Chapter 6 Permutations And Combinations

Given:

The word is ROSE.

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

We need to form 4-letter words.

Repetition of letters is not allowed.

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To Find:

The number of possible 4-letter words.

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Solution:

The word ROSE has 4 distinct letters: R, O, S, E.

We need to form a 4-letter word using these 4 letters without repetition.

This is equivalent to arranging 4 distinct letters in 4 distinct positions.

We can think of this as filling 4 empty boxes, one for each position in the word.

There are 4 choices for the first position (any of R, O, S, E).

Once the first position is filled, there are 3 letters remaining, so there are 3 choices for the second position.

After filling the first two positions, there are 2 letters remaining, so there are 2 choices for the third position.

Finally, there is only 1 letter remaining, so there is 1 choice for the fourth position.

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The total number of ways to fill these positions is the product of the number of choices for each position.

Number of words $= 4 \times 3 \times 2 \times 1$

This product is denoted as 4 factorial, or $4!$.

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

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Alternatively, this problem is a permutation of 4 distinct objects taken 4 at a time, denoted as $P(4, 4)$.

The formula for permutation is $P(n, r) = \frac{n!}{(n-r)!}$.

Here, $n = 4$ (total number of distinct letters) and $r = 4$ (number of letters to be used to form the word).

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

Since $0! = 1$, we have:

$P(4, 4) = \frac{4 \times 3 \times 2 \times 1}{1} = 24$

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

Given:

We have 4 flags of different colours.

A signal requires the use of 2 flags placed one below the other.

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To Find:

The number of different signals that can be generated.

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Solution:

We need to select 2 flags out of 4 distinct flags and arrange them in a specific order (one above the other) to form a signal.

Let's consider the positions for the two flags in the signal:

Position 1 (Top flag)

Position 2 (Bottom flag)

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For the first position (top flag), we have 4 choices (any of the 4 different coloured flags).

Once a flag is chosen for the top position, we have 3 flags remaining.

For the second position (bottom flag), we have 3 choices (any of the remaining 3 flags).

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

Number of signals $= 4 \times 3 = 12$

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Alternatively, this problem is a permutation of 4 distinct objects taken 2 at a time, denoted as $P(4, 2)$.

The formula for permutation is $P(n, r) = \frac{n!}{(n-r)!}$.

Here, $n = 4$ (total number of distinct flags) and $r = 2$ (number of flags used to form a signal).

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

$P(4, 2) = \frac{4 \times 3 \times 2 \times 1}{2 \times 1} = \frac{24}{2} = 12$

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Thus, the number of different signals that can be generated using 2 flags out of 4 of different colours is 12.

Given:

We need to form 2-digit even numbers.

The available digits are 1, 2, 3, 4, 5.

Repetition of digits is allowed.

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To Find:

The number of possible 2-digit even numbers.

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Solution:

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

Units Place | Tens Place

For a number to be even, the digit in the units place must be an even digit.

From the given digits {1, 2, 3, 4, 5}, the even digits are 2 and 4.

So, there are 2 choices for the units place (either 2 or 4).

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For the tens place, we can use any of the given digits {1, 2, 3, 4, 5}.

Since repetition is allowed, there are 5 choices for the tens place.

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The total number of 2-digit even numbers is the product of the number of choices for each place.

Number of 2-digit even numbers = (Number of choices for Units Place) $\times$ (Number of choices for Tens Place)

Number of 2-digit even numbers $= 2 \times 5 = 10$

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

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Thus, 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:

We have 5 different flags.

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

"At least 2 flags" means the signal can be formed using 2 flags, 3 flags, 4 flags, or 5 flags.

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To Find:

The total number of different signals that can be generated.

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Solution:

We need to find the number of permutations of 5 distinct flags taken $r$ at a time, where $r$ can be 2, 3, 4, or 5.

