set of $n$ distinct objects.

We can think of this as filling $r$ positions with distinct objects chosen from the $n$ available objects.

For the first position, there are $n$ choices (any of the $n$ objects).

For the second position, since the object chosen for the first position cannot be reused, there are $n-1$ choices remaining.

For the third position, there are $n-2$ choices remaining.

This continues for each of the $r$ positions.

For the $r$-th position, there are $n - (r-1) = n - r + 1$ choices remaining.

By the Fundamental Principle of Counting, the total number of ways to fill these $r$ positions is the product of the number of choices for each position.

$P(n, r) = n \times (n-1) \times (n-2) \times \dots \times (n-r+1)$

Now, we need to show that this product is equal to $\frac{n!}{(n-r)!}$.

Recall that $n! = n \times (n-1) \times \dots \times (n-r+1) \times (n-r) \times \dots \times 1$.

And $(n-r)! = (n-r) \times (n-r-1) \times \dots \times 1$.

Consider the expression $\frac{n!}{(n-r)!}$:

$\frac{n!}{(n-r)!} = \frac{n \times (n-1) \times (n-2) \times \dots \times (n-r+1) \times \cancel{(n-r)!}}{\cancel{(n-r)!}}$

Assuming $0 \le r \le n$, $(n-r)!$ is well-defined. For $r=n$, $(n-r)! = 0! = 1$. For $r < n$, $(n-r)!$ is a positive integer factorial.

Cancelling the $(n-r)!$ term from the numerator and denominator, we get:

$\frac{n!}{(n-r)!} = n \times (n-1) \times (n-2) \times \dots \times (n-r+1)$

This product is exactly the same as the product we obtained for $P(n, r)$ using the Fundamental Principle of Counting.

Thus, $P(n, r) = n \times (n-1) \times (n-2) \times \dots \times (n-r+1) = \frac{n!}{(n-r)!}$.

Let's consider the edge case $r=0$.

$P(n, 0)$: Number of ways to arrange 0 objects from $n$ is 1 (the empty arrangement).

Formula: $\frac{n!}{(n-0)!} = \frac{n!}{n!} = 1$. The formula holds.

Let's consider the edge case $r=n$.

$P(n, n)$: Number of ways to arrange $n$ objects from $n$ is $n!$.

Formula: $\frac{n!}{(n-n)!} = \frac{n!}{0!} = \frac{n!}{1} = n!$. The formula holds.

The formula $P(n, r) = \frac{n!}{(n-r)!}$ is valid for $0 \le r \le n$ and non-negative integer $n, r$.

This completes the proof.

Given:

$\binom{n}{r}$ represents the number of combinations of $n$ distinct things taken $r$ at a time.

A combination is a selection where the order of the selected objects does not matter.

$P(n, r)$ represents the number of permutations of $n$ distinct things taken $r$ at a time.

A permutation is an arrangement where the order of the selected objects matters.

The formula for permutations is $P(n, r) = \frac{n!}{(n-r)!}$.

To Prove:

$\binom{n}{r} = \frac{n!}{r!(n-r)!}$ for $0 \le r \le n$, where $n$ and $r$ are non-negative integers.

Proof:

Consider the task of forming a permutation of $r$ objects from a set of $n$ distinct objects, $P(n, r)$. This task can be thought of as a two-step process:

Step 1: Select $r$ objects from the $n$ distinct objects. The number of ways to do this is the number of combinations, $\binom{n}{r}$. The order of selection does not matter in this step.

Step 2: Arrange the $r$ selected objects in order. Since we have selected $r$ distinct objects, the number of ways to arrange these $r$ objects is $r!$ (the number of permutations of $r$ distinct objects).

By the Fundamental Principle of Counting, the total number of permutations of $n$ objects taken $r$ at a time, $P(n, r)$, is the product of the number of ways to perform Step 1 and Step 2.

$P(n, r) = (\text{Number of ways to select } r \text{ objects}) \times (\text{Number of ways to arrange } r \text{ objects})$

$P(n, r) = \binom{n}{r} \times r!$

We already know the formula for $P(n, r)$:

Substitute the formula for $P(n, r)$ from (i) into the equation $P(n, r) = \binom{n}{r} \times r!$:

$\frac{n!}{(n-r)!} = \binom{n}{r} \times r!$

To find the formula for $\binom{n}{r}$, we divide both sides of this equation by $r!$. Since $r \ge 0$, $r!$ is always a positive integer.

$\frac{1}{r!} \times \frac{n!}{(n-r)!} = \frac{1}{r!} \times \left( \binom{n}{r} \times r! \right)$

$\frac{n!}{r!(n-r)!} = \binom{n}{r} \times \frac{r!}{r!}$

$\frac{n!}{r!(n-r)!} = \binom{n}{r} \times 1$

$\binom{n}{r} = \frac{n!}{r!(n-r)!}$

This formula is valid for non-negative integers $n$ and $r$ such that $0 \le r \le n$. When $r=0$ or $r=n$, the definition relies on $0! = 1$.

This completes the proof.

Given:

The word is 'KOLKATA'.

To Find:

The number of distinct 5-letter words that can be formed using the letters of 'KOLKATA'.

Solution:

First, let's list the letters in the word 'KOLKATA' and count the frequency of each distinct letter:

• K: appears 2 times
• O: appears 1 time
• L: appears 1 time
• K: (already counted)
• A: appears 2 times
• T: appears 1 time
• A: (already counted)

The letters available are K, K, A, A, O, L, T. There are 7 letters in total.

We need to form 5-letter words using these letters. This involves selecting 5 letters from the available 7 letters and arranging them. The number of distinct words depends on the composition of the 5 selected letters regarding repetitions.

We can solve this by considering the possible cases for the types of letters chosen for the 5-letter word.

The available letters and their counts are: K (2), A (2), O (1), L (1), T (1).

We need to select 5 letters from these 7 and arrange them. Let's analyze the possible compositions of the 5 letters based on the repeated letters (K and A).

Case 1: The 5 letters include both pairs of identical letters (K, K and A, A).

We select the 2 K's and the 2 A's. This accounts for 4 letters.

We need one more letter from the remaining single letters {O, L, T}. The number of ways to choose this 1 letter from 3 is $\binom{3}{1} = 3$.

The possible sets of 5 letters are: {K, K, A, A, O}, {K, K, A, A, L}, {K, K, A, A, T}.

For each set, we find the number of permutations. For a set like {K, K, A, A, O} with letters of the form $x, x, y, y, z$, the number of permutations is $\frac{5!}{2!2!1!}$.

Number of permutations for each set = $\frac{5!}{2!2!} = \frac{120}{4} = 30$.

Since there are 3 such sets, the total number of words in this case = $3 \times 30 = 90$.

Case 2: The 5 letters include exactly one pair of identical letters (either K, K or A, A) and three distinct letters.

We choose one type of letter to be repeated: $\binom{2}{1} = 2$ ways (choose K or A).

Subcase 2a: We choose the pair {K, K}. The remaining letters available for selection are A, A, O, L, T. We need to choose 3 distinct letters from this remaining pool such that the 5 selected letters contain {K, K} and 3 distinct ones. The distinct letters available from {A, A, O, L, T} are {A, O, L, T}. We need to choose 3 distinct letters from these 4: $\binom{4}{3} = 4$. The sets of 5 letters are: {K, K, A, O, L}, {K, K, A, O, T}, {K, K, A, L, T}, {K, K, O, L, T}.

For each set (e.g., {K, K, A, O, L}), the letters are of the form $x, x, y, z, w$. The number of permutations is $\frac{5!}{2!1!1!1!} = \frac{120}{2} = 60$.

Number of words in Subcase 2a = $4 \times 60 = 240$.

Subcase 2b: We choose the pair {A, A}. The remaining letters available for selection are K, K, O, L, T. We need to choose 3 distinct letters from this remaining pool such that the 5 selected letters contain {A, A} and 3 distinct ones. The distinct letters available from {K, K, O, L, T} are {K, O, L, T}. We need to choose 3 distinct letters from these 4: $\binom{4}{3} = 4$. The sets of 5 letters are: {A, A, K, O, L}, {A, A, K, O, T}, {A, A, K, L, T}, {A, A, O, L, T}.

For each set (e.g., {A, A, K, O, L}), the letters are of the form $x, x, y, z, w$. The number of permutations is $\frac{5!}{2!1!1!1!} = 60$.

Number of words in Subcase 2b = $4 \times 60 = 240$.

Total number of words in Case 2 = $240 + 240 = 480$. (Correction: This summation is incorrect from the thought block. Let's recalculate based on selecting the pair first). Let's retry Case 2 calculation:

Case 2: One identical pair, three distinct letters.

Step 1: Choose which letter forms the pair: K or A ($\binom{2}{1}=2$ ways).

Step 2: Choose the three distinct letters from the pool of available single letters. Available single letters are O, L, T (3 letters). Also, the letter that was *not* chosen for the pair is available as a single (e.g., if K was chosen for the pair, A is available as a single from the remaining A). The set of available *distinct* letters is {K, A, O, L, T} (5 distinct letters). If we choose {K, K}, the remaining distinct letters are {A, O, L, T} (4 letters). We need to choose 3 distinct letters from these 4: $\binom{4}{3}=4$ ways. The sets formed are {K, K, A, O, L}, {K, K, A, O, T}, {K, K, A, L, T}, {K, K, O, L, T}. If we choose {A, A}, the remaining distinct letters are {K, O, L, T} (4 letters). We need to choose 3 distinct letters from these 4: $\binom{4}{3}=4$ ways. The sets formed are {A, A, K, O, L}, {A, A, K, O, T}, {A, A, K, L, T}, {A, A, O, L, T}. Total number of *sets* of letters with one pair and three singles is $4+4=8$. No, this is not correct either.

