Chapter 7 Permutations And Combinations

The given word is ROSE.

The letters in the word ROSE are R, O, S, E. There are 4 distinct letters.

We need to form 4-letter words using these letters.

The repetition of letters is not allowed.

Since the order of the letters matters and repetition is not allowed, this is a problem of permutations.

We need to find the number of permutations of 4 distinct letters taken all at a time ($n=4$, $r=4$).

The number of permutations of $n$ distinct objects taken $r$ at a time is given by the formula: $P(n, r) = \frac{n!}{(n-r)!}$

In this case, $n=4$ (number of distinct letters) and $r=4$ (number of letters in the word to be formed).

Number of 4-letter words = $P(4, 4)$

$P(4, 4) = \frac{4!}{(4-4)!} = \frac{4!}{0!}$

We know that $0! = 1$.

So, $P(4, 4) = \frac{4!}{1} = 4!$

Now, calculate the factorial of 4:

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

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

$4! = 24 \times 1$

$4! = 24$

Therefore, the number of 4-letter words that can be formed from the letters of the word ROSE without repetition is 24.

We have 4 flags of different colours.

A signal is generated by using 2 flags, placed one below the other.

This problem involves selecting 2 flags out of 4 and arranging them in a specific order (one below the other). The order matters (the top flag is different from the bottom flag).

This is a problem of permutations of $n$ distinct objects taken $r$ at a time.

Here, $n$ is the total number of distinct flags, so $n=4$.

$r$ is the number of flags used in a signal, so $r=2$.

The number of permutations of $n$ distinct objects taken $r$ at a time is given by the formula: $P(n, r) = \frac{n!}{(n-r)!}$

We need to calculate $P(4, 2)$:

Now, calculate the factorials:

Substitute these values back into the formula:

Alternatively, we can think of this using the Fundamental Principle of Counting.

For the top position, there are 4 choices of flags.

After choosing a flag for the top position, there are 3 remaining flags for the bottom position.

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

Total number of signals = $4 \times 3 = 12$

Therefore, the number of different signals that can be generated is 12.

We need to form 2-digit even numbers using the digits 1, 2, 3, 4, 5.

A 2-digit number has two places: the tens place and the units place.

For the number to be even, the digit in the units place must be an even digit from the given set of digits.

The even digits in the set {1, 2, 3, 4, 5} are 2 and 4.

Let the 2-digit number be represented as $\text{_}\text{_}$, where the first blank is the tens place and the second blank is the units place.

Consider the units place:

The digit in the units place must be even. The available even digits are 2 and 4.

Number of choices for the units place = 2 (either 2 or 4).

Consider the tens place:

The digits can be repeated, so any of the given digits {1, 2, 3, 4, 5} can be used for the tens place.

Number of choices for the tens place = 5 (1, 2, 3, 4, or 5).

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

Total number of even numbers = (Choices for tens place) $\times$ (Choices for units place)

Total number of even numbers = $5 \times 2 = 10$

The 10 possible numbers are: 12, 14, 22, 24, 32, 34, 42, 44, 52, 54.

Therefore, the number of 2-digit even numbers that can be formed from the digits 1, 2, 3, 4, 5 with repetition allowed is 10.

Given:

Number of different flags available = 5.

A signal is to be generated by arranging at least 2 flags in order on a vertical staff.

To Find:

The total number of different signals that can be generated.

Solution:

The problem requires arranging flags "in order," which means the sequence of flags matters. This indicates that we should use the concept of permutations.

The condition "at least 2 flags" means we need to consider the following cases:

1. Signals made with exactly 2 flags.

2. Signals made with exactly 3 flags.

3. Signals made with exactly 4 flags.

4. Signals made with exactly 5 flags.

The total number of signals will be the sum of the number of signals from each of these cases.

The formula for the number of permutations of arranging $r$ items from a set of $n$ distinct items is given by:

Here, $n=5$ (the total number of available flags).

Case 1: Signals using 2 flags

We need to find the number of arrangements of 5 flags taken 2 at a time. Here, $n=5$ and $r=2$.

Number of signals = $^5P_2 = \frac{5!}{(5-2)!} = \frac{5!}{3!} = \frac{5 \times 4 \times 3!}{3!} = 5 \times 4 = 20$.

Case 2: Signals using 3 flags

We need to find the number of arrangements of 5 flags taken 3 at a time. Here, $n=5$ and $r=3$.

Number of signals = $^5P_3 = \frac{5!}{(5-3)!} = \frac{5!}{2!} = \frac{5 \times 4 \times 3 \times 2!}{2!} = 5 \times 4 \times 3 = 60$.

Case 3: Signals using 4 flags

We need to find the number of arrangements of 5 flags taken 4 at a time. Here, $n=5$ and $r=4$.

Number of signals = $^5P_4 = \frac{5!}{(5-4)!} = \frac{5!}{1!} = 5 \times 4 \times 3 \times 2 = 120$.

Case 4: Signals using 5 flags

We need to find the number of arrangements of 5 flags taken 5 at a time. Here, $n=5$ and $r=5$.

Number of signals = $^5P_5 = \frac{5!}{(5-5)!} = \frac{5!}{0!} = 5 \times 4 \times 3 \times 2 \times 1 = 120$. (Since $0! = 1$)

Total Number of Signals

To find the total number of signals, we add the number of signals from all the cases.

Total signals = (Signals with 2 flags) + (Signals with 3 flags) + (Signals with 4 flags) + (Signals with 5 flags)

Total signals = $20 + 60 + 120 + 120 = 320$.

Hence, the total number of different signals that can be generated is 320.

Alternate Solution (Using the Multiplication Principle)

We can solve this problem by considering the number of choices for each position on the staff for each case.

Case 1: Signal with 2 flags

The first (top) position can be filled in 5 ways.

The second position can be filled in 4 ways (since one flag is already used).

Number of signals = $5 \times 4 = 20$.

Case 2: Signal with 3 flags

Number of choices for the three positions are 5, 4, and 3 respectively.

Number of signals = $5 \times 4 \times 3 = 60$.

Case 3: Signal with 4 flags

Number of choices for the four positions are 5, 4, 3, and 2 respectively.

Number of signals = $5 \times 4 \times 3 \times 2 = 120$.

Case 4: Signal with 5 flags

Number of choices for the five positions are 5, 4, 3, 2, and 1 respectively.

Number of signals = $5 \times 4 \times 3 \times 2 \times 1 = 120$.

Total Number of Signals

The total number of signals is the sum of signals from all possible cases.

Total number of signals = $20 + 60 + 120 + 120 = 320$.

Given:

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

Total number of available digits = 5.

We need to form 3-digit numbers.

(i) Repetition of the digits is allowed

A 3-digit number has three places to fill: hundreds, tens, and units.

Hundreds place: We can use any of the 5 available digits. So, there are 5 choices.

Tens place: Since repetition is allowed, we can again use any of the 5 available digits. So, there are 5 choices.

Units place: Similarly, since repetition is allowed, we can use any of the 5 available digits. So, there are 5 choices.

By the Fundamental Principle of Multiplication, the total number of 3-digit numbers is the product of the number of choices for each place.

Total number of 3-digit numbers = (Choices for hundreds place) $\times$ (Choices for tens place) $\times$ (Choices for units place)

Total number of 3-digit numbers = $5 \times 5 \times 5 = 125$.

Therefore, 125 three-digit numbers can be formed when repetition is allowed.

(ii) Repetition of the digits is not allowed

A 3-digit number has three places to fill: hundreds, tens, and units.

Hundreds place: We can use any of the 5 available digits. So, there are 5 choices.

Tens place: Since repetition is not allowed, one digit has been used for the hundreds place. This leaves us with $5-1=4$ digits. So, there are 4 choices.

Units place: Two distinct digits have now been used for the hundreds and tens places. This leaves us with $5-2=3$ digits. So, there are 3 choices.

By the Fundamental Principle of Multiplication, the total number of 3-digit numbers is the product of the number of choices for each place.

Total number of 3-digit numbers = $5 \times 4 \times 3 = 60$.

Alternate Method (Using Permutations):

This is equivalent to finding the number of arrangements (permutations) of 5 distinct digits taken 3 at a time. The formula for permutations is $^nP_r = \frac{n!}{(n-r)!}$.

Here, $n=5$ and $r=3$.

Number of 3-digit numbers = $^5P_3 = \frac{5!}{(5-3)!} = \frac{5!}{2!} \ $$ = \frac{5 \times 4 \times 3 \times 2!}{2!} \ $$ = 5 \times 4 \times 3 = 60$.

Therefore, 60 three-digit numbers can be formed when repetition is not allowed.

Given:

The set of available digits is {1, 2, 3, 4, 5, 6}.

Total number of available digits = 6.

We need to form 3-digit even numbers, and repetition of digits is allowed.

To Find:

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

Solution:

A 3-digit number has three places to fill: hundreds, tens, and units.

For a number to be even, its units digit must be an even number.

From the given set of digits {1, 2, 3, 4, 5, 6}, the even digits are {2, 4, 6}.

We will fill the places based on the constraints, starting with the units place.

Units place: The digit must be even. So, the choices are 2, 4, or 6. There are 3 choices for the units place.

Tens place: Since repetition is allowed, any of the 6 given digits can be used. There are 6 choices for the tens place.