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

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

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Case 1: Signals using exactly 2 flags ($r=2$)

Number of signals using 2 flags $= P(5, 2) = \frac{5!}{(5-2)!} = \frac{5!}{3!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1} = 5 \times 4 = 20$

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Case 2: Signals using exactly 3 flags ($r=3$)

Number of signals using 3 flags $= P(5, 3) = \frac{5!}{(5-3)!} = \frac{5!}{2!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 5 \times 4 \times 3 = 60$

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Case 3: Signals using exactly 4 flags ($r=4$)

Number of signals using 4 flags $= P(5, 4) = \frac{5!}{(5-4)!} = \frac{5!}{1!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{1} = 5 \times 4 \times 3 \times 2 = 120$

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Case 4: Signals using exactly 5 flags ($r=5$)

Number of signals using 5 flags $= P(5, 5) = \frac{5!}{(5-5)!} = \frac{5!}{0!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{1} = 120$

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The total number of different signals that can be generated is the sum of the number of signals from each case.

Total number of signals = (Signals using 2 flags) + (Signals using 3 flags) + (Signals using 4 flags) + (Signals using 5 flags)

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

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Thus, the number of different signals that can be generated by arranging at least 2 flags out of 5 different flags is 320.

Given:

The available digits are 1, 2, 3, 4, and 5.

We need to form 3-digit numbers.

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To Find:

The number of 3-digit numbers formed under two conditions:

(i) Repetition of digits is allowed.

(ii) Repetition of digits is not allowed.

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Solution:

A 3-digit number consists of three places: the hundreds place, the tens place, and the units place.

Digits available: {1, 2, 3, 4, 5}. Total number of available digits = 5.

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(i) Repetition of the digits is allowed:

For the hundreds place, there are 5 choices (any of the digits 1, 2, 3, 4, 5).

For the tens place, since repetition is allowed, there are also 5 choices (any of the digits 1, 2, 3, 4, 5).

For the units place, since repetition is allowed, there are again 5 choices (any of the digits 1, 2, 3, 4, 5).

By the Multiplication Principle, the total number of 3-digit numbers that can be formed with repetition allowed is:

Number of numbers $= 5 \times 5 \times 5 = 125$

Thus, the number of 3-digit numbers that can be formed with repetition allowed is 125.

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(ii) Repetition of the digits is not allowed:

For the hundreds place, there are 5 choices (any of the digits 1, 2, 3, 4, 5).

For the tens place, since repetition is not allowed, one digit has already been used for the hundreds place. So, there are $5 - 1 = 4$ choices remaining.

For the units place, two digits have already been used for the hundreds and tens places. So, there are $5 - 2 = 3$ choices remaining.

By the Multiplication Principle, the total number of 3-digit numbers that can be formed without repetition is:

Number of numbers $= 5 \times 4 \times 3 = 60$

Alternatively, this problem is a permutation of 5 distinct objects taken 3 at a time, which is denoted by $P(5, 3)$.

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

Thus, the number of 3-digit numbers that can be formed without repetition is 60.

Given:

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

We need to form 3-digit even numbers.

Repetition of digits is allowed.

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To Find:

The number of possible 3-digit even numbers.

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Solution:

A 3-digit number has three places: the hundreds place, the tens place, and the units place.

Available digits: {1, 2, 3, 4, 5, 6}. Total number of available digits = 6.

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For the number to be even, the units place must be filled with an even digit.

The even digits from the available set {1, 2, 3, 4, 5, 6} are 2, 4, and 6.

So, there are 3 choices for the units place (2, 4, or 6).

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For the tens place, any of the 6 available digits can be used, since repetition is allowed.

So, there are 6 choices for the tens place.

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For the hundreds place, any of the 6 available digits can be used, since repetition is allowed.

So, there are 6 choices for the hundreds place.

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By the Fundamental Principle of Counting (Multiplication Principle), the total number of 3-digit even numbers that can be formed with repetition allowed is the product of the number of choices for each place.