Let's use the refined Case B method from the thought block:

Case B: One identical pair, three distinct letters.

- Choose the letter for the pair (K or A): $\binom{2}{1}=2$.

- Suppose K is chosen for the pair {K,K}. The remaining letters are {A, A, O, L, T}. We need 3 distinct letters from these. The distinct options are A (as a single instance), O, L, T. We need to choose 3 from these 4 distinct options: $\binom{4}{3}=4$. Sets: {K,K,A,O,L}, {K,K,A,O,T}, {K,K,A,L,T}, {K,K,O,L,T}. Permutations $\frac{5!}{2!}=60$ each. Total = $4 \times 60 = 240$.

- Suppose A is chosen for the pair {A,A}. The remaining letters are {K, K, O, L, T}. We need 3 distinct letters from these. The distinct options are K (as a single instance), O, L, T. We need to choose 3 from these 4 distinct options: $\binom{4}{3}=4$. Sets: {A,A,K,O,L}, {A,A,K,O,T}, {A,A,K,L,T}, {A,A,O,L,T}. Permutations $\frac{5!}{2!}=60$ each. Total = $4 \times 60 = 240$.

Total for Case B = $240 + 240 = 480$. Still getting 480. Let me re-check the sets and compositions.

Available: K(2), A(2), O(1), L(1), T(1).

Case 1: KKAAY (Y from O,L,T). $\binom{3}{1}=3$ sets. {KKA AO}, {KKAAL}, {KKAAT}. Perms: $\frac{5!}{2!2!} = 30$. Total: $3 \times 30 = 90$. This is correct.

Case 2: Two Ks, one A, two from OLT. Select A: $\binom{1}{1}=1$. Select 2 from OLT: $\binom{3}{2}=3$. Sets: {KKAOL}, {KKAOT}, {KKALT}. Perms $\frac{5!}{2!} = 60$. Total $3 \times 60 = 180$. Correct.

Case 3: Two As, one K, two from OLT. Select K: $\binom{1}{1}=1$. Select 2 from OLT: $\binom{3}{2}=3$. Sets: {AAKOL}, {AAKOT}, {AAKLT}. Perms $\frac{5!}{2!} = 60$. Total $3 \times 60 = 180$. Correct.

Case 4: One K, one A, three from OLT. Select K: $\binom{1}{1}=1$. Select A: $\binom{1}{1}=1$. Select 3 from OLT: $\binom{3}{3}=1$. Set: {KAOLT}. Perms $5! = 120$. Total $1 \times 120 = 120$. Correct.

Total = $90 + 180 + 180 + 120 = 570$. The previous sum 480 was just a mistake in copying the thought process.

Case 3: All 5 letters are distinct.

The distinct letters available in 'KOLKATA' are K, A, O, L, T. There are 5 distinct letters.

We need to choose 5 distinct letters from these 5 distinct letters. The number of ways to do this is $\binom{5}{5} = 1$. The set is {K, A, O, L, T}.

The number of permutations of these 5 distinct letters is $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.

Number of words in this case = $1 \times 120 = 120$.

The total number of distinct 5-letter words is the sum of the number of words formed in each disjoint case.

Total words = (Words from Case 1) + (Words from Case 2) + (Words from Case 3)

Total words = $90 + 180 + 180 + 120 = 570$. (Summing up the corrected individual case totals: 90 + 180 + 180 + 120)

Therefore, 570 distinct 5-letter words can be formed using the letters of the word 'KOLKATA'.

The final answer is $\mathbf{570}$.

Given:

Total number of gentlemen available = 8.

Total number of ladies available = 5.

Size of the committee to be formed = 6.

Requirement for the committee: exactly 4 gentlemen.

To Find:

The number of ways to form a committee of 6 with exactly 4 gentlemen from the given group.

Solution:

Forming a committee involves selecting a group of individuals, and the order in which they are selected does not matter. Therefore, this problem requires the use of combinations.

The committee must have exactly 4 gentlemen. Since the total committee size is 6, the remaining members must be ladies.

Number of ladies in the committee = Total committee size - Number of gentlemen

Number of ladies in the committee = $6 - 4 = 2$.

So, the committee must consist of exactly 4 gentlemen and exactly 2 ladies.

We need to perform two independent selections:

1. Select exactly 4 gentlemen from the available 8 gentlemen.

2. Select exactly 2 ladies from the available 5 ladies.

The number of ways to select 4 gentlemen from 8 gentlemen is given by $\binom{8}{4}$.

$\binom{8}{4} = \frac{8!}{4!(8-4)!} = \frac{8!}{4!4!}$

$\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5 \times 4!}{(4 \times 3 \times 2 \times 1) \times 4!} = \frac{8 \times 7 \times 6 \times 5}{24}$

Simplify the expression:

$\binom{8}{4} = \frac{\cancel{8}^1 \times 7 \times \cancel{6}^1 \times 5}{\cancel{24}^3 \times \cancel{1}} = \frac{1 \times 7 \times 1 \times 5}{1} = 7 \times 5 = 35$

Alternatively, $\frac{8 \times 7 \times 6 \times 5}{24} = \frac{1680}{24} = 70$. Let me re-calculate the simplification.

$\binom{8}{4} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = \frac{\cancel{8}^1}{\cancel{4} \times \cancel{2}} \times \frac{\cancel{6}^2}{\cancel{3}} \times 7 \times 5 = 1 \times 2 \times 7 \times 5 = 70$. My mental math simplification was wrong. $\frac{1680}{24} = 70$.

So, there are 70 ways to select 4 gentlemen from 8 gentlemen.

The number of ways to select 2 ladies from 5 ladies is given by $\binom{5}{2}$.

$\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!}$

$\binom{5}{2} = \frac{5 \times 4 \times 3!}{(2 \times 1) \times 3!} = \frac{5 \times 4}{2} = \frac{20}{2} = 10$

There are 10 ways to select 2 ladies from 5 ladies.

To find the total number of committees with exactly 4 gentlemen AND exactly 2 ladies, we multiply the number of ways to select the gentlemen by the number of ways to select the ladies. This is based on the Fundamental Principle of Counting, as the selection of gentlemen and the selection of ladies are independent events that occur together to form the committee.

Total number of committees = (Number of ways to select gentlemen) $\times$ (Number of ways to select ladies)

Total number of committees = $\binom{8}{4} \times \binom{5}{2}$

Total number of committees = $70 \times 10$

$70 \times 10 = 700$

Therefore, there are 700 ways to form a committee of 6 consisting of exactly 4 gentlemen from a group of 8 gentlemen and 5 ladies.

The final answer is $\mathbf{700}$.

The given word is 'ENGINE'.

Let's list the letters in the word and count their frequencies:

• E: appears 2 times
• N: appears 2 times
• G: appears 1 time
• I: appears 1 time

The total number of letters in the word 'ENGINE' is $2 + 2 + 1 + 1 = 6$.

We want to find the number of arrangements such that the two 'E's are not together.

We can solve this by finding the total number of distinct arrangements and subtracting the number of arrangements where the two 'E's are together.

Part 1: Total number of distinct arrangements

This is the number of permutations of 6 objects where E is repeated 2 times and N is repeated 2 times.

The formula is $\frac{n!}{n_1! n_2! \dots n_k!}$.

Total distinct arrangements = $\frac{6!}{2! 2! 1! 1!} = \frac{6!}{2!2!}$

$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$

$2! = 2 \times 1 = 2$

Total distinct arrangements = $\frac{720}{2 \times 2} = \frac{720}{4} = 180$

Part 2: Number of arrangements where the two Es are together

To find the number of arrangements where the two 'E's are together, we treat the two 'E's as a single block or unit (EE).

Now, we are arranging this block (EE) and the remaining letters (N, N, G, I).

So, the objects to arrange are: (EE), N, N, G, I.

There are 5 objects to arrange, and the letter 'N' is repeated 2 times.

The number of arrangements of these objects is $\frac{5!}{2! 1! 1! 1!} = \frac{5!}{2!}$.

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

$2! = 2 \times 1 = 2$

Number of arrangements with Es together = $\frac{120}{2} = 60$

(Note: Within the (EE) block, the two E's are identical, so there is only $\frac{2!}{2!} = 1$ way to arrange them within the block).

Part 3: Number of arrangements where the two Es are not together

This is found by subtracting the number of arrangements where the Es are together from the total number of distinct arrangements.

Number of arrangements (Es not together) = (Total distinct arrangements) - (Arrangements with Es together)

Number of arrangements (Es not together) = $180 - 60 = 120$

Therefore, there are 120 arrangements of the letters of the word 'ENGINE' such that the two Es are not together.

The final answer is $\mathbf{120}$.

Given:

Available coin denominations: $\textsf{₹}1, \textsf{₹}2, \textsf{₹}5, \textsf{₹}10$. There are 4 different types of coins.

Supply of each coin type is unlimited.

Number of coins to be selected in a collection = 5.

To Find:

The number of different collections of 5 coins.

Solution:

This problem is about finding the number of combinations with repetition allowed, where we select a total of 5 items from 4 distinct categories (coin types) with an unlimited supply in each category.

Let $x_1$, $x_2$, $x_3$, and $x_4$ be the number of $\textsf{₹}1$, $\textsf{₹}2$, $\textsf{₹}5$, and $\textsf{₹}10$ coins selected, respectively.