Hundreds place: Since repetition is allowed, any of the 6 given digits can be used. There are 6 choices for the hundreds place.

By the Fundamental Principle of Multiplication, the total number of 3-digit even numbers is the product of the number of choices for each place.

Total number of 3-digit even numbers = (Choices for hundreds place) $\times$ (Choices for tens place) $\times$ (Choices for units place)

Total number of 3-digit even numbers = $6 \times 6 \times 3 = 108$.

Therefore, 108 three-digit even numbers can be formed.

Given:

The first 10 letters of the English alphabet are available: {A, B, C, D, E, F, G, H, I, J}.

Total number of available letters = 10.

We need to form a 4-letter code, and no letter can be repeated.

To Find:

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

Solution:

Since the order of letters in a code matters and repetition is not allowed, this is a permutation problem.

We are arranging 4 letters out of 10 available distinct letters.

Method 1: Using the Fundamental Principle of Multiplication

A 4-letter code has four positions to fill.

First position: There are 10 choices.

Second position: Since one letter is used and not repeated, there are $10-1=9$ choices.

Third position: Two letters are used, leaving $10-2=8$ choices.

Fourth position: Three letters are used, leaving $10-3=7$ choices.

Total number of codes = $10 \times 9 \times 8 \times 7 = 5040$.

Method 2: Using the Permutation Formula

The number of permutations of $n$ distinct items taken $r$ at a time is given by $^nP_r = \frac{n!}{(n-r)!}$.

Here, $n=10$ (total letters) and $r=4$ (letters in the code).

Number of codes = $^{10}P_4 = \frac{10!}{(10-4)!} = \frac{10!}{6!}$

$^{10}P_4 = \frac{10 \times 9 \times 8 \times 7 \times 6!}{6!} = 10 \times 9 \times 8 \times 7 = 5040$.

Therefore, 5040 different 4-letter codes can be formed.

Given:

Available digits are {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Total of 10 digits.

A 5-digit telephone number is to be constructed.

Condition 1: The number must start with the digits 67.

Condition 2: No digit appears more than once (no repetition).

To Find:

The total number of such 5-digit telephone numbers.

Solution:

A 5-digit telephone number has 5 places to fill.

_ _ _ _ _

The first two places are fixed with the digits 6 and 7.

6 7 _ _ _

So, the first two digits are already chosen. The digits 6 and 7 are now used and cannot be repeated.

Number of available digits remaining = Total digits - Digits used = $10 - 2 = 8$.

The remaining available digits are {0, 1, 2, 3, 4, 5, 8, 9}.

We need to fill the remaining 3 places of the telephone number using these 8 available digits, without repetition.

Third place: We have 8 choices from the remaining digits.

Fourth place: After filling the third place, we have $8-1=7$ choices remaining.

Fifth place: After filling the third and fourth places, we have $8-2=6$ choices remaining.

By the Fundamental Principle of Multiplication, the number of ways to fill the last three places is:

Number of ways = $8 \times 7 \times 6 = 336$.

Since the first two places are fixed, the total number of telephone numbers is 336.

Alternate Method (Using Permutations):

The first two positions are fixed. The remaining 3 positions must be filled from the remaining 8 digits without repetition. This is an arrangement of 8 items taken 3 at a time.

Number of ways = $^8P_3 = \frac{8!}{(8-3)!} = \frac{8!}{5!} = \frac{8 \times 7 \times 6 \times 5!}{5!} = 8 \times 7 \times 6 = 336$.

Therefore, 336 such 5-digit telephone numbers can be constructed.

Given:

A coin is tossed 3 times.

To Find:

The total number of possible outcomes.

Solution:

When a single coin is tossed, there are two possible outcomes: a Head (H) or a Tail (T).

We are tossing the coin 3 times. Each toss is an independent event.

We can use the Fundamental Principle of Multiplication to find the total number of outcomes.

First toss: There are 2 possible outcomes (H or T).

Second toss: There are 2 possible outcomes (H or T).

Third toss: There are 2 possible outcomes (H or T).

Total number of possible outcomes = (Outcomes of 1st toss) $\times$ (Outcomes of 2nd toss) $\times$ (Outcomes of 3rd toss)

Total number of possible outcomes = $2 \times 2 \times 2 = 2^3 = 8$.

The 8 possible outcomes can be listed as follows:

• HHH
• HHT
• HTH
• THH
• HTT
• THT
• TTH
• TTT

Therefore, there are 8 possible outcomes.

Given:

Number of available flags of different colours = 5.

Each signal requires the use of 2 flags, arranged one below the other.

To Find:

The total number of different signals that can be generated.

Solution:

Since the flags are arranged one below the other, the order in which the flags are chosen matters. For example, a red flag above a blue flag is a different signal from a blue flag above a red flag. As the flags are of different colours, they are distinct, and a flag cannot be used in both positions simultaneously (no repetition).

This is a problem of arranging 2 items from a set of 5 distinct items, which is a permutation.

Method 1: Using the Fundamental Principle of Multiplication

A signal has two positions to fill:

Top position: We have 5 choices for the flag in the top position.

Bottom position: After choosing a flag for the top position, we are left with $5-1=4$ flags. So, there are 4 choices for the bottom position.

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

Total number of signals = $5 \times 4 = 20$.

Method 2: Using the Permutation Formula

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

Here, $n=5$ (total flags) and $r=2$ (flags per signal).

Number of signals = $^5P_2 = \frac{5!}{(5-2)!} = \frac{5!}{3!} = \frac{5 \times 4 \times 3!}{3!} = 5 \times 4 = 20$.

Therefore, 20 different signals can be generated.

To Evaluate:

(i) $5!$

(ii) $7!$

(iii) $7! - 5!$

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

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

(i) 5!

We calculate the product of integers from 1 to 5.

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

$5! = 20 \times 6 = 120$.

Therefore, $5! = 120$.

(ii) 7!

We calculate the product of integers from 1 to 7.

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

We can also write $7!$ as $7 \times 6 \times 5!$. Since we already know $5! = 120$ from part (i):

$7! = 7 \times 6 \times 120$

$7! = 42 \times 120 = 5040$.

Therefore, $7! = 5040$.

(iii) 7! – 5!

Using the values we calculated above:

$7! - 5! = 5040 - 120 = 4920$.

Alternate Method:

We can factor out the smaller factorial, $5!$.

$7! - 5! = (7 \times 6 \times 5!) - (1 \times 5!)$

$7! - 5! = 5! \times (7 \times 6 - 1)$

$7! - 5! = 5! \times (42 - 1)$

$7! - 5! = 5! \times 41$

Since $5! = 120$, we have:

$7! - 5! = 120 \times 41 = 4920$.

Therefore, $7! - 5! = 4920$.

To Compute:

(i) $\frac{7!}{5!}$

(ii) $\frac{12!}{(10!)(2!)}$

(i) $\frac{7!}{5!}$

We can expand the larger factorial, $7!$, until we get the smaller factorial, $5!$.

$7! = 7 \times 6 \times 5!$

Now, substitute this into the expression:

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

Cancel out the common term $5!$ from the numerator and denominator:

$\frac{7 \times 6 \times \cancel{5!}}{\cancel{5!}} = 7 \times 6 = 42$.

Therefore, $\frac{7!}{5!} = 42$.

(ii) $\frac{12!}{(10!)(2!)}$

We expand the largest factorial, $12!$, until we get the next largest factorial, $10!$.

$12! = 12 \times 11 \times 10!$

We also compute $2! = 2 \times 1 = 2$.

Now, substitute these into the expression:

$\frac{12!}{(10!)(2!)} = \frac{12 \times 11 \times 10!}{(10!)(2)}$

Cancel out the common term $10!$:

$\frac{12 \times 11 \times \cancel{10!}}{(\cancel{10!})(2)} = \frac{12 \times 11}{2}$

Simplify the expression:

$\frac{132}{2} = 66$.

Therefore, $\frac{12!}{(10!)(2!)} = 66$.

Given:

The expression $\frac{n!}{r!(n-r)!}$ with $n=5$ and $r=2$.

To Evaluate:

The value of the given expression.

Solution:

First, substitute the values of $n=5$ and $r=2$ into the expression:

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

Simplify the term inside the parenthesis:

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

Now, we expand the largest factorial, $5!$, until we reach the next largest factorial in the denominator, which is $3!$.

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

Substitute this back into the expression:

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

Cancel out the common term $3!$:

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

We know that $2! = 2 \times 1 = 2$.

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

This expression is also the formula for combinations, $^nC_r$, which represents the number of ways to choose $r$ items from a set of $n$ items without regard to order. In this case, $^5C_2 = 10$.

Therefore, the value of the expression is 10.

Given:

The equation $\frac{1}{8!} + \frac{1}{9!} = \frac{x}{10!}$.

To Find:

The value of $x$.

Solution:

The main idea is to make the denominators on the left-hand side equal. The smallest factorial is $8!$. We can express $9!$ in terms of $8!$.

$9! = 9 \times 8!$

Now, substitute this into the left-hand side (LHS) of the equation:

LHS = $\frac{1}{8!} + \frac{1}{9 \times 8!}$

To add these fractions, we find a common denominator, which is $9 \times 8!$ (or $9!$).