Number of 3-digit even numbers = (Choices for Hundreds Place) $\times$ (Choices for Tens Place) $\times$ (Choices for Units Place)

Number of 3-digit even numbers $= 6 \times 6 \times 3 = 36 \times 3 = 108$

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Thus, the number of 3-digit even numbers that can be formed from the digits 1, 2, 3, 4, 5, 6 with repetition allowed is 108.

Given:

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

Total number of distinct letters ($n$) = 10.

We need to form a 4-letter code.

Number of letters to be used in the code ($r$) = 4.

Repetition of letters is not allowed.

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To Find:

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

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Solution:

A code is an arrangement of letters where the order matters (e.g., 'ABCD' is a different code from 'ACBD'). Since the letters must be distinct (no repetition allowed), this problem is a permutation problem.

We need to find the number of permutations of 10 distinct objects taken 4 at a time, which is denoted by $P(10, 4)$.

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

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

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Substitute the given values $n=10$ and $r=4$ into the formula:

$P(10, 4) = \frac{10!}{(10-4)!} = \frac{10!}{6!}$

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Now, calculate the factorial values and simplify:

$P(10, 4) = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1}$

$P(10, 4) = 10 \times 9 \times 8 \times 7$

$10 \times 9 = 90$

$8 \times 7 = 56$

$90 \times 56 = 5040$

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Thus, the number of 4-letter codes that can be formed using the first 10 letters of the English alphabet without repetition is 5040.

Given:

We need to form 5-digit telephone numbers.

The available digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

Total number of distinct available digits = 10.

Each number must start with 67.

No digit appears more than once (no repetition is allowed).

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To Find:

The number of possible 5-digit telephone numbers.

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Solution:

A 5-digit telephone number has five places:

Place 1 | Place 2 | Place 3 | Place 4 | Place 5

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The first two places are fixed; they must be 6 and 7, respectively.

Place 1: 6 (1 choice)

Place 2: 7 (1 choice)

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Since no digit can appear more than once, the digits 6 and 7 are already used for the first two places.

The available digits are {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.

The digits remaining for the remaining places are all digits except 6 and 7.

Remaining available digits = {0, 1, 2, 3, 4, 5, 8, 9}.

Number of remaining distinct available digits = 8.

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We need to fill the remaining three places (Place 3, Place 4, and Place 5) using the 8 remaining distinct digits without repetition.

For Place 3, there are 8 choices (any of the remaining 8 digits).

For Place 4, since one more digit is used, there are $8 - 1 = 7$ choices remaining.

For Place 5, since two more digits are used, there are $8 - 2 = 6$ choices remaining.

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By the Fundamental Principle of Counting (Multiplication Principle), the total number of 5-digit telephone numbers that can be constructed is the product of the number of choices for each place.

Number of telephone numbers = (Choices for Place 1) $\times$ (Choices for Place 2) $\times$ (Choices for Place 3) $\times$ (Choices for Place 4) $\times$ (Choices for Place 5)

Number of telephone numbers $= 1 \times 1 \times 8 \times 7 \times 6$

Number of telephone numbers $= 8 \times 7 \times 6 = 56 \times 6 = 336$

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Alternatively, we are arranging 3 out of the remaining 8 distinct digits for the last three places. This is a permutation problem $P(8, 3)$.

$P(8, 3) = \frac{8!}{(8-3)!} = \frac{8!}{5!} = \frac{8 \times 7 \times 6 \times 5!}{5!} = 8 \times 7 \times 6 = 336$

Since the first two digits are fixed in 1 way, the total number of telephone numbers is $1 \times P(8, 3) = 336$.

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Thus, the number of 5-digit telephone numbers that can be constructed is 336.

Given:

A coin is tossed 3 times.

The outcomes of each toss are recorded.

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To Find:

The total number of possible outcomes.

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Solution:

For a single toss of a coin, there are two possible outcomes: Head (H) or Tail (T).

The outcome of