We need to find the number of non-negative integer solutions to the equation:

$x_1 + x_2 + x_3 + x_4 = 5$

This can be solved using the "stars and bars" method. The number of non-negative integer solutions to the equation $x_1 + x_2 + \dots + x_k = n$ is given by the formula $\binom{n+k-1}{k-1}$ or $\binom{n+k-1}{n}$.

In this problem:

• $n$ is the total number of items to select (the sum in the equation), which is 5.
• $k$ is the number of categories (the number of variables in the equation), which is 4 (for the 4 coin types).

Using the formula $\binom{n+k-1}{k-1}$, the number of different collections is:

$\binom{5+4-1}{4-1} = \binom{8}{3}$

Now, we calculate the value of $\binom{8}{3}$ using the combination formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$:

$\binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!}$

Expand the factorials. We can partially expand $8!$ to cancel out $5!$:

$8! = 8 \times 7 \times 6 \times 5!$

$3! = 3 \times 2 \times 1 = 6$

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

Substitute these into the expression for $\binom{8}{3}$:

$\binom{8}{3} = \frac{8 \times 7 \times 6 \times 5!}{3! \times 5!}$

Cancel out the $5!$ term from the numerator and the denominator:

$\binom{8}{3} = \frac{8 \times 7 \times 6}{3!}$

$\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1}$

$\binom{8}{3} = \frac{8 \times 7 \times 6}{6}$

Perform the multiplication and division:

$\binom{8}{3} = \frac{336}{6} = 56$

Therefore, there are 56 different collections of 5 coins that can be made from an unlimited supply of $\textsf{₹}1, \textsf{₹}2, \textsf{₹}5,$ and $\textsf{₹}10$ coins.

The final answer is $\mathbf{56}$.

We need to evaluate the expression $\binom{10}{4} + \binom{10}{5}$.

We can use the combination formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ to calculate each term separately and then add them.

First, calculate $\binom{10}{4}$. Here $n=10$ and $r=4$.

$\binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!}$

$\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7 \times 6!}{(4 \times 3 \times 2 \times 1) \times 6!} = \frac{10 \times 9 \times 8 \times 7}{24}$

$\binom{10}{4} = \frac{\cancel{10}^5 \times \cancel{9}^3 \times \cancel{8}^1 \times 7}{\cancel{4}_1 \times \cancel{3}_1 \times \cancel{2}_1 \times 1} = 5 \times 3 \times 1 \times 7 = 105$. Let me redo the cancellation carefully.

$\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = \frac{10 \times \cancel{9}^3 \times \cancel{8}^1 \times 7}{\cancel{4} \times \cancel{3}_1 \times \cancel{2}_1 \times 1} = 10 \times 3 \times 1 \times 7 / 4$. Still wrong.

$\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = \frac{\cancel{10}^5 \times \cancel{9}^3 \times \cancel{8}^1 \times 7}{\cancel{4} \times \cancel{3} \times \cancel{2}_1 \times 1}$. No, $\cancel{4} \times \cancel{3} \times \cancel{2}$ is 24. The cancellation should be:

$\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = \frac{10 \times \cancel{9}^3 \times \cancel{8}^{4/2} \times 7}{\cancel{4}_1 \times \cancel{3}_1 \times \cancel{2}_1 \times 1}$. This is complicated. Let's do it step-by-step.

$\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{24}$.

$\cancel{24}$ and $\cancel{8}$ gives 3 in the denominator. $\frac{10 \times 9 \times \cancel{8}^1 \times 7}{\cancel{24}_3} = \frac{10 \times 9 \times 7}{3}$.

$\cancel{9}$ and $\cancel{3}$ gives 3 in the numerator. $\frac{10 \times \cancel{9}^3 \times 7}{\cancel{3}_1} = 10 \times 3 \times 7 = 210$.

So, $\binom{10}{4} = 210$.

Next, calculate $\binom{10}{5}$. Here $n=10$ and $r=5$.

$\binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10!}{5!5!}$

$\binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{(5 \times 4 \times 3 \times 2 \times 1) \times 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{120}$

$\binom{10}{5} = \frac{\cancel{10}^1 \times \cancel{9}^3 \times \cancel{8}^{2} \times 7 \times \cancel{6}^1}{\cancel{120}_{1 \times 4 \times 1 \times 1 \times 1}}$. $\frac{120}{10 \times 9 \times 8 \times 7 \times 6}$. $5! = 120$.

$\binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{120}$.

$\frac{\cancel{10}^1 \times \cancel{9}^3 \times \cancel{8}^1 \times 7 \times \cancel{6}^1}{\cancel{120}_{\cancel{12} \times \cancel{1} \times 1}}$. Let's use the denominator value 120 directly.

$\binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{120} = \frac{\cancel{10} \times \cancel{9} \times \cancel{8} \times 7 \times 6}{120}$

$\frac{10 \times 9 \times 8 \times 7 \times 6}{120} = \frac{30240}{120} = \frac{3024}{12}$.

$3024 \div 12 = 252$.

So, $\binom{10}{5} = 252$.

Alternatively, using the identity $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$.

The given expression is $\binom{10}{4} + \binom{10}{5}$.

Here, $n=10$, $r=4$. The second term has the lower index $r+1 = 4+1 = 5$.

Using the identity, $\binom{10}{4} + \binom{10}{5} = \binom{10+1}{4+1} = \binom{11}{5}$.

Now, calculate $\binom{11}{5}$. Here $n=11$ and $r=5$.

$\binom{11}{5} = \frac{11!}{5!(11-5)!} = \frac{11!}{5!6!}$

$\binom{11}{5} = \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6!}{(5 \times 4 \times 3 \times 2 \times 1) \times 6!} = \frac{11 \times 10 \times 9 \times 8 \times 7}{120}$

$\binom{11}{5} = \frac{11 \times \cancel{10}^1 \times \cancel{9}^3 \times \cancel{8}^1 \times 7}{\cancel{120}_{1 \times \cancel{4} \times \cancel{3} \times \cancel{2} \times 1} } $. The denominator is $5! = 120$.

$\binom{11}{5} = \frac{11 \times 10 \times 9 \times 8 \times 7}{120} = \frac{362880 \text{ (incorrect)}}{120}$. No, $11 \times 10 \times 9 \times 8 \times 7 = 110 \times 72 \times 7 = 110 \times 504 = 55440$. Still not easy.

$\binom{11}{5} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1}$.

$\binom{11}{5} = \frac{11 \times \cancel{10}^1 \times \cancel{9}^3 \times \cancel{8}^2 \times 7}{\cancel{5}_1 \times \cancel{4}_1 \times \cancel{3}_1 \times \cancel{2}_1 \times 1} = 11 \times 1 \times 3 \times 2 \times 7 = 11 \times 42 = 462$. This is better.

So, $\binom{11}{5} = 462$.

Let's check the sum of the individual calculations: $210 + 252 = 462$.

Both methods give the same result.

Therefore, $\binom{10}{4} + \binom{10}{5} = 462$.

The final answer is $\mathbf{462}$.

We are given the equation $P(n, 4) = 12 P(n, 2)$.

The formula for the number of permutations of $n$ distinct objects taken $r$ at a time is $P(n, r) = \frac{n!}{(n-r)!}$.

Using the formula, we can write the expressions for $P(n, 4)$ and $P(n, 2)$:

$P(n, 4) = \frac{n!}{(n-4)!}$

$P(n, 2) = \frac{n!}{(n-2)!}$

Substitute these expressions into the given equation:

$\frac{n!}{(n-4)!} = 12 \cdot \frac{n!}{(n-2)!}$

For the permutations to be defined, $n$ must be a non-negative integer, and $n \ge r$. In this equation, we have $r=4$ and $r=2$. Therefore, we must have $n \ge 4$. This also implies $n! \ne 0$.

Divide both sides of the equation by $n!$ (since $n! \ne 0$ for $n \ge 4$):

$\frac{1}{(n-4)!} = 12 \cdot \frac{1}{(n-2)!}$

Rearrange the equation to isolate the factorials:

$(n-2)! = 12 \cdot (n-4)!$

Expand the larger factorial, $(n-2)!$, in terms of the smaller factorial, $(n-4)!$. We know that $k! = k \times (k-1) \times (k-2) \times \dots \times m!$ where $m < k$.

$(n-2)! = (n-2) \times ((n-2)-1)! = (n-2) \times (n-3)!$

$(n-3)! = (n-3) \times ((n-3)-1)! = (n-3) \times (n-4)!$

So, $(n-2)! = (n-2) \times (n-3) \times (n-4)!$

Substitute this expansion back into the equation $(n-2)! = 12 \cdot (n-4)!$:

$(n-2)(n-3)(n-4)! = 12 \cdot (n-4)!$

Since we established that $n \ge 4$, we have $n-4 \ge 0$. Thus $(n-4)!$ is defined and non-zero (unless $n=4$, in which case $(n-4)! = 0! = 1$). We can divide both sides by $(n-4)!$:

$(n-2)(n-3) = 12$

Expand the left side of the equation:

$n^2 - 3n - 2n + 6 = 12$}

$n^2 - 5n + 6 = 12$}

Rearrange the equation into a quadratic form by moving all terms to one side:

$n^2 - 5n + 6 - 12 = 0$}

$n^2 - 5n - 6 = 0$}

Now, we need to solve this quadratic equation for $n$. We can factor the quadratic expression. We look for two numbers that multiply to -6 and add up to -5. These numbers are -6 and 1.

So, we can factor the equation as:

$(n - 6)(n + 1) = 0$

This gives two possible solutions for $n$:

$n - 6 = 0 \implies n = 6$

$n + 1 = 0 \implies n = -1$

In the context of permutations $P(n, r)$, $n$ must be a non-negative integer and $n \ge r$. In this equation, we have $r=4$. Therefore, we must have $n \ge 4$.