LHS = $\frac{1}{8!} \times \frac{9}{9} + \frac{1}{9 \times 8!}$

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

Now, our original equation becomes:

To solve for $x$, we can multiply both sides by $10!$:

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

We know that $10! = 10 \times 9!$. Substitute this into the equation:

$x = \frac{10}{9!} \times (10 \times 9!)$

Cancel out the common term $9!$:

$x = 10 \times 10 = 100$.

Alternate Method:

Start with the original equation:

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

Multiply the entire equation by the largest factorial, which is $10!$:

$10! \left(\frac{1}{8!} + \frac{1}{9!}\right) = 10! \left(\frac{x}{10!}\right)$

$\frac{10!}{8!} + \frac{10!}{9!} = x$

Now, simplify each term:

$\frac{10 \times 9 \times 8!}{8!} + \frac{10 \times 9!}{9!} = x$

$(10 \times 9) + 10 = x$

$90 + 10 = x$

$100 = x$

Therefore, the value of $x = 100$.

To Evaluate:

(i) $8!$

(ii) $4! - 3!$

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

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

(i) 8!

We calculate the product of integers from 1 to 8.

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

$8! = 56 \times 6 \times 120$

$8! = 336 \times 120 = 40320$.

Therefore, $8! = 40320$.

(ii) 4! – 3!

First, we evaluate each factorial separately.

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

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

Now, we perform the subtraction:

$4! - 3! = 24 - 6 = 18$.

Alternate Method:

We can factor out the smaller factorial, $3!$.

$4! - 3! = (4 \times 3!) - (1 \times 3!) = 3! \times (4 - 1) = 3! \times 3 = 6 \times 3 = 18$.

Therefore, $4! - 3! = 18$.

To Determine:

Whether the equation $3! + 4! = 7!$ is true or false.

Solution:

We will evaluate the Left-Hand Side (LHS) and the Right-Hand Side (RHS) of the equation separately and compare them.

Left-Hand Side (LHS):

LHS = $3! + 4!$

First, calculate the values of the factorials:

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

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

Now, add them:

LHS = $6 + 24 = 30$.

Right-Hand Side (RHS):

RHS = $7!$

Calculate the value of $7!$:

$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$.

Comparison:

LHS = 30

RHS = 5040

Since $30 \neq 5040$, the statement $3! + 4! = 7!$ is false.

Therefore, the answer is No.

To Compute:

The value of the expression $\frac{8!}{6! \times 2!}$.

Solution:

We can simplify the expression by expanding the largest factorial, $8!$, until we get the next largest factorial in the denominator, which is $6!$.

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

We also compute the value of $2!$:

$2! = 2 \times 1 = 2$.

Now, substitute these into the original expression:

$\frac{8!}{6! \times 2!} = \frac{8 \times 7 \times 6!}{6! \times 2}$

Cancel out the common term $6!$ from the numerator and denominator:

$\frac{8 \times 7}{2}$

Simplify the expression:

$\frac{56}{2} = 28$.

This expression is also the formula for combinations, $^8C_2$, which has the same value.

Therefore, the value of the expression is 28.

Given:

The equation $\frac{1}{6!} + \frac{1}{7!} = \frac{x}{8!}$.

To Find:

The value of $x$.

Solution:

We will simplify the left-hand side (LHS) of the equation by finding a common denominator.

LHS = $\frac{1}{6!} + \frac{1}{7!}$

We know that $7! = 7 \times 6!$. The common denominator is $7!$.

LHS = $\frac{1}{6!} \times \frac{7}{7} + \frac{1}{7!} = \frac{7}{7!} + \frac{1}{7!} = \frac{7+1}{7!} = \frac{8}{7!}$

Now the equation is:

To solve for $x$, multiply both sides by $8!$:

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

Since $8! = 8 \times 7!$, we have:

$x = \frac{8}{7!} \times (8 \times 7!)$

Cancel out the common term $7!$:

$x = 8 \times 8 = 64$.

Alternate Method:

Multiply the entire equation by the largest factorial, $8!$:

$8! \left(\frac{1}{6!} + \frac{1}{7!}\right) = 8! \left(\frac{x}{8!}\right)$

$\frac{8!}{6!} + \frac{8!}{7!} = x$

Now, simplify the terms on the LHS:

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

$(8 \times 7) + 8 = x$

$56 + 8 = x$

$64 = x$

Therefore, the value of $x = 64$.

The expression $\frac{n!}{(n-r)!}$ is the formula for permutations, denoted as $^nP_r$.

(i) when n = 6, r = 2

We need to evaluate $\frac{6!}{(6-2)!}$.

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

Expand the larger factorial, $6!$, until you reach the smaller factorial, $4!$.

$\frac{6 \times 5 \times 4!}{4!}$

Cancel out the common term $4!$:

$6 \times 5 = 30$.

Therefore, when $n=6, r=2$, the value is 30.

(ii) when n = 9, r = 5

We need to evaluate $\frac{9!}{(9-5)!}$.

$\frac{9!}{(9-5)!} = \frac{9!}{4!}$

Expand the larger factorial, $9!$, until you reach the smaller factorial, $4!$.

$\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4!}$

Cancel out the common term $4!$:

$9 \times 8 \times 7 \times 6 \times 5$

Calculate the product:

$(9 \times 8) \times (7 \times 6) \times 5 = 72 \times 42 \times 5 = 72 \times 210 = 15120$.

Therefore, when $n=9, r=5$, the value is 15120.

Given:

The word is ALLAHABAD.

To Find:

The total number of different permutations (arrangements) of the letters of the word.

Solution:

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

Total number of letters, $n = 9$.

The letters are A, L, H, B, D.

• The letter 'A' occurs 4 times.
• The letter 'L' occurs 2 times.
• The letter 'H' occurs 1 time.
• The letter 'B' occurs 1 time.
• The letter 'D' occurs 1 time.

The number of permutations of $n$ objects, where there are $p_1$ objects of one kind, $p_2$ of a second kind, ..., $p_k$ of a $k$-th kind, is given by the formula:

In this case, $n=9$, $p_1=4$ (for A's), and $p_2=2$ (for L's). The other letters appear only once, so their factorials are $1! = 1$, which does not affect the calculation.

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

Now, we compute the value:

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

Cancel out the $4!$ term:

$\frac{9 \times 8 \times 7 \times 6 \times 5}{2} = 9 \times 4 \times 7 \times 6 \times 5$

$= 36 \times 210 = 7560$.

Therefore, there are 7560 different permutations of the letters of the word ALLAHABAD.

Given:

The set of available digits is {1, 2, 3, 4, 5, 6, 7, 8, 9}.

Total number of available digits = 9.

We need to form 4-digit numbers without repetition.

To Find:

The total number of 4-digit numbers that can be formed.

Solution:

Since the order of digits in a number matters and repetition is not allowed, this is a permutation problem.

Method 1: Using the Fundamental Principle of Multiplication

A 4-digit number has four places to fill.

Thousands place: 9 choices.

Hundreds place: 8 choices remaining.

Tens place: 7 choices remaining.

Units place: 6 choices remaining.

Total number of 4-digit numbers = $9 \times 8 \times 7 \times 6 = 3024$.

Method 2: Using the Permutation Formula

We need to find the number of arrangements of 9 distinct digits taken 4 at a time ($^9P_4$).

$^9P_4 = \frac{9!}{(9-4)!} = \frac{9!}{5!} = \frac{9 \times 8 \times 7 \times 6 \times 5!}{5!} = 9 \times 8 \times 7 \times 6 = 3024$.

Therefore, 3024 different 4-digit numbers can be formed.

Given:

The set of available digits is {0, 1, 2, 3, 4, 5}. Total of 6 digits.

The numbers must lie between 100 and 1000, which means they must be 3-digit numbers.

Repetition of digits is not allowed.

To Find:

The total number of such 3-digit numbers.

Solution:

We need to form 3-digit numbers. A 3-digit number cannot start with 0.

Hundreds place: The digit cannot be 0. So, the choices are {1, 2, 3, 4, 5}. There are 5 choices.

Tens place: One digit has been used for the hundreds place. The remaining number of digits is $6-1=5$. The digit 0 can now be used. So, there are 5 choices.

Units place: Two distinct digits have been used. The remaining number of digits is $6-2=4$. So, there are 4 choices.

By the Fundamental Principle of Multiplication:

Total number of 3-digit numbers = $5 \times 5 \times 4 = 100$.

Therefore, 100 such numbers can be formed.

(i) Find n for $^nP_5 = 42 \cdot ^nP_3$, with $n>4$.

Using the permutation formula $^nP_r = \frac{n!}{(n-r)!}$, we write the equation as:

$\frac{n!}{(n-5)!} = 42 \cdot \frac{n!}{(n-3)!}$

Since $n>4$, $n!$ is not zero, so we can divide both sides by $n!$:

$\frac{1}{(n-5)!} = \frac{42}{(n-3)!}$

Rearranging the terms, we get:

$(n-3)! = 42 \cdot (n-5)!$

$(n-3)(n-4)(n-5)! = 42 \cdot (n-5)!$

$(n-3)(n-4) = 42$

$n^2 - 7n + 12 = 42$

$n^2 - 7n - 30 = 0$

Factoring the quadratic equation:

$(n-10)(n+3) = 0$

The possible values for $n$ are $n=10$ or $n=-3$. Since the condition is $n>4$, the only valid solution is $n=10$.

(ii) Find n for $\frac{^nP_4}{^{n-1}P_4} = \frac{5}{3}$, with $n>4$.

Using the permutation formula:

$\frac{\frac{n!}{(n-4)!}}{\frac{(n-1)!}{((n-1)-4)!}} = \frac{5}{3}$

$\frac{n!}{(n-4)!} \cdot \frac{(n-5)!}{(n-1)!} = \frac{5}{3}$

Expand the factorials:

$\frac{n \cdot (n-1)!}{(n-4) \cdot (n-5)!} \cdot \frac{(n-5)!}{(n-1)!} = \frac{5}{3}$

Cancel the common terms:

$\frac{n}{n-4} = \frac{5}{3}$

Cross-multiply:

$3n = 5(n-4)$

$3n = 5n - 20$

$20 = 2n$

$n=10$

This satisfies the condition $n>4$. Therefore, $n=10$.

Given:

The equation $5 \cdot ^4P_r = 6 \cdot ^5P_{r-1}$.

To Find:

The value of $r$.

Solution:

First, we establish the conditions on $r$. For $^4P_r$, we need $4 \ge r$. For $^5P_{r-1}$, we need $5 \ge r-1$ (which is $r \le 6$) and $r-1 \ge 0$ (which is $r \ge 1$). Combining these, we need $1 \le r \le 4$.

Using the permutation formula $^nP_r = \frac{n!}{(n-r)!}$:

$5 \cdot \frac{4!}{(4-r)!} = 6 \cdot \frac{5!}{(5-(r-1))!}$

$5 \cdot \frac{4!}{(4-r)!} = 6 \cdot \frac{5 \cdot 4!}{(6-r)!}$

Divide both sides by $5 \cdot 4!$:

$\frac{1}{(4-r)!} = \frac{6}{(6-r)!}$

$(6-r)! = 6 \cdot (4-r)!$

$(6-r)(5-r)(4-r)! = 6 \cdot (4-r)!$

$(6-r)(5-r) = 6$

$30 - 11r + r^2 = 6$

$r^2 - 11r + 24 = 0$

Factoring the quadratic equation:

$(r-3)(r-8) = 0$

The possible values for $r$ are $r=3$ or $r=8$.

Based on our condition $1 \le r \le 4$, the only valid solution is $r=3$.

Given:

The word is DAUGHTER. It has 8 distinct letters.

Vowels: {A, U, E} (3 vowels)

Consonants: {D, G, H, T, R} (5 consonants)

Total arrangements of the 8 distinct letters = $8! = 40320$.

(i) all vowels occur together

Treat the 3 vowels {A, U, E} as a single block. Now we have 6 items to arrange: the vowel block and the 5 consonants.

Number of ways to arrange these 6 items = $6! = 720$.

Within the vowel block, the 3 vowels can be arranged in $3! = 6$ ways.

Total arrangements with vowels together = (Arrangements of blocks) $\times$ (Arrangements within the block)

Total = $6! \times 3! = 720 \times 6 = 4320$.

So, there are 4320 arrangements where all vowels occur together.

(ii) all vowels do not occur together

This is the complement of the case where all vowels occur together.

Number of arrangements (vowels not together) = (Total arrangements) - (Arrangements with vowels together)

Number = $8! - (6! \times 3!)$

Number = $40320 - 4320 = 36000$.

So, there are 36000 arrangements where all vowels do not occur together.

Given:

Number of red discs = 4 (indistinguishable)

Number of yellow discs = 3 (indistinguishable)

Number of green discs = 2 (indistinguishable)

Total number of discs to be arranged, $n = 4 + 3 + 2 = 9$.

To Find:

The total number of different arrangements.

Solution:

This is a problem of permutations with repetitions. The formula is $\frac{n!}{p_1! p_2! \dots p_k!}$.

Here, $n=9$, $p_1=4$ (red), $p_2=3$ (yellow), and $p_3=2$ (green).

Number of arrangements = $\frac{9!}{4! \cdot 3! \cdot 2!}$

Number of arrangements = $\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4! \cdot (3 \times 2 \times 1) \cdot (2 \times 1)}$

Number of arrangements = $\frac{9 \times 8 \times 7 \times 6 \times 5}{6 \times 2}$

Number of arrangements = $9 \times 4 \times 7 \times 5 = 1260$.

Therefore, there are 1260 ways to arrange the discs.

The word is INDEPENDENCE. Total letters = 12.

Letter frequencies: E (4), N (3), D (2), I (1), P (1), C (1).

Total number of arrangements = $\frac{12!}{4! \cdot 3! \cdot 2!} = 1,663,200$.

(i) do the words start with P

Fix 'P' at the start. We are left with 11 letters to arrange: E(4), N(3), D(2), I(1), C(1).

Number of arrangements = $\frac{11!}{4! \cdot 3! \cdot 2!} = 138,600$.

(ii) do all the vowels always occur together

The vowels are {I, E, E, E, E} (5 vowels). Treat them as a single block.

Arrangements within the vowel block = $\frac{5!}{4!} = 5$.

The remaining 7 consonants are {N, N, N, D, D, P, C}. We now arrange 8 items (the vowel block + 7 consonants).

Arrangements of these 8 items = $\frac{8!}{3! \cdot 2!} = \frac{40320}{12} = 3360$.

Total arrangements with vowels together = $3360 \times 5 = 16,800$.

(iii) do the vowels never occur together

This is the total arrangements minus the arrangements where vowels are together.

Number = $1,663,200 - 16,800 = 1,646,400$.

(iv) do the words begin with I and end in P?

Fix 'I' at the start and 'P' at the end. We are left with 10 letters to arrange: E(4), N(3), D(2), C(1).

Number of arrangements = $\frac{10!}{4! \cdot 3! \cdot 2!} = \frac{3,628,800}{24 \cdot 6 \cdot 2} = \frac{3,628,800}{288} = 12,600$.

Given:

The set of available digits is {1, 2, 3, 4, 5, 6, 7, 8, 9}. Total of 9 digits.

We need to form 3-digit numbers.

Condition: No digit is repeated.

To Find:

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

Solution:

Since the order of digits matters and repetition is not allowed, this is a permutation problem.

Method 1: Using the Fundamental Principle of Multiplication

A 3-digit number has three places to fill: hundreds, tens, and units.

Hundreds place: We can use any of the 9 available digits. There are 9 choices.

Tens place: Since one digit is used and not repeated, there are $9-1=8$ choices remaining. There are 8 choices.

Units place: Two digits are used, leaving $9-2=7$ choices. There are 7 choices.

Total number of 3-digit numbers = $9 \times 8 \times 7 = 504$.

Method 2: Using the Permutation Formula

We need to find the number of arrangements (permutations) of 9 distinct digits taken 3 at a time ($^9P_3$).

$^9P_3 = \frac{9!}{(9-3)!} = \frac{9!}{6!} = \frac{9 \times 8 \times 7 \times 6!}{6!} = 9 \times 8 \times 7 = 504$.

Therefore, 504 three-digit numbers can be formed.

Given:

The available digits for a number are {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. Total of 10 digits.

We need to form 4-digit numbers.

Condition: No digit is repeated.

To Find:

The total number of 4-digit numbers with no digit repeated.

Solution:

A 4-digit number has four places to fill. A 4-digit number cannot start with 0.

Thousands place: The digit cannot be 0. So, the choices are from {1, 2, ..., 9}. There are 9 choices.

Hundreds place: One non-zero digit has been used. The remaining number of digits is $10-1=9$. The digit 0 can now be used. There are 9 choices.

Tens place: Two distinct digits have been used. The remaining number of digits is $10-2=8$. There are 8 choices.

Units place: Three distinct digits have been used. The remaining number of digits is $10-3=7$. There are 7 choices.

By the Fundamental Principle of Multiplication:

Total number of 4-digit numbers = $9 \times 9 \times 8 \times 7 = 4536$.

Therefore, there are 4536 four-digit numbers with no digit repeated.

Given:

The available digits are {1, 2, 3, 4, 6, 7}. Total of 6 digits.

We need to form 3-digit even numbers.

Condition: No digit is repeated.

To Find:

The total number of such 3-digit even numbers.

Solution:

A 3-digit number has three places. For the number to be even, the units place must be an even digit.

The even digits available are {2, 4, 6}.

Units place: There are 3 choices (2, 4, or 6).

Hundreds place: One even digit is used for the units place. We are left with $6-1=5$ digits. There are 5 choices.

Tens place: Two distinct digits are used. We are left with $6-2=4$ digits. There are 4 choices.

By the Fundamental Principle of Multiplication:

Total number of 3-digit even numbers = (Choices for hundreds) $\times$ (Choices for tens) $\times$ (Choices for units)

Total = $5 \times 4 \times 3 = 60$.

Therefore, 60 such numbers can be formed.

Given:

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

We need to form 4-digit numbers with no digit repeated.

Part 1: Total number of 4-digit numbers

This is a permutation of 5 digits taken 4 at a time ($^5P_4$).

Total numbers = $5 \times 4 \times 3 \times 2 = 120$.

Therefore, 120 four-digit numbers can be formed.