The value $n = -1$ is not a valid solution because $n$ must be non-negative and $n \ge 4$.

The value $n = 6$ is a valid solution because it is a non-negative integer and $6 \ge 4$.

Therefore, the value of $n$ is 6.

The final answer is $\mathbf{n=6}$.

Given:

The number is 720.

To Find:

The number of factors (divisors) of 720.

Solution:

To find the number of factors of an integer, we first need to find its prime factorization.

We find the prime factorization of 720:

$\begin{array}{c|cc} 2 & 720 \\ \hline 2 & 360 \\ \hline 2 & 180 \\ \hline 2 & 90 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$

The prime factorization of 720 is $2^4 \times 3^2 \times 5^1$.

If the prime factorization of a number $N$ is given by $N = p_1^{a_1} \times p_2^{a_2} \times \dots \times p_k^{a_k}$, where $p_1, p_2, \dots, p_k$ are distinct prime numbers and $a_1, a_2, \dots, a_k$ are non-negative integers, then the number of factors of $N$ is given by the product of one more than each exponent:

Number of factors = $(a_1 + 1)(a_2 + 1)\dots(a_k + 1)$

In the prime factorization of 720, $2^4 \times 3^2 \times 5^1$, the prime factors and their exponents are:

• Prime $p_1 = 2$, exponent $a_1 = 4$.
• Prime $p_2 = 3$, exponent $a_2 = 2$.
• Prime $p_3 = 5$, exponent $a_3 = 1$.

Using the formula for the number of factors:

Number of factors = $(a_1 + 1)(a_2 + 1)(a_3 + 1)$

Number of factors = $(4 + 1)(2 + 1)(1 + 1)$

Number of factors = $(5)(3)(2)$

Number of factors = $5 \times 3 \times 2 = 15 \times 2 = 30$

Therefore, the number of factors of 720 is 30.

The final answer is $\mathbf{30}$.

The given word is 'FACTORIAL'.

Let's list the letters in the word and count their frequencies:

• F: appears 1 time
• A: appears 2 times
• C: appears 1 time
• T: appears 1 time
• O: appears 1 time
• R: appears 1 time
• I: appears 1 time
• L: appears 1 time

The total number of letters in the word 'FACTORIAL' is $1 + 2 + 1 + 1 + 1 + 1 + 1 + 1 = 9$.

This is a problem of finding the number of distinct permutations of a set of objects where some objects are identical.

The formula for the number of distinct permutations of $n$ objects, where $n_1$ are of one type, $n_2$ are of a second type, ..., $n_k$ are of a $k$-th type, and $n_1 + n_2 + \dots + n_k = n$, is given by:

$\frac{n!}{n_1! n_2! \dots n_k!}$

In this case, we have:

• Total number of letters, $n = 9$.
• Number of 'A's, $n_1 = 2$.
• The other letters (F, C, T, O, R, I, L) each appear 1 time. So, $n_2=1, n_3=1, \dots, n_8=1$.

The number of distinct arrangements that can be formed is:

$\frac{9!}{2! 1! 1! 1! 1! 1! 1! 1!} = \frac{9!}{2!}$

Calculate the factorials:

$9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362880$

$2! = 2 \times 1 = 2$}

Substitute the factorial values into the formula:

$\frac{362880}{2}$

$362880 \div 2 = 181440$

Therefore, 181440 different arrangements can be made from the letters of the word 'FACTORIAL'.

The final answer is $\mathbf{181440}$.

Solution:

We are given the digits 0, 1, 2, 3, 4, 5. We need to form 4-digit numbers without repetition.

Part 1: Total number of 4-digit numbers.

A 4-digit number has four places: Thousands, Hundreds, Tens, and Units.

Digits available: ${0, 1, 2, 3, 4, 5}$ (total 6 digits).

The thousands digit cannot be 0.

Number of choices for the Thousands place:

It can be any of the digits 1, 2, 3, 4, 5. There are 5 choices.

Number of choices for the Hundreds place:

One digit is used for the thousands place. We have 5 digits remaining (including 0). There are 5 choices.

Number of choices for the Tens place:

Two digits have been used (thousands and hundreds). We have 4 digits remaining. There are 4 choices.

Number of choices for the Units place:

Three digits have been used. We have 3 digits remaining. There are 3 choices.

The total number of 4-digit numbers is the product of the number of choices for each place:

$5 (\text{Thousands}) \times 5 (\text{Hundreds}) \times 4 (\text{Tens}) \times 3 (\text{Units})$

$= 5 \times 5 \times 4 \times 3$

$= 25 \times 12$

$= 300$

So, a total of 300 four-digit numbers can be formed.

Part 2: Number of even 4-digit numbers.

For a number to be even, its units digit must be an even number. The even digits in the given set $\{0, 1, 2, 3, 4, 5\}$ are ${0, 2, 4}$.

We consider two cases based on the units digit:

Case 1: The units digit is 0.

Number of choices for the Units place: 1 (must be 0).

Number of choices for the Thousands place:

Since 0 is used in the units place, the thousands place cannot be 0 (which is already satisfied) and cannot be the digit used for units. The available digits are now {1, 2, 3, 4, 5}. There are 5 choices.

Number of choices for the Hundreds place:

Two digits are used (units and thousands). We have 4 digits remaining. There are 4 choices.

Number of choices for the Tens place:

Three digits are used. We have 3 digits remaining. There are 3 choices.

Number of even numbers ending in 0:

$5 (\text{Thousands}) \times 4 (\text{Hundreds}) \times 3 (\text{Tens}) \times 1 (\text{Units})$

$= 5 \times 4 \times 3 \times 1$

$= 60$

Case 2: The units digit is a non-zero even digit (2 or 4).

Number of choices for the Units place: 2 (can be 2 or 4).

Number of choices for the Thousands place:

It cannot be 0, and it cannot be the digit used for the units place. One digit is used for units (either 2 or 4). Out of the remaining 5 digits, one is 0. So we exclude 0. There are $5 - 1 = 4$ choices.

Number of choices for the Hundreds place:

Two digits are used (units and thousands). We have 4 digits remaining (including 0). There are 4 choices.

Number of choices for the Tens place:

Three digits are used. We have 3 digits remaining. There are 3 choices.

Number of even numbers ending in 2 or 4:

$4 (\text{Thousands}) \times 4 (\text{Hundreds}) \times 3 (\text{Tens}) \times 2 (\text{Units})$

$= 4 \times 4 \times 3 \times 2$

$= 16 \times 6$

$= 96$

Total number of even 4-digit numbers is the sum of the numbers from Case 1 and Case 2:

Total even numbers $= (\text{Numbers ending in 0}) + (\text{Numbers ending in 2 or 4})$

$= 60 + 96$

$= 156$

So, there are 156 even four-digit numbers.

Final Answer:

Total number of 4-digit numbers that can be formed: 300

Number of these 4-digit numbers that are even: 156

Solution:

The given word is 'PERMUTATIONS'.

The word contains 12 letters.

Let's list the letters and their frequencies:

P: 1, E: 1, R: 1, M: 1, U: 1, T: 2, A: 1, I: 1, O: 1, N: 1, S: 1.

The letter 'T' is repeated 2 times. All other 10 letters are distinct.

The total number of distinct arrangements of the letters in 'PERMUTATIONS' is $\frac{12!}{2!}$.

Part 1: The words start with P and end with S.

The first position is fixed with 'P' and the last position (12th) is fixed with 'S'.

We need to arrange the remaining $12 - 2 = 10$ letters in the remaining $12 - 2 = 10$ positions (from position 2 to position 11).

The remaining letters are E, R, M, U, T, A, I, O, N, T.

Among these 10 letters, the letter 'T' is repeated 2 times.

The number of ways to arrange these 10 letters is given by the formula for permutations of objects with repetition:

Number of arrangements $= \frac{(\text{Number of remaining letters})!}{(\text{Frequency of repeated letters})!}$

$= \frac{10!}{2!}$

Calculation:

$10! = 3,628,800$

$2! = 2 \times 1 = 2$

$\frac{10!}{2!} = \frac{3,628,800}{2} = 1,814,400$

Thus, the number of ways the letters of the word 'PERMUTATIONS' can be arranged such that the words start with P and end with S is 1,814,400.

Part 2: There are always 4 letters between P and S.

Let the 12 positions for the letters be numbered 1 to 12.

We need to place P and S such that there are exactly 4 letters between them.

This means P and S are separated by 5 positions (e.g., P _ _ _ _ S).

The possible pairs of positions $(i, j)$ for (P, S) such that there are 4 letters between them are when $|i - j| = 5$.

The pairs of positions are:

(1, 6), (2, 7), (3, 8), (4, 9), (5, 10), (6, 11), (7, 12).

There are 7 such pairs of positions.

For each pair of positions $(i, j)$, P can be in position $i$ and S in position $j$, or S can be in position $i$ and P in position $j$.

So, the number of ways to place P and S in the 12 positions is $7 \times 2 = 14$.

Once P and S are placed, the remaining 10 letters must be arranged in the remaining 10 positions.

The remaining 10 letters are E, R, M, U, T, A, I, O, N, T. Among these, 'T' is repeated 2 times.

The number of ways to arrange these 10 letters in the remaining 10 positions is $\frac{10!}{2!}$.

$\frac{10!}{2!} = 1,814,400$ (from Part 1).