Part 2: Number of even 4-digit numbers

For the number to be even, the units digit must be even. The available even digits are {2, 4}.

Units place: There are 2 choices (2 or 4).

Thousands place: One even digit is used. We are left with $5-1=4$ digits. There are 4 choices.

Hundreds place: Two digits are used. We are left with $5-2=3$ digits. There are 3 choices.

Tens place: Three digits are used. We are left with $5-3=2$ digits. There are 2 choices.

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

Total = $4 \times 3 \times 2 \times 2 = 48$.

Therefore, 48 of these numbers will be even.

Given:

A committee of 8 persons.

We need to choose a chairman and a vice chairman.

Condition: One person cannot hold more than one position.

To Find:

The number of ways to choose the two positions.

Solution:

Since the positions (chairman, vice chairman) are distinct, the order of selection matters. This is a permutation problem.

Method 1: Using the Fundamental Principle of Multiplication

Chairman position: There are 8 choices.

Vice Chairman position: After choosing the chairman, there are $8-1=7$ persons left. There are 7 choices.

Total ways = $8 \times 7 = 56$.

Method 2: Using the Permutation Formula

We are arranging 2 people from a group of 8 into specific roles ($^8P_2$).

$^8P_2 = \frac{8!}{(8-2)!} = \frac{8!}{6!} = 8 \times 7 = 56$.

Therefore, there are 56 ways to choose a chairman and a vice chairman.

Given:

The ratio $^{n-1}P_3 : ^nP_4 = 1 : 9$.

To Find:

The value of $n$.

Solution:

We can write the ratio as a fraction:

$\frac{^{n-1}P_3}{^nP_4} = \frac{1}{9}$

Using the permutation formula $^kP_m = \frac{k!}{(k-m)!}$:

$\frac{\frac{(n-1)!}{(n-1-3)!}}{\frac{n!}{(n-4)!}} = \frac{1}{9}$

$\frac{\frac{(n-1)!}{(n-4)!}}{\frac{n!}{(n-4)!}} = \frac{1}{9}$

$\frac{(n-1)!}{(n-4)!} \times \frac{(n-4)!}{n!} = \frac{1}{9}$

$\frac{(n-1)!}{n!} = \frac{1}{9}$

Since $n! = n \times (n-1)!$, we have:

$\frac{(n-1)!}{n \times (n-1)!} = \frac{1}{9}$

$\frac{1}{n} = \frac{1}{9}$

Therefore, $n=9$. We can verify that for $n=9$, both permutations are defined ($9-1 \ge 3$ and $9 \ge 4$).

(i) Find r if $^5P_r = 2 \cdot ^6P_{r-1}$

First, we establish the conditions on $r$. For $^5P_r$, we need $5 \ge r$. For $^6P_{r-1}$, we need $r-1 \ge 0 \implies r \ge 1$ and $6 \ge r-1 \implies r \le 7$. Combining these, we need $1 \le r \le 5$.

Using the permutation formula $^nP_r = \frac{n!}{(n-r)!}$:

$\frac{5!}{(5-r)!} = 2 \cdot \frac{6!}{(6-(r-1))!}$

$\frac{5!}{(5-r)!} = 2 \cdot \frac{6 \cdot 5!}{(7-r)!}$

Divide both sides by $5!$:

$\frac{1}{(5-r)!} = \frac{12}{(7-r)!}$

$(7-r)! = 12 \cdot (5-r)!$

$(7-r)(6-r)(5-r)! = 12 \cdot (5-r)!$

$(7-r)(6-r) = 12$

$42 - 13r + r^2 = 12$

$r^2 - 13r + 30 = 0$

Factoring the quadratic equation:

$(r-3)(r-10) = 0$

The possible values for $r$ are $r=3$ or $r=10$.

Based on our condition $1 \le r \le 5$, the only valid solution is $r=3$.

(ii) Find r if $^5P_r = ^6P_{r-1}$

The condition on $r$ is still $1 \le r \le 5$.

Using the permutation formula:

$\frac{5!}{(5-r)!} = \frac{6!}{(6-(r-1))!}$

$\frac{5!}{(5-r)!} = \frac{6 \cdot 5!}{(7-r)!}$

Divide both sides by $5!$:

$\frac{1}{(5-r)!} = \frac{6}{(7-r)!}$

$(7-r)! = 6 \cdot (5-r)!$

$(7-r)(6-r)(5-r)! = 6 \cdot (5-r)!$

$(7-r)(6-r) = 6$

$42 - 13r + r^2 = 6$

$r^2 - 13r + 36 = 0$

Factoring the quadratic equation:

$(r-4)(r-9) = 0$

The possible values for $r$ are $r=4$ or $r=9$.

Based on our condition $1 \le r \le 5$, the only valid solution is $r=4$.

Given:

The word is EQUATION.

To Find:

The number of words that can be formed using all the letters, each used exactly once.

Solution:

The word EQUATION has 8 letters: E, Q, U, A, T, I, O, N.

All 8 letters are distinct (no letter is repeated).

We need to find the number of arrangements (permutations) of these 8 distinct letters.

The number of permutations of $n$ distinct objects is given by $n!$.

Here, $n=8$.

Number of words = $8!$

$8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40,320$.

Therefore, 40,320 words can be formed.

The word is MONDAY. It has 6 distinct letters.

Vowels: {O, A} (2 vowels)

Consonants: {M, N, D, Y} (4 consonants)

(i) 4 letters are used at a time

This is a permutation of 6 distinct letters taken 4 at a time ($^6P_4$).

$^6P_4 = \frac{6!}{(6-4)!} = \frac{6!}{2!} = 6 \times 5 \times 4 \times 3 = 360$.

So, 360 words can be formed.

(ii) all letters are used at a time

This is a permutation of all 6 distinct letters.

Number of words = $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$.

So, 720 words can be formed.

(iii) all letters are used but first letter is a vowel

We need to form a 6-letter word.

First position: Must be a vowel. There are 2 choices (O or A).

Remaining 5 positions: After filling the first position, we have 5 letters left to arrange in the remaining 5 positions. This can be done in $5!$ ways.

$5! = 120$.

Total number of words = (Choices for first position) $\times$ (Arrangements for remaining positions)

Total = $2 \times 5! = 2 \times 120 = 240$.

So, 240 such words can be formed.

The word is MISSISSIPPI. Total letters = 11.

Letter frequencies: M(1), I(4), S(4), P(2).

Step 1: Find the total number of distinct permutations.

Total permutations = $\frac{11!}{4! \cdot 4! \cdot 2!} = \frac{39,916,800}{24 \cdot 24 \cdot 2} = \frac{39,916,800}{1152} = 34,650$.

Step 2: Find the number of permutations where the four I's come together.

Treat the four I's {IIII} as a single block. Now we have 8 items to arrange: {M, S, S, S, S, P, P, (IIII)}.

The repetitions are S(4) and P(2).

Number of arrangements = $\frac{8!}{4! \cdot 2!} = \frac{40,320}{24 \cdot 2} = \frac{40,320}{48} = 840$.

Step 3: Find the number of permutations where the four I's do not come together.

This is the difference between the total permutations and the permutations where the I's are together.

Number = (Total permutations) - (Permutations with I's together)

Number = $34,650 - 840 = 33,810$.

Therefore, there are 33,810 permutations where the four I's do not come together.

The word is PERMUTATIONS. Total letters = 12. The only repeated letter is 'T' (2 times).

Vowels: {E, U, A, I, O} (5 distinct vowels).

(i) words start with P and end with S

Fix 'P' at the start and 'S' at the end. We are left with 10 letters to arrange in the middle: {E, R, M, U, T, A, T, I, O, N}.

In these 10 letters, 'T' is repeated twice.

Number of arrangements = $\frac{10!}{2!} = \frac{3,628,800}{2} = 1,814,400$.

(ii) vowels are all together

Treat the 5 vowels {E, U, A, I, O} as a single block. The arrangements within this block are $5! = 120$.

We are now arranging 8 items: the vowel block and the 7 consonants {P, R, M, T, T, N, S}.

In these 8 items, 'T' is repeated twice.

Number of arrangements of these items = $\frac{8!}{2!} = \frac{40,320}{2} = 20,160$.

Total arrangements with vowels together = $20,160 \times 120 = 2,419,200$.

(iii) there are always 4 letters between P and S

Consider placing P and S. The possible positions for (P, S) are (1,6), (2,7), ..., (7,12). This gives 7 ways. Similarly, S can come before P in 7 ways. Total ways to place P and S = $7+7 = 14$.

After placing P and S, we are left with 10 empty positions and 10 letters to arrange: {E, R, M, U, T, A, T, I, O, N}.

The number of ways to arrange these remaining 10 letters (with 'T' repeated twice) is $\frac{10!}{2!} = 1,814,400$.

Total arrangements = (Ways to place P and S) $\times$ (Ways to arrange other letters)

Total = $14 \times 1,814,400 = 25,401,600$.

Given:

The equation $^nC_9 = ^nC_8$.

To Find:

The value of $^{n}C_{17}$.

Solution:

We use the property of combinations which states that if $^nC_a = ^nC_b$, then either $a=b$ or $a+b=n$.

In the given equation, $a=9$ and $b=8$.