The total number of arrangements where there are always 4 letters between P and S is the product of the number of ways to place P and S and the number of ways to arrange the remaining letters:

Total arrangements = (Number of ways to place P and S) $\times$ (Number of ways to arrange the remaining 10 letters)

$= 14 \times \frac{10!}{2!}$

$= 14 \times 1,814,400$

$= 25,401,600$

Thus, the number of ways the letters of the word 'PERMUTATIONS' can be arranged such that there are always 4 letters between P and S is 25,401,600.

Part 3: All vowels are together.

The vowels in the word 'PERMUTATIONS' are E, U, A, I, O.

There are 5 vowels, and they are all distinct.

Treat these 5 vowels as a single block. Let's call this the Vowel Block (V).

The remaining letters are the consonants: P, R, M, T, T, N, S.

There are 7 consonants. Note that the consonant 'T' is repeated 2 times.

Now, we consider the arrangement of the Vowel Block (V) and the 7 consonants.

We are arranging a total of $1 (\text{Vowel Block}) + 7 (\text{Consonants}) = 8$ units.

The 8 units are: V, P, R, M, T, T, N, S.

Among these 8 units, the consonant 'T' is repeated 2 times.

The number of ways to arrange these 8 units is $\frac{8!}{2!}$.

Calculation:

$8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40,320$

$2! = 2 \times 1 = 2$

$\frac{8!}{2!} = \frac{40,320}{2} = 20,160$

Now, we need to consider the arrangements of the vowels within the Vowel Block.

The vowels are E, U, A, I, O. They are 5 distinct vowels.

The number of ways to arrange these 5 distinct vowels within the block is $5!$.

Calculation:

$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

The total number of arrangements where all vowels are together is the product of the number of ways to arrange the 8 units and the number of ways to arrange the vowels within the block.

Total arrangements = (Arrangements of 8 units) $\times$ (Arrangements of 5 vowels within the block)

$= \frac{8!}{2!} \times 5!$

$= 20,160 \times 120$

Calculation:

$= 2,419,200$

Thus, the number of ways the letters of the word 'PERMUTATIONS' can be arranged such that all vowels are together is 2,419,200.

Given:

Number of men in the group = 7

Number of women in the group = 6

Size of the committee to be formed = 5

To Find:

The number of ways a committee of 5 can be formed if it is to consist of at least 3 men.

Solution:

The committee must consist of at least 3 men. This means the number of men in the committee can be 3, 4, or 5.

Since the total size of the committee is 5, the number of women will be $5 - (\text{number of men})$.

We consider the possible cases:

Case 1: Committee has exactly 3 men and 2 women.

Number of ways to choose 3 men from 7 = $\binom{7}{3}$

Number of ways to choose 2 women from 6 = $\binom{6}{2}$

The number of ways to form the committee in this case is the product of the number of ways to choose men and women:

Number of ways = $\binom{7}{3} \times \binom{6}{2}$

Using the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$:

$\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 7 \times 5 = 35$

$\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} = \frac{6 \times 5}{2 \times 1} = 3 \times 5 = 15$

Number of ways in Case 1 = $35 \times 15 = 525$

Case 2: Committee has exactly 4 men and 1 woman.

Number of ways to choose 4 men from 7 = $\binom{7}{4}$

Number of ways to choose 1 woman from 6 = $\binom{6}{1}$

The number of ways to form the committee in this case is the product:

Number of ways = $\binom{7}{4} \times \binom{6}{1}$

$\binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 7 \times 5 = 35$

$\binom{6}{1} = \frac{6!}{1!(6-1)!} = \frac{6!}{1!5!} = 6$

Number of ways in Case 2 = $35 \times 6 = 210$

Case 3: Committee has exactly 5 men and 0 women.

Number of ways to choose 5 men from 7 = $\binom{7}{5}$

Number of ways to choose 0 women from 6 = $\binom{6}{0}$

The number of ways to form the committee in this case is the product:

Number of ways = $\binom{7}{5} \times \binom{6}{0}$

$\binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7!}{5!2!} = \frac{7 \times 6}{2 \times 1} = 21$

$\binom{6}{0} = 1$ (There is only one way to choose zero items)

Number of ways in Case 3 = $21 \times 1 = 21$

The total number of ways to form the committee with at least 3 men is the sum of the ways in these three cases:

Total ways = (Ways in Case 1) + (Ways in Case 2) + (Ways in Case 3)

Total ways = $525 + 210 + 21$

Total ways = $735 + 21 = 756$

Thus, the number of ways to form a committee of 5 with at least 3 men is 756.

We want to prove the identity $\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}$ using a combinatorial argument.

A combinatorial proof works by showing that both sides of the equation count the same set of objects, but in different ways.

Consider the right-hand side of the equation, $\binom{n+1}{k}$.

This term represents the number of ways to choose a subset of $k$ distinct elements from a set of $n+1$ distinct elements.

Now, let's consider the left-hand side. We need to show that $\binom{n}{k} + \binom{n}{k-1}$ also counts the number of ways to choose a subset of $k$ elements from a set of $n+1$ elements.

Let our set be $S = \{1, 2, \dots, n, n+1\}$. We want to choose a subset of $k$ elements from $S$. Let's single out a particular element from the set, say the element $(n+1)$.

Any subset of $k$ elements chosen from $S$ must either contain the element $(n+1)$ or not contain the element $(n+1)$. These two possibilities form two disjoint cases that cover all possible subsets of size $k$.

Case 1: The subset of $k$ elements includes the element $(n+1)$.

If the subset must contain $(n+1)$, then we need to choose the remaining $k-1$ elements from the remaining $n$ elements in the set $\{1, 2, \dots, n\}$.

The number of ways to choose $k-1$ elements from a set of $n$ elements is given by the binomial coefficient $\binom{n}{k-1}$.

Case 2: The subset of $k$ elements does NOT include the element $(n+1)$.

If the subset must not contain $(n+1)$, then we need to choose all $k$ elements from the remaining $n$ elements in the set $\{1, 2, \dots, n\}$.

The number of ways to choose $k$ elements from a set of $n$ elements is given by the binomial coefficient $\binom{n}{k}$.

Since these two cases are mutually exclusive and exhaustive, the total number of ways to choose a subset of $k$ elements from the set $S$ (which has $n+1$ elements) is the sum of the number of ways in Case 1 and Case 2.

Total number of ways = (Number of ways in Case 1) + (Number of ways in Case 2)

Total number of ways = $\binom{n}{k-1} + \binom{n}{k}$.

We have shown that the number of ways to choose $k$ elements from a set of $n+1$ elements can be counted in two ways:

1. Directly: $\binom{n+1}{k}$

2. By splitting into cases based on a specific element: $\binom{n}{k} + \binom{n}{k-1}$

Since both expressions count the same set of objects (subsets of size $k$ from a set of $n+1$ elements), they must be equal.

Therefore, $\binom{n+1}{k} = \binom{n}{k} + \binom{n}{k-1}$.

This concludes the combinatorial proof of Pascal's Identity.

We need to find the number of distinct permutations of the letters in the word 'MISSISSIPPI' and the number of permutations where the four 'I's do not appear together.

First, let's analyze the letters in the word 'MISSISSIPPI' and count the frequency of each distinct letter.

The word 'MISSISSIPPI' has a total of $11$ letters.

The distinct letters and their frequencies are:

M: $1$ time

I: $4$ times

S: $4$ times

P: $2$ times

Part 1: Total number of distinct permutations.

The formula for the number of distinct permutations of $n$ objects where there are $n_1$ identical objects of type 1, $n_2$ identical objects of type 2, ..., $n_k$ identical objects of type $k$ is given by:

$\frac{n!}{n_1! n_2! \dots n_k!}$

In this case, $n=11$, $n_M=1$, $n_I=4$, $n_S=4$, $n_P=2$.

Total number of distinct permutations = $\frac{11!}{1! \times 4! \times 4! \times 2!}$

Total number of distinct permutations = $\frac{39916800}{1 \times 24 \times 24 \times 2}$

Total number of distinct permutations = $\frac{39916800}{1152}$

Total number of distinct permutations = $34650$.

Part 2: Number of permutations where the four I's do not come together.

To find the number of permutations where the four I's do not come together, we can subtract the number of permutations where the four I's do come together from the total number of distinct permutations.

First, let's calculate the number of permutations where the four I's come together. We can treat the four I's as a single block or unit (IIII).

Now, the items to permute are M, (IIII), S, S, S, S, P, P.

The total number of these new items is $1$ (M) + $1$ (block of I's) + $4$ (S) + $2$ (P) = $8$ items.

Among these $8$ items, the distinct items and their frequencies are:

M: $1$ time

(IIII): $1$ time

S: $4$ times

P: $2$ times

The number of distinct permutations of these $8$ items is:

Number of permutations with I's together = $\frac{8!}{1! \times 1! \times 4! \times 2!}$

Number of permutations with I's together = $\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{1 \times 1 \times (4 \times 3 \times 2 \times 1) \times (2 \times 1)}$

Number of permutations with I's together = $\frac{8 \times 7 \times 6 \times 5}{2}$

Number of permutations with I's together = $4 \times 7 \times 6 \times 5$

Number of permutations with I's together = $840$.

Finally, the number of permutations where the four I's do not come together is:

Number of permutations with I's not together = (Total distinct permutations) - (Number of permutations with I's together)

Number of permutations with I's not together = $34650 - 840$

Number of permutations with I's not together = $33810$.

The total number of distinct permutations of the letters of the word 'MISSISSIPPI' is $34650$.

The number of permutations in which the four I's do not come together is $33810$.

We need to find the total number of distinct permutations of the letters in the word 'CALCUTTA', and then the number of permutations where the two 'T's are together and where they are not together.