Clearly, $9 \neq 8$. Therefore, we must use the second condition:

$n = a + b$

$n = 9 + 8$

$n = 17$

Now that we have found the value of $n$, we need to evaluate $^{n}C_{17}$.

Substitute $n=17$ into the expression:

$^{17}C_{17}$

We use another property of combinations which states that $^nC_n = 1$ for any non-negative integer $n$.

Therefore,

$^{17}C_{17} = 1$.

Alternate Method (using formula):

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

$^{17}C_{17} = \frac{17!}{17!(17-17)!} = \frac{17!}{17! \cdot 0!}$

Since $0! = 1$, we have:

$^{17}C_{17} = \frac{17!}{17! \cdot 1} = 1$.

The value of $^nC_{17}$ is 1.

Given:

A group of 2 men and 3 women. Total persons = $2+3 = 5$.

A committee of 3 persons is to be formed.

Part 1: In how many ways can the committee be formed?

We need to select 3 persons from a total of 5. Since the order of selection for a committee does not matter, this is a combination problem.

The number of ways to choose 3 persons from 5 is given by $^5C_3$.

$^5C_3 = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4 \times 3!}{3! \times (2 \times 1)} = \frac{5 \times 4}{2} = 10$.

Therefore, the committee can be constituted in 10 ways.

Part 2: How many committees consist of 1 man and 2 women?

This requires two separate selections:

1. Selecting 1 man from the available 2 men.

2. Selecting 2 women from the available 3 women.

Number of ways to select 1 man from 2 men = $^2C_1 = \frac{2!}{1!(2-1)!} = \frac{2}{1} = 2$.

Number of ways to select 2 women from 3 women = $^3C_2 = \frac{3!}{2!(3-2)!} = \frac{3!}{2!1!} = 3$.

By the Fundamental Principle of Multiplication, the total number of ways to form such a committee is the product of the number of ways for each selection.

Total ways = (Ways to select 1 man) $\times$ (Ways to select 2 women)

Total ways = $2 \times 3 = 6$.

Therefore, 6 of these committees would consist of 1 man and 2 women.

Total number of ways of choosing 4 cards from 52:

This is a combination of 52 items taken 4 at a time.

$^{52}C_4 = \frac{52!}{4!(52-4)!} = \frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1} = 270,725$.

(i) four cards are of the same suit

There are 4 suits. First, we choose one suit ($^4C_1$ ways). Then, we choose 4 cards from the 13 cards of that suit ($^{13}C_4$ ways).

Number of ways = $^4C_1 \times ^{13}C_4 = 4 \times \frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1} = 4 \times 715 = 2,860$.

(ii) four cards belong to four different suits

We must choose one card from each of the four suits.

Number of ways = $^{13}C_1 \times ^{13}C_1 \times ^{13}C_1 \times ^{13}C_1 \ $$ = 13 \times 13 \times 13 \times 13 \ $$ = 13^4 = 28,561$.

(iii) are face cards

There are 12 face cards (Jack, Queen, King of each of the 4 suits).

We need to choose 4 cards from these 12 face cards.

Number of ways = $^{12}C_4 = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495$.

(iv) two are red cards and two are black cards

There are 26 red cards and 26 black cards.

Number of ways = (Ways to choose 2 red cards) $\times$ (Ways to choose 2 black cards)

Number of ways = $^{26}C_2 \times ^{26}C_2 = \left(\frac{26 \times 25}{2}\right) \times \left(\frac{26 \times 25}{2}\right) \ $$ = 325 \times 325 \ $$ = 105,625$.

(v) cards are of the same colour

This means either all 4 cards are red OR all 4 cards are black.

Ways to choose 4 red cards = $^{26}C_4 = \frac{26 \times 25 \times 24 \times 23}{4 \times 3 \times 2 \times 1} = 14,950$.

Ways to choose 4 black cards = $^{26}C_4 = 14,950$.

Total ways = $14,950 + 14,950 = 29,900$.

Given:

The equation $^nC_8 = ^nC_2$.

To Find:

The value of $^nC_2$.

Solution:

We use the property of combinations which states that if $^nC_a = ^nC_b$, then either $a=b$ or $a+b=n$.

Here, $a=8$ and $b=2$. Since $8 \neq 2$, we must have:

$n = 8 + 2 = 10$.

Now we need to find the value of $^nC_2$, which is $^{10}C_2$.

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

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

$^{10}C_2 = \frac{10 \times 9 \times 8!}{2 \times 1 \times 8!} = \frac{10 \times 9}{2} = 45$.

Therefore, the value of $^nC_2$ is 45.

(i) Determine n if $^{2n}C_3 : ^nC_3 = 12 : 1$

We write the ratio as an equation:

$\frac{^{2n}C_3}{^nC_3} = 12$

Using the formula $^kC_m = \frac{k(k-1)(k-2)...(k-m+1)}{m!}$:

$\frac{\frac{2n(2n-1)(2n-2)}{3!}}{\frac{n(n-1)(n-2)}{3!}} = 12$

$\frac{2n(2n-1)2(n-1)}{n(n-1)(n-2)} = 12$

For $n \ge 3$, we can cancel $n$ and $(n-1)$:

$\frac{4(2n-1)}{n-2} = 12$

$\frac{2n-1}{n-2} = 3$

$2n-1 = 3(n-2) = 3n-6$

$5 = n$

Therefore, $n=5$.

(ii) Determine n if $^{2n}C_3 : ^nC_3 = 11 : 1$

Using the simplified expression from part (i):

$\frac{4(2n-1)}{n-2} = 11$

$4(2n-1) = 11(n-2)$

$8n-4 = 11n-22$

$18 = 3n$

$n=6$

Therefore, $n=6$.

Given:

There are 21 points on a circle.

To Find:

The number of chords that can be drawn.

Solution:

A chord is a line segment that connects two distinct points on a circle. To form a chord, we need to choose any 2 points from the 21 available points.

Since the order of choosing the two points does not matter (a chord from point A to B is the same as from B to A), this is a combination problem.

We need to find the number of ways to choose 2 points from 21, which is given by $^{21}C_2$.

$^{21}C_2 = \frac{21!}{2!(21-2)!} = \frac{21!}{2!19!} = \frac{21 \times 20}{2 \times 1} = 210$.

Therefore, 210 chords can be drawn.

Given:

Available: 5 boys and 4 girls.

To be selected: A team of 3 boys and 3 girls.

To Find:

The total number of ways to form the team.

Solution:

This is a two-step selection process. The order of selection does not matter, so we use combinations.

Step 1: Select 3 boys from 5 boys.

Number of ways = $^5C_3 = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10$.

Step 2: Select 3 girls from 4 girls.

Number of ways = $^4C_3 = \frac{4!}{3!1!} = 4$.

By the Fundamental Principle of Multiplication, the total number of ways to form the team is the product of the ways in each step.

Total ways = $10 \times 4 = 40$.

Therefore, the team can be selected in 40 ways.

Given:

Available balls: 6 red, 5 white, 5 blue.

Selection required: 3 balls of each colour (total 9 balls).

To Find:

The total number of ways to make this selection.

Solution:

This is a three-step selection process using combinations.

Step 1: Select 3 red balls from 6.

Number of ways = $^6C_3 = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.

Step 2: Select 3 white balls from 5.

Number of ways = $^5C_3 = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10$.

Step 3: Select 3 blue balls from 5.

Number of ways = $^5C_3 = 10$.

By the Fundamental Principle of Multiplication, the total number of ways is the product of the ways in each step.

Total ways = $20 \times 10 \times 10 = 2000$.

Therefore, there are 2000 ways to make the selection.

Given:

A standard deck of 52 playing cards.

We need to form 5-card combinations.

Condition: There must be exactly one ace in each combination.

To Find:

The total number of such 5-card combinations.

Solution:

A standard deck has 4 aces and $52-4=48$ non-ace cards.

To form the required 5-card combination, we need to perform two independent selections:

Step 1: Select exactly 1 ace from the 4 available aces.

The number of ways to do this is given by the combination $^4C_1$.

$^4C_1 = \frac{4!}{1!(4-1)!} = \frac{4!}{1!3!} = 4$.

Step 2: Select the remaining $5-1=4$ cards from the 48 non-ace cards.

The number of ways to do this is given by the combination $^{48}C_4$.

$^{48}C_4 = \frac{48!}{4!(48-4)!} = \frac{48!}{4!44!} = \frac{48 \times 47 \times 46 \times 45}{4 \times 3 \times 2 \times 1}$

$^{48}C_4 = 2 \times 47 \times 46 \times 45 = 194,580$.

By the Fundamental Principle of Multiplication, the total number of combinations is the product of the ways in each step.

Total ways = (Ways to select 1 ace) $\times$ (Ways to select 4 non-aces)

Total ways = $^4C_1 \times ^{48}C_4 = 4 \times 194,580 = 778,320$.

Therefore, there are 778,320 ways to form such a 5-card combination.

Given:

Total players available = 17.

Number of players who can bowl (bowlers) = 5.

Number of players who cannot bowl (batsmen/others) = $17 - 5 = 12$.

Team size to be selected = 11 players.

Condition: The team must include exactly 4 bowlers.

To Find:

The total number of ways to select the team.

Solution:

To form the team, we need to make two selections:

Step 1: Select exactly 4 bowlers from the 5 available bowlers.

Number of ways = $^5C_4 = \frac{5!}{4!1!} = 5$.

Step 2: Select the remaining players for the team.

Remaining players needed = Total team size - Number of bowlers = $11 - 4 = 7$.

These 7 players must be selected from the 12 non-bowlers.

Number of ways = $^{12}C_7 = \frac{12!}{7!(12-7)!} = \frac{12!}{7!5!} = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = 792$.

By the Fundamental Principle of Multiplication, the total number of ways to form the team is the product of the ways in each step.

Total ways = $^5C_4 \times ^{12}C_7 = 5 \times 792 = 3960$.

Therefore, the team can be selected in 3,960 ways.

Given:

A bag with 5 black balls and 6 red balls.

We need to select 2 black balls and 3 red balls.

To Find:

The total number of ways to make this selection.

Solution:

Since the order of selection does not matter, we use combinations.

Step 1: Select 2 black balls from 5.

Number of ways = $^5C_2 = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$.

Step 2: Select 3 red balls from 6.

Number of ways = $^6C_3 = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.

By the Fundamental Principle of Multiplication, the total number of ways is the product of the ways in each step.

Total ways = $10 \times 20 = 200$.

Therefore, there are 200 ways to make the selection.

Given:

Total courses available = 9.

Number of courses to choose = 5.

Condition: 2 specific courses are compulsory.

To Find:

The number of ways the student can choose the programme.

Solution:

Since 2 courses are compulsory, the student has no choice regarding them. They are already part of the programme.

Number of courses left to choose = Total courses to choose - Compulsory courses = $5 - 2 = 3$.

Number of courses available to choose from = Total available courses - Compulsory courses = $9 - 2 = 7$.

The problem reduces to selecting 3 courses from the remaining 7 available courses. Since the order of choosing courses does not matter, we use combinations.

Number of ways = $^7C_3 = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.

Therefore, the student can choose the programme in 35 ways.

Given:

The word is INVOLUTE. All 8 letters are distinct.

Vowels: {I, O, U, E} (4 vowels)

Consonants: {N, V, L, T} (4 consonants)

We need to form 5-letter words with 3 vowels and 2 consonants.

To Find:

The total number of such words.

Solution:

This is a two-step process: selection and then arrangement.

Step 1: Select the letters.

We need to select 3 vowels from the 4 available vowels. The number of ways is $^4C_3$.

$^4C_3 = \frac{4!}{3!1!} = 4$.

We need to select 2 consonants from the 4 available consonants. The number of ways is $^4C_2$.

$^4C_2 = \frac{4!}{2!2!} = \frac{4 \times 3}{2 \times 1} = 6$.

Total number of ways to select the group of 5 letters (3 vowels and 2 consonants) = $^4C_3 \times ^4C_2 = 4 \times 6 = 24$.

Step 2: Arrange the selected letters.

Each selected group has 5 distinct letters. These 5 letters can be arranged to form a word in $5!$ ways.

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

By the Fundamental Principle of Multiplication, the total number of words is the product of the number of ways to select the letters and the number of ways to arrange them.

Total words = (Number of selections) $\times$ (Number of arrangements)

Total words = $24 \times 120 = 2880$.

Therefore, 2,880 words can be formed.

Given:

A group of 4 girls and 7 boys. Total people = 11.

A team of 5 members is to be selected.

(i) no girl

If there are no girls, the team must consist of 5 boys. We need to select 5 boys from the 7 available boys.

Number of ways = $^7C_5 = ^7C_2 = \frac{7 \times 6}{2 \times 1} = 21$.

(ii) at least one boy and one girl

This is the complement of selecting a team with either all boys or all girls.

Total ways to select 5 members from 11 = $^{11}C_5 = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462$.

Ways to select a team of all boys (no girls) = $^7C_5 = 21$.

Ways to select a team of all girls (no boys) = $^4C_5 = 0$ (impossible).

Number of ways = Total ways - (Ways with no girls + Ways with no boys)

Number of ways = $462 - (21 + 0) = 441$.

(iii) at least 3 girls

This means we can have either 3 girls or 4 girls in the team.

Case 1: 3 girls and 2 boys

Ways = (Select 3 girls from 4) $\times$ (Select 2 boys from 7) = $^4C_3 \times ^7C_2 = 4 \times 21 = 84$.

Case 2: 4 girls and 1 boy

Ways = (Select 4 girls from 4) $\times$ (Select 1 boy from 7) = $^4C_4 \times ^7C_1 = 1 \times 7 = 7$.

Total ways = Ways for Case 1 + Ways for Case 2 = $84 + 7 = 91$.

The word is AGAIN. Total letters = 5. Letter 'A' is repeated twice.

Part 1: Total number of words

Number of words = $\frac{5!}{2!} = \frac{120}{2} = 60$.

Part 2: 50^th word in dictionary order

The letters in alphabetical order are A, G, I, N.

1. Words starting with A: Fix 'A' at the start. The remaining letters are {G, I, N, A}. These are 4 distinct letters. Number of arrangements = $4! = 24$. (Words 1 to 24)

2. Words starting with G: Fix 'G' at the start. The remaining letters are {A, A, I, N}. The number of arrangements = $\frac{4!}{2!} = 12$. (Words 25 to 36)

3. Words starting with I: Fix 'I' at the start. The remaining letters are {A, A, G, N}. The number of arrangements = $\frac{4!}{2!} = 12$. (Words 37 to 48)

The first 48 words start with A, G, or I. The 49th and 50th words must start with N.

4. Words starting with N: Fix 'N' at the start. The remaining letters are {A, A, G, I}. We need to find the 2nd arrangement of these letters in alphabetical order.

The arrangements of {A, A, G, I} in alphabetical order start with 'A'.

- Starting with NA... Remaining letters {A, G, I}. Arrangements: AGI, AIG, GAI, GIA, IAG, IGA.

- Words starting with NAA... Remaining letters {G, I}. Arrangements: GI, IG.

- 49th word: NAAGI

- 50th word: NAAIG

Therefore, the 50^th word is NAAIG.

Given:

The available digits are {1, 2, 0, 2, 4, 2, 4}. Total of 7 digits.

Frequencies: '2' (3 times), '4' (2 times), '1' (1 time), '0' (1 time).

We need to form numbers greater than 1,000,000.

Solution:

Any 7-digit number formed with these digits will be greater than 1,000,000, provided the first digit is not 0.

Step 1: Find the total number of arrangements of the 7 digits.

Total arrangements = $\frac{7!}{3! \cdot 2!} = \frac{5040}{6 \cdot 2} = \frac{5040}{12} = 420$.

Step 2: Find the number of arrangements that start with 0.

If we fix '0' at the start, we are left with 6 digits to arrange: {1, 2, 2, 2, 4, 4}.

Repetitions: '2' (3 times), '4' (2 times).

Number of arrangements starting with 0 = $\frac{6!}{3! \cdot 2!} = \frac{720}{6 \cdot 2} = \frac{720}{12} = 60$.

Step 3: Find the number of valid 7-digit numbers.

This is the total arrangements minus the arrangements that start with 0.

Number of valid numbers = $420 - 60 = 360$.

Therefore, 360 numbers greater than 1,000,000 can be formed.

Given:

5 girls and 3 boys to be seated in a row.

Condition: No two boys are together.

To Find:

The total number of such seating arrangements.

Solution:

To ensure no two boys are together, we first place the girls and then place the boys in the gaps between the girls.

Step 1: Arrange the 5 girls.

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

$5! = 120$.

Step 2: Place the 3 boys.

Arranging the 5 girls creates 6 possible spaces where the boys can be seated (one at each end and four between the girls).

_ G _ G _ G _ G _ G _

We need to choose 3 of these 6 spaces for the 3 boys. Since the boys are distinct, the order of placing them matters. This is a permutation of 6 spaces taken 3 at a time.

Number of ways to place the boys = $^6P_3 = \frac{6!}{(6-3)!} = \frac{6!}{3!} = 6 \times 5 \times 4 = 120$.

By the Fundamental Principle of Multiplication, the total number of arrangements is the product of the ways in each step.

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

Total ways = $5! \times ^6P_3 = 120 \times 120 = 14,400$.

Therefore, there are 14,400 ways to seat them so that no two boys are together.

Given:

The word is DAUGHTER. All 8 letters are distinct.

Vowels: {A, U, E} (3 vowels)

Consonants: {D, G, H, T, R} (5 consonants)

We need to form 5-letter words with 2 vowels and 3 consonants.

To Find:

The total number of such words.

Solution:

This is a two-step process: selection and then arrangement.

Step 1: Select the letters.

Number of ways to select 2 vowels from 3 = $^3C_2 = \frac{3!}{2!1!} = 3$.

Number of ways to select 3 consonants from 5 = $^5C_3 = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10$.

Total number of ways to select the group of 5 letters = $3 \times 10 = 30$.

Step 2: Arrange the selected letters.

Each selected group has 5 distinct letters. These can be arranged in $5!$ ways.

$5! = 120$.

Total number of words = (Number of selections) $\times$ (Number of arrangements)

Total words = $30 \times 120 = 3600$.

Therefore, 3,600 words can be formed.

Given:

The word is EQUATION. All 8 letters are distinct.