First, let's analyze the letters in the word 'CALCUTTA' and count the frequency of each distinct letter.

The word 'CALCUTTA' has a total of $8$ letters.

The distinct letters and their frequencies are:

C: $2$ times

A: $2$ times

L: $1$ time

U: $1$ time

T: $2$ times

Total number of letters, $n = 8$. The frequencies of the repeating letters are $n_C=2$, $n_A=2$, and $n_T=2$.

Part 1: Total number of distinct permutations.

The formula for the number of distinct permutations of $n$ objects with repetitions is $\frac{n!}{n_1! n_2! \dots n_k!}$.

Total number of distinct permutations = $\frac{8!}{2! \times 2! \times 1! \times 1! \times 2!}$

Total number of distinct permutations = $\frac{8!}{2! \times 2! \times 2!}$

Total number of distinct permutations = $\frac{40320}{2 \times 2 \times 2}$

Total number of distinct permutations = $\frac{40320}{8}$

Total number of distinct permutations = $5040$.

Part 2: Number of words where the two Ts are together.

To find the number of permutations where the two 'T's are together, we can treat the pair 'TT' as a single block or unit.

Now, the items to permute are C, A, L, U, (TT), C, A.

The total number of these new items is $7$ (C, A, L, U, (TT), C, A).

Among these $7$ items, the distinct items and their frequencies are:

C: $2$ times

A: $2$ times

L: $1$ time

U: $1$ time

(TT): $1$ time

The number of distinct permutations of these $7$ items is:

Number of permutations with Ts together = $\frac{7!}{2! \times 2! \times 1! \times 1! \times 1!}$

Number of permutations with Ts together = $\frac{7!}{2! \times 2!}$

Number of permutations with Ts together = $\frac{5040}{2 \times 2}$

Number of permutations with Ts together = $\frac{5040}{4}$

Number of permutations with Ts together = $1260$.

Part 3: Number of words where the two Ts are not together.

The number of permutations where the two 'T's are not together is equal to the total number of distinct permutations minus the number of permutations where the two 'T's are together.

Number of permutations with Ts not together = (Total distinct permutations) - (Number of permutations with Ts together)

Number of permutations with Ts not together = $5040 - 1260$

Number of permutations with Ts not together = $3780$.

Summary:

Total number of distinct words formed from the letters of 'CALCUTTA' is $5040$.

Number of words where the two Ts are together is $1260$.

Number of words where the two Ts are not together is $3780$.

The bag contains 5 red, 6 white, and 7 black balls. We need to select a total of 5 balls.

The selection must satisfy one of the following two conditions:

Condition 1: Exactly 2 white balls.

Condition 2: At least 4 black balls.

Let's calculate the number of ways for each condition separately and then consider the possibility of overlap.

Condition 1: Exactly 2 white balls.

If we must select exactly 2 white balls from the 6 available white balls, the number of ways to do this is $\binom{6}{2}$.

The remaining $5 - 2 = 3$ balls must be selected from the non-white balls, which are the red and black balls. There are $5$ red balls and $7$ black balls, making a total of $5 + 7 = 12$ non-white balls.

The number of ways to select 3 balls from these 12 non-white balls is $\binom{12}{3}$.

The number of ways to select 5 balls with exactly 2 white balls is the product of the number of ways to choose the white balls and the number of ways to choose the remaining non-white balls.

Number of ways for Condition 1 = $\binom{6}{2} \times \binom{12}{3}$

Calculating the combination values:

$\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6!}{2!4!} = \frac{6 \times 5}{2 \times 1} = 15$

$\binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12!}{3!9!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 2 \times 11 \times 10 = 220$

Number of ways for Condition 1 = $15 \times 220 = 3300$.

Condition 2: At least 4 black balls.

This means the selection of 5 balls must contain either exactly 4 black balls or exactly 5 black balls.

Case 2a: Exactly 4 black balls.

If we select exactly 4 black balls from the 7 available black balls, the number of ways is $\binom{7}{4}$.

The remaining $5 - 4 = 1$ ball must be selected from the non-black balls, which are the red and white balls. There are $5$ red balls and $6$ white balls, making a total of $5 + 6 = 11$ non-black balls.

The number of ways to select 1 ball from these 11 non-black balls is $\binom{11}{1}$.

Number of ways for Case 2a = $\binom{7}{4} \times \binom{11}{1}$

Calculating the combination values:

$\binom{7}{4} = \frac{7!}{4!(7-4)!} = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$

$\binom{11}{1} = \frac{11!}{1!(11-1)!} = \frac{11!}{1!10!} = 11$

Number of ways for Case 2a = $35 \times 11 = 385$.

Case 2b: Exactly 5 black balls.

If we select exactly 5 black balls from the 7 available black balls, the number of ways is $\binom{7}{5}$.

The remaining $5 - 5 = 0$ balls must be selected from the non-black balls (11 total).

The number of ways to select 0 balls from these 11 non-black balls is $\binom{11}{0}$.

Number of ways for Case 2b = $\binom{7}{5} \times \binom{11}{0}$

Calculating the combination values:

$\binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7!}{5!2!} = \frac{7 \times 6}{2 \times 1} = 21$

$\binom{11}{0} = \frac{11!}{0!(11-0)!} = \frac{11!}{0!11!} = 1$ (since $0! = 1$)

Number of ways for Case 2b = $21 \times 1 = 21$.

Total number of ways for Condition 2 (at least 4 black balls) is the sum of the ways for Case 2a and Case 2b.

Number of ways for Condition 2 = $385 + 21 = 406$.

Checking for Overlap:

We need to find the number of selections that satisfy both Condition 1 (exactly 2 white balls) AND Condition 2 (at least 4 black balls) simultaneously.

Let the number of red, white, and black balls selected be R, W, and B, respectively. We are selecting 5 balls, so $R+W+B = 5$.

Condition 1 requires $W=2$. Substituting this into the sum, we get $R + 2 + B = 5$, which simplifies to $R + B = 3$.

Condition 2 requires $B \geq 4$.

We need to find the number of selections where $W=2$, $R+B=3$, and $B \geq 4$.

If $B=4$, then $R+4=3$, which means $R = -1$. This is not possible as the number of red balls selected cannot be negative.

If $B=5$, then $R+5=3$, which means $R = -2$. Not possible.

For any value of $B \geq 4$, if $R+B=3$, $R$ will be negative. Since $R$ must be non-negative (we can select 0 or more red balls), there are no selections that can simultaneously have exactly 2 white balls and at least 4 black balls.

Therefore, the two conditions are mutually exclusive, and there is no overlap between the sets of selections satisfying Condition 1 and Condition 2.

Total Number of Ways:

Since the conditions are mutually exclusive, the total number of ways to make a selection such that there are exactly 2 white balls OR at least 4 black balls is the sum of the number of ways for Condition 1 and the number of ways for Condition 2.

Total number of ways = (Number of ways for Condition 1) + (Number of ways for Condition 2)

Total number of ways = $3300 + 406 = 3706$.

Thus, there are $3706$ ways to select 5 balls such that there are exactly 2 white balls, or at least 4 black balls.

Given: The word is 'ASSASSINATION'.

To find:

1. The total number of arrangements of the letters in 'ASSASSINATION'.
2. The number of arrangements where the two A's come together.

Solution:

First, let's count the total number of letters in the word 'ASSASSINATION' and the frequency of each letter.

The word 'ASSASSINATION' has 13 letters.

The letters and their frequencies are:

• A: 2
• S: 4
• I: 2
• N: 2
• T: 1
• O: 1

Part 1: Total number of arrangements

The total number of arrangements of $n$ objects where there are $n_1$ identical objects of type 1, $n_2$ identical objects of type 2, ..., $n_k$ identical objects of type k is given by the formula:

$ \frac{n!}{n_1! n_2! \cdots n_k!} $

In this case, $n = 13$, and the frequencies of the repeated letters are:

• A: $n_1 = 2$
• S: $n_2 = 4$
• I: $n_3 = 2$
• N: $n_4 = 2$

The total number of arrangements is:

$ \frac{13!}{2! \cdot 4! \cdot 2! \cdot 2!} $

Calculating the factorials:

• $13! = 6,227,020,800$
• $2! = 2$
• $4! = 4 \times 3 \times 2 \times 1 = 24$

So, the total number of arrangements is:

$ \frac{6,227,020,800}{2 \times 24 \times 2 \times 2} = \frac{6,227,020,800}{192} $

Now, we perform the division:

$\frac{6,227,020,800}{192} = 32,432,400$

Therefore, the total number of arrangements of the letters of the word 'ASSASSINATION' is 32,432,400.

Part 2: Number of arrangements where the two A's come together

To find the number of arrangements where the two A's come together, we treat the two A's as a single unit or block (AA).

Now, we have effectively 12 units to arrange:

• (AA): 1 unit
• S: 4 units
• I: 2 units
• N: 2 units
• T: 1 unit
• O: 1 unit

The total number of these units is $1 + 4 + 2 + 2 + 1 + 1 = 11$.

Wait, let's recheck the units. We have the block (AA) which counts as one item. The remaining letters are S, S, S, S, I, I, N, N, T, O. So, the total number of items to arrange is 1 (for AA) + 4 (for S) + 2 (for I) + 2 (for N) + 1 (for T) + 1 (for O) = 11 items.

The repeated items among these 11 units are:

• S: 4 times
• I: 2 times
• N: 2 times

The number of arrangements where the two A's come together is given by the formula for permutations with repetitions:

$ \frac{11!}{4! \cdot 2! \cdot 2!} $

Calculating the factorials:

• $11! = 39,916,800$
• $4! = 24$
• $2! = 2$

So, the number of arrangements with AA together is:

$ \frac{39,916,800}{24 \times 2 \times 2} = \frac{39,916,800}{96} $

Now, we perform the division:

$\frac{39,916,800}{96} = 415,800$

Therefore, there are 415,800 arrangements where the two A's come together.

Summary:

The total number of arrangements of the letters in 'ASSASSINATION' is 32,432,400.

The number of arrangements where the two A's come together is 415,800.

To form a 5-digit number, we need to determine the number of choices for each of the five positions (ten thousands, thousands, hundreds, tens, and units). The digits available are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

Condition for a 5-digit number: The first digit (ten thousands place) cannot be 0.

Condition for divisibility by 5: A number is divisible by 5 if its units digit is either 0 or 5.

Condition for repetition of digits: Repetition of digits is allowed, meaning each digit can be used multiple times.

Let's consider the number of choices for each digit's place:

1. Units Digit (5th digit):

For the number to be divisible by 5, the units digit must be either 0 or 5. So, there are 2 choices for the units digit.

2. Ten Thousands Digit (1st digit):

The first digit of a 5-digit number cannot be 0. Since repetition is allowed, we can choose any digit from 1 to 9 for the first position. So, there are 9 choices for the ten thousands digit.

3. Thousands Digit (2nd digit):

Since repetition of digits is allowed, any of the 10 digits (0 to 9) can be used for the thousands place. So, there are 10 choices for the thousands digit.

4. Hundreds Digit (3rd digit):

Similarly, since repetition is allowed, any of the 10 digits (0 to 9) can be used for the hundreds place. So, there are 10 choices for the hundreds digit.

5. Tens Digit (4th digit):

Again, since repetition is allowed, any of the 10 digits (0 to 9) can be used for the tens place. So, there are 10 choices for the tens digit.

To find the total number of such 5-digit numbers, we multiply the number of choices for each position:

Total number of 5-digit numbers = (Choices for 1st digit) $\times$ (Choices for 2nd digit) $\times$ (Choices for 3rd digit) $\times$ (Choices for 4th digit) $\times$ (Choices for 5th digit)

Total number of 5-digit numbers = $9 \times 10 \times 10 \times 10 \times 2$

Total number of 5-digit numbers = $9 \times 1000 \times 2$

Total number of 5-digit numbers = $18000$

Therefore, there are 18,000 such 5-digit numbers that can be formed using the digits 0 to 9, with repetition allowed, and are divisible by 5.

The problem requires us to find the number of ways to form a cricket team of 11 players from a set of 15 players, with specific conditions on the composition of the team.

Given:

Total number of players = 15

Number of bowlers = 5

Number of non-bowlers = Total players - Number of bowlers = $15 - 5 = 10$

Size of the cricket team to be chosen = 11

Condition: The team must include exactly 4 bowlers.

To form a team of 11 players with exactly 4 bowlers, we need to select:

1. Exactly 4 bowlers from the 5 available bowlers.

2. The remaining players ( $11 - 4 = 7$ players) from the non-bowlers.

Step 1: Select 4 bowlers from 5 bowlers.

The number of ways to choose 4 bowlers from 5 is given by the combination formula: $\binom{n}{k} = \frac{n!}{k!(n-k)!}$, where $n$ is the total number of items to choose from, and $k$ is the number of items to choose.

Number of ways to choose 4 bowlers from 5 = $\binom{5}{4}$

$\binom{5}{4} = \frac{5!}{4!(5-4)!} = \frac{5!}{4!1!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{(4 \times 3 \times 2 \times 1) \times 1} = 5$

Step 2: Select 7 non-bowlers from 10 non-bowlers.

The number of ways to choose the remaining 7 players from the 10 non-bowlers is given by the combination formula:

Number of ways to choose 7 non-bowlers from 10 = $\binom{10}{7}$

$\binom{10}{7} = \frac{10!}{7!(10-7)!} = \frac{10!}{7!3!} = \frac{10 \times 9 \times 8 \times 7!}{7! \times (3 \times 2 \times 1)} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120$

To Find:

The total number of ways to form the cricket team of 11 players with exactly 4 bowlers.

Solution:

Since the selection of bowlers and non-bowlers are independent events, the total number of ways to form the team is the product of the number of ways to perform each step.

Total number of ways = (Number of ways to choose bowlers) $\times$ (Number of ways to choose non-bowlers)

Total number of ways = $\binom{5}{4} \times \binom{10}{7}$

Total number of ways = $5 \times 120$

Total number of ways = 600

Thus, there are 600 ways to choose a cricket team of 11 players from a set of 15 players (including 5 bowlers) such that the team includes exactly 4 bowlers.

This problem involves finding the number of straight lines and triangles formed by a set of points, with a special condition about collinear points.

Given:

Total number of points in the plane = 10

Number of points that are collinear = 4

No other three points are collinear.

To Find:

1. The number of straight lines that can be formed.

2. The number of triangles that can be formed.

Part 1: Number of Straight Lines

To form a straight line, we need to choose any 2 points. The general formula for the number of lines formed by $n$ non-collinear points is $\binom{n}{2}$.

If all 10 points were non-collinear, the total number of lines would be:

Total lines (if all points were non-collinear) = $\binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10!}{2!8!} = \frac{10 \times 9}{2 \times 1} = 45$

However, we have 4 points that are collinear. These 4 collinear points, if considered non-collinear, would form $\binom{4}{2}$ lines. But since they are collinear, they form only 1 straight line.

Number of lines formed by the 4 collinear points = $\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6$

The difference between the lines formed by these 4 points if they were not collinear (6 lines) and the single line they actually form (1 line) needs to be accounted for. We subtract the extra lines that were counted and add back the single line that these collinear points form.

Number of straight lines = (Total lines if all points were non-collinear) - (Lines formed by collinear points as if they were non-collinear) + (Actual line formed by collinear points)

Number of straight lines = $\binom{10}{2} - \binom{4}{2} + 1$

Number of straight lines = $45 - 6 + 1 = 40$

Part 2: Number of Triangles

To form a triangle, we need to choose any 3 points. The general formula for the number of triangles formed by $n$ non-collinear points is $\binom{n}{3}$.

If all 10 points were non-collinear, the total number of triangles would be:

Total triangles (if all points were non-collinear) = $\binom{10}{3} = \frac{10!}{3!(10-3)!} = \frac{10!}{3!7!} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120$

However, the 4 collinear points cannot form a triangle. If we choose any 3 points from these 4 collinear points, they will lie on a straight line and thus will not form a triangle. So, we must subtract the number of combinations of 3 points chosen from the 4 collinear points.

Number of combinations of 3 points from the 4 collinear points = $\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = 4$

Number of triangles = (Total triangles if all points were non-collinear) - (Triangles formed by choosing 3 points from the 4 collinear points)

Number of triangles = $\binom{10}{3} - \binom{4}{3}$

Number of triangles = $120 - 4 = 116$

Conclusion:

The number of straight lines that can be formed is 40.

The number of triangles that can be formed is 116.

This question asks for the number of arrangements of the letters in the word 'COMMITTEE' under different conditions.

1. Total number of arrangements of the letters in 'COMMITTEE'

First, let's identify the letters and their frequencies in the word 'COMMITTEE':

C - 1

O - 1

M - 2

I - 1

T - 2

E - 2

Total number of letters = 9

There are repeated letters: M (2 times), T (2 times), E (2 times).

The formula for the number of permutations of $n$ objects where there are $n_1$ identical objects of type 1, $n_2$ identical objects of type 2, ..., $n_k$ identical objects of type k is:

$\text{Total Arrangements} = \frac{n!}{n_1! n_2! \dots n_k!}$

In this case, $n=9$, $n_M=2$, $n_T=2$, $n_E=2$.

Total number of arrangements = $\frac{9!}{2!2!2!}$

Total number of arrangements = $\frac{362880}{(2 \times 1) \times (2 \times 1) \times (2 \times 1)} = \frac{362880}{8}$

Total number of arrangements = 45360

2. Number of arrangements with all vowels together

The vowels in 'COMMITTEE' are O, I, E, E. There are 4 vowels.

The consonants are C, M, M, T, T, E. There are 6 consonants.

Let's re-check the letters: C(1), O(1), M(2), I(1), T(2), E(2). Total 9 letters.

Vowels: O, I, E, E (4 vowels)

Consonants: C, M, M, T, T (5 consonants)

Let's correct the breakdown: C-1, O-1, M-2, I-1, T-2, E-2. Total 9 letters.

Vowels: O, I, E, E (4 vowels). Frequencies: I(1), O(1), E(2).

Consonants: C, M, M, T, T (5 consonants). Frequencies: C(1), M(2), T(2).

To arrange the letters such that all vowels are together, we treat the group of vowels as a single unit.

The group of vowels: (O I E E)

The consonants are: C, M, M, T, T

Now we have these 5 consonants and the single block of vowels to arrange. This gives us a total of $5 + 1 = 6$ units to arrange.

Arrangements of the 6 units (5 consonants + 1 vowel block):

Among the consonants, M is repeated 2 times, and T is repeated 2 times. The vowel block is treated as a single unit.

Number of ways to arrange these 6 units = $\frac{6!}{2!2!}$

$\frac{6!}{2!2!} = \frac{720}{2 \times 2} = \frac{720}{4} = 180$

Now, we need to consider the arrangements within the vowel group (O I E E).

There are 4 vowels: O, I, E, E. The letter E is repeated 2 times.

Number of ways to arrange the vowels within their group = $\frac{4!}{2!}$

$\frac{4!}{2!} = \frac{24}{2} = 12$

To get the total number of arrangements with all vowels together, we multiply the number of ways to arrange the units by the number of ways to arrange the vowels within their group.

Arrangements with vowels together = (Arrangements of units) $\times$ (Arrangements within vowel group)

Arrangements with vowels together = $180 \times 12 = 2160$

3. Number of arrangements with all consonants together

The consonants are C, M, M, T, T (5 consonants). Frequencies: C(1), M(2), T(2).

The vowels are O, I, E, E (4 vowels). Frequencies: O(1), I(1), E(2).

To arrange the letters such that all consonants are together, we treat the group of consonants as a single unit.

The group of consonants: (C M M T T)

The vowels are: O, I, E, E

Now we have these 4 vowels and the single block of consonants to arrange. This gives us a total of $4 + 1 = 5$ units to arrange.

Arrangements of the 5 units (4 vowels + 1 consonant block):

Among the vowels, E is repeated 2 times. The consonant block is treated as a single unit.

Number of ways to arrange these 5 units = $\frac{5!}{2!}$

$\frac{5!}{2!} = \frac{120}{2} = 60$

Now, we need to consider the arrangements within the consonant group (C M M T T).

There are 5 consonants: C, M, M, T, T. The letter M is repeated 2 times, and T is repeated 2 times.

Number of ways to arrange the consonants within their group = $\frac{5!}{2!2!}$

$\frac{5!}{2!2!} = \frac{120}{(2 \times 1) \times (2 \times 1)} = \frac{120}{4} = 30$

To get the total number of arrangements with all consonants together, we multiply the number of ways to arrange the units by the number of ways to arrange the consonants within their group.

Arrangements with consonants together = (Arrangements of units) $\times$ (Arrangements within consonant group)

Arrangements with consonants together = $60 \times 30 = 1800$

Summary:

The total number of ways to arrange the letters of the word 'COMMITTEE' is 45360.

The number of these arrangements where all the vowels are together is 2160.

The number of these arrangements where all the consonants are together is 1800.

This problem involves combinations, as the order in which the cards are chosen does not matter.

Given:

Total number of cards in a standard deck = 52

Number of Aces in a deck = 4

Number of non-Ace cards in a deck = $52 - 4 = 48$

Number of cards to be chosen = 5

Condition: The selection must contain exactly 3 Aces.

To Find:

The number of ways to choose 5 cards from a deck of 52 such that exactly 3 Aces are selected.

To satisfy the condition of exactly 3 Aces in a 5-card selection, we need to:

1. Choose exactly 3 Aces from the 4 available Aces.

2. Choose the remaining $5 - 3 = 2$ cards from the non-Ace cards.

Step 1: Choose 3 Aces from 4 Aces.

The number of ways to choose 3 Aces from 4 is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$:

Number of ways to choose 3 Aces from 4 = $\binom{4}{3}$

$\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = \frac{4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1) \times 1} = 4$

Step 2: Choose 2 non-Ace cards from 48 non-Ace cards.

The number of ways to choose 2 cards from the remaining 48 non-Ace cards is:

Number of ways to choose 2 non-Ace cards from 48 = $\binom{48}{2}$

$\binom{48}{2} = \frac{48!}{2!(48-2)!} = \frac{48!}{2!46!} = \frac{48 \times 47}{2 \times 1} = 24 \times 47$

Calculation of $24 \times 47$:

47

x 24

-----

188 ($47 \times 4$)

940 ($47 \times 20$)

-----

1128

So, $\binom{48}{2} = 1128$

Solution:

To find the total number of ways to choose the 5-card hand with exactly 3 Aces, we multiply the number of ways to perform each step (by the multiplication principle).

Total number of ways = (Ways to choose 3 Aces) $\times$ (Ways to choose 2 non-Ace cards)

Total number of ways = $\binom{4}{3} \times \binom{48}{2}$

Total number of ways = $4 \times 1128$

Calculation of $4 \times 1128$:

1128

x 4

-----

4512

Total number of ways = 4512

Therefore, there are 4512 ways to choose 5 cards from a deck of 52 playing cards such that the selection contains exactly 3 Aces.

This problem is about arrangements (permutations) with a restriction that no two women sit together.

Given:

Number of men = 7

Number of women = 4

Total number of chairs = 11

Condition: No two women sit together.

To Find:

The number of ways to seat 7 men and 4 women in a row of 11 chairs such that no two women sit together.

To ensure that no two women sit together, we first seat the men. This creates spaces between the men where the women can be seated.

Step 1: Arrange the 7 men.

There are 7 men to be seated in 7 specific chairs (out of the 11 chairs). Since the chairs are distinct and the men are distinct, the number of ways to arrange the 7 men is $P(7, 7)$ or $7!$.

Number of ways to arrange the 7 men = $7!$

$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$

When the 7 men are seated, they create spaces where the women can sit. Let's represent the men by 'M' and the spaces by '_':

_ M _ M _ M _ M _ M _ M _ M _

There are 7 men, creating $7 + 1 = 8$ possible spaces where the women can be seated (before the first man, between any two men, and after the last man).

Step 2: Place the 4 women in the available spaces.

We need to choose 4 of these 8 spaces for the 4 women. Since the women are distinct, the order in which they are placed in these spaces matters. This is a permutation problem.

We need to select 4 spaces from 8 and arrange the 4 women in these selected spaces. The number of ways to do this is $P(8, 4)$.

Number of ways to seat 4 women in 8 spaces = $P(8, 4) = \frac{8!}{(8-4)!} = \frac{8!}{4!}$

$P(8, 4) = 8 \times 7 \times 6 \times 5$

$8 \times 7 = 56$

$56 \times 6 = 336$

$336 \times 5 = 1680$

Solution:

To find the total number of ways to seat the 7 men and 4 women such that no two women sit together, we multiply the number of ways to arrange the men by the number of ways to arrange the women in the spaces created by the men.

Total number of ways = (Ways to arrange men) $\times$ (Ways to seat women)

Total number of ways = $7! \times P(8, 4)$

Total number of ways = $5040 \times 1680$

Calculation of $5040 \times 1680$:

5040

x 1680

------

0000 (5040 x 0)

403200 (5040 x 80)

3024000 (5040 x 600)

5040000 (5040 x 1000)

--------

8467200

Total number of ways = 8,467,200

Therefore, there are 8,467,200 ways to seat 7 men and 4 women in a row of 11 chairs so that no two women sit together.

This question involves solving a combinatorial equation and then evaluating a binomial coefficient.

Part 1: Solve for $n$ in the equation $\binom{n}{4} = \binom{n}{3}$

We know the property of binomial coefficients that if $\binom{n}{r} = \binom{n}{k}$, then either $r=k$ or $r+k=n$.

In this case, we have $r=4$ and $k=3$. Clearly, $r \neq k$ (since $4 \neq 3$).

Therefore, we must use the second condition: $r+k=n$.

$4 + 3 = n$

$n = 7$

Let's verify this using the formula for binomial coefficients: $\binom{n}{k} = \frac{n!}{k!(n-k)!}$

$\binom{n}{4} = \frac{n!}{4!(n-4)!}$

$\binom{n}{3} = \frac{n!}{3!(n-3)!}$

Setting them equal:

$\frac{n!}{4!(n-4)!} = \frac{n!}{3!(n-3)!}$

Since $n!$ is on both sides (and for the combinations to be defined, $n$ must be at least 4), we can cancel $n!$:

$\frac{1}{4!(n-4)!} = \frac{1}{3!(n-3)!}$

Cross-multiply:

$3!(n-3)! = 4!(n-4)!$

We know that $4! = 4 \times 3!$ and $(n-3)! = (n-3) \times (n-4)!$. Substitute these into the equation:

$3! \times (n-3) \times (n-4)! = (4 \times 3!) \times (n-4)!$

Now, we can cancel $3!$ and $(n-4)!$ from both sides (assuming $n>4$ for $(n-4)!$ to be defined, which is consistent with $\binom{n}{4}$):

$n-3 = 4$

Solving for $n$:

$n = 4 + 3$

$n = 7$

Part 2: Find the value of $\binom{n}{n-1}$

We found that $n=7$. Now we need to find the value of $\binom{n}{n-1}$, which is $\binom{7}{7-1} = \binom{7}{6}$.

Using the property $\binom{n}{k} = \binom{n}{n-k}$, we have:

$\binom{7}{6} = \binom{7}{7-6} = \binom{7}{1}$

The value of $\binom{n}{1}$ is always $n$. Therefore:

$\binom{7}{1} = 7$

Alternatively, using the formula directly:

$\binom{7}{6} = \frac{7!}{6!(7-6)!} = \frac{7!}{6!1!} = \frac{7 \times 6!}{6! \times 1} = 7$

Conclusion:

The value of $n$ is 7.

The value of $\binom{n}{n-1}$ is 7.

This problem involves combinations, as the order of selection of team members does not matter. We need to calculate the number of ways to form a team of 5 members with different conditions on the number of girls.

Given:

Number of girls = 4

Number of boys = 7

Total number of people in the group = $4 + 7 = 11$

Size of the team to be selected = 5

i) Number of ways to select a team with no girl:

If the team has no girl, it means all 5 members must be boys.

We need to select 5 boys from the 7 available boys.

Number of ways = $\binom{7}{5}$

$\binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7!}{5!2!} = \frac{7 \times 6}{2 \times 1} = 21$

ii) Number of ways to select a team with at least one girl:

The condition "at least one girl" means the team can have 1 girl, 2 girls, 3 girls, or 4 girls (since there are only 4 girls available).

We can solve this in two ways:

Method 1: Sum of cases**