Vowels: {E, Q, U, A, T, I, O, N}. (Correction: Vowels are {E, U, A, I, O}, 5 vowels)

Consonants: {Q, T, N} (3 consonants).

Condition: Vowels and consonants must occur together.

Solution:

We treat the group of 5 vowels as one block and the group of 3 consonants as another block.

Step 1: Arrange the letters within each block.

The 5 distinct vowels can be arranged in $5! = 120$ ways.

The 3 distinct consonants can be arranged in $3! = 6$ ways.

Step 2: Arrange the two blocks.

The two blocks (vowel block and consonant block) can be arranged in $2! = 2$ ways (vowels first or consonants first).

By the Fundamental Principle of Multiplication, the total number of words is:

Total ways = (Arrangements of vowels) $\times$ (Arrangements of consonants) $\times$ (Arrangements of blocks)

Total ways = $5! \times 3! \times 2! = 120 \times 6 \times 2 = 1440$.

Therefore, 1,440 such words can be formed.

Given:

A group of 9 boys and 4 girls. Total people = 13.

A committee of 7 members is to be formed.

(i) exactly 3 girls

If there are 3 girls, there must be $7-3=4$ boys.

Ways = (Select 3 girls from 4) $\times$ (Select 4 boys from 9) = $^4C_3 \times ^9C_4$.

$^4C_3 = 4$.

$^9C_4 = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$.

Total ways = $4 \times 126 = 504$.

(ii) at least 3 girls

This means either 3 girls or 4 girls.

Case 1 (3 girls, 4 boys): From part (i), there are 504 ways.

Case 2 (4 girls, 3 boys): Ways = $^4C_4 \times ^9C_3 = 1 \times \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 1 \times 84 = 84$.

Total ways = $504 + 84 = 588$.

(iii) at most 3 girls

This means 0, 1, 2, or 3 girls.

Case 1 (0 girls, 7 boys): $^4C_0 \times ^9C_7 = 1 \times ^9C_2 = 1 \times \frac{9 \times 8}{2} = 36$.

Case 2 (1 girl, 6 boys): $^4C_1 \times ^9C_6 = 4 \times ^9C_3 = 4 \times 84 = 336$.

Case 3 (2 girls, 5 boys): $^4C_2 \times ^9C_5 = 6 \times ^9C_4 = 6 \times 126 = 756$.

Case 4 (3 girls, 4 boys): From part (i), there are 504 ways.

Total ways = $36 + 336 + 756 + 504 = 1632$.

Given:

The word is EXAMINATION. Total letters = 11.

Frequencies: A(2), I(2), N(2), E(1), X(1), M(1), T(1), O(1).

To Find:

The number of words that appear before the first word starting with 'E' in a dictionary listing.

Solution:

In a dictionary, words are arranged in alphabetical order. The distinct letters of EXAMINATION in alphabetical order are A, E, I, M, N, O, T, X.

The words that come before the first word starting with 'E' are all the words that start with 'A'.

We need to find the number of permutations that start with 'A'.

Fix 'A' at the first position. We are left with 10 letters to arrange:

{E, X, M, I, N, A, T, I, O, N}

The frequencies of these remaining letters are: I(2), N(2), and all others appear once.

The number of ways to arrange these 10 letters is given by the formula for permutations with repetitions:

Number of words = $\frac{10!}{2! \cdot 2!} = \frac{3,628,800}{2 \cdot 2} = \frac{3,628,800}{4} = 907,200$.

Therefore, there are 907,200 words before the first word starting with E.

Given:

Available digits: {0, 1, 3, 5, 7, 9}. Total of 6 digits.

We need to form 6-digit numbers.

Condition 1: The number must be divisible by 10.

Condition 2: No digit is repeated.

Solution:

A number is divisible by 10 if its units digit is 0.

So, for our 6-digit number, the last (units) place is fixed with the digit 0.

_ _ _ _ _ 0

Now, we need to fill the first 5 places using the remaining 5 distinct digits: {1, 3, 5, 7, 9}.

The number of ways to arrange these 5 distinct digits in the 5 available places is a permutation of 5 items, which is $5!$.

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

Therefore, 120 such 6-digit numbers can be formed.

Given:

Number of vowels = 5.

Number of consonants = 21.

We need to form words with 2 different vowels and 2 different consonants.

Solution:

This is a two-step process: selection and then arrangement.

Step 1: Select the letters.

Number of ways to select 2 different vowels from 5 = $^5C_2 = \frac{5 \times 4}{2} = 10$.

Number of ways to select 2 different consonants from 21 = $^{21}C_2 = \frac{21 \times 20}{2} = 210$.

Total ways to select the group of 4 letters = $10 \times 210 = 2100$.

Step 2: Arrange the selected letters.

Each selected group has 4 distinct letters. These can be arranged in $4!$ ways.

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

By the Fundamental Principle of Multiplication:

Total number of words = (Number of selections) $\times$ (Number of arrangements)

Total words = $2100 \times 24 = 50,400$.

Therefore, 50,400 words can be formed.

Given:

Part I has 5 questions.

Part II has 7 questions.

Total questions to attempt = 8.

Condition: At least 3 questions must be selected from each part.

To Find:

The total number of ways to select the questions.

Solution:

Let the number of questions selected from Part I be $n_1$ and from Part II be $n_2$.

We have $n_1 + n_2 = 8$, with $n_1 \ge 3$ and $n_2 \ge 3$. Also, $n_1 \le 5$ and $n_2 \le 7$.

The possible combinations of selections are:

Case 1: 3 questions from Part I and 5 questions from Part II.

Number of ways = $^5C_3 \times ^7C_5 = 10 \times 21 = 210$.

Case 2: 4 questions from Part I and 4 questions from Part II.

Number of ways = $^5C_4 \times ^7C_4 = 5 \times 35 = 175$.

Case 3: 5 questions from Part I and 3 questions from Part II.

Number of ways = $^5C_5 \times ^7C_3 = 1 \times 35 = 35$.

The total number of ways is the sum of the ways for all possible cases.

Total ways = $210 + 175 + 35 = 420$.

Therefore, the student can select the questions in 420 ways.

Given:

A standard deck of 52 cards.

A 5-card combination is to be selected.

Condition: Exactly one king must be in the selection.

Solution:

There are 4 kings and 48 non-king cards in a deck.

To meet the condition, we must perform two selections:

Step 1: Select exactly 1 king from the 4 kings.

Number of ways = $^4C_1 = 4$.

Step 2: Select the remaining 4 cards from the 48 non-king cards.

Number of ways = $^{48}C_4 = \frac{48 \times 47 \times 46 \times 45}{4 \times 3 \times 2 \times 1} = 194,580$.

By the Fundamental Principle of Multiplication, the total number of combinations is:

Total ways = $4 \times 194,580 = 778,320$.

Therefore, there are 778,320 ways.

Given:

5 men and 4 women to be seated in a row of 9 positions.

Condition: Women must occupy the even places.

Solution:

In a row of 9 seats, the even-numbered positions are 2, 4, 6, and 8. There are 4 even places.

The odd-numbered positions are 1, 3, 5, 7, and 9. There are 5 odd places.

Step 1: Arrange the 4 women in the 4 even places.

The number of ways to arrange 4 distinct women in 4 distinct places is $4!$.

$4! = 24$.

Step 2: Arrange the 5 men in the 5 odd places.

The number of ways to arrange 5 distinct men in the 5 remaining odd places is $5!$.

$5! = 120$.

By the Fundamental Principle of Multiplication, the total number of arrangements is:

Total ways = (Ways to arrange women) $\times$ (Ways to arrange men) = $4! \times 5! = 24 \times 120 = 2880$.

Therefore, 2,880 arrangements are possible.

Given:

Total students = 25. Party size = 10.

A group of 3 students has a condition: either all 3 join or none join.

Solution:

We have two mutually exclusive cases.

Case 1: All 3 special students join the party.

The 3 students are already chosen. We need to choose the remaining $10 - 3 = 7$ students from the other $25 - 3 = 22$ students.

Number of ways = $^{22}C_7 = \frac{22!}{7!15!} = 170,544$.

Case 2: None of the 3 special students join the party.

We must choose all 10 students from the other 22 students.

Number of ways = $^{22}C_{10} = \frac{22!}{10!12!} = 646,646$.

Total number of ways is the sum of the ways in both cases.

Total ways = $170,544 + 646,646 = 817,190$.

Therefore, the party can be chosen in 817,190 ways.

Given:

The word is ASSASSINATION. Total letters = 13.

Letter frequencies: A(3), S(4), I(2), N(2), T(1), O(1).

Condition: All 4 'S's must be together.

Solution:

To satisfy the condition, we treat the four 'S's as a single block (SSSS).

Now we are arranging 10 items: the (SSSS) block and the remaining 9 letters {A, A, A, I, I, N, N, T, O}.

Total items to arrange = 10.

Repetitions among these items: A(3), I(2), N(2).

The number of arrangements is given by the formula for permutations with repetitions:

Number of arrangements = $\frac{10!}{3! \cdot 2! \cdot 2!} = \frac{3,628,800}{6 \cdot 2 \cdot 2} = \frac{3,628,800}{24} = 151,200$.

(Note: Since all 'S's are identical, there is only 1 way to arrange them within their block).

Therefore, there are 151,200 ways to arrange the letters so that all S's are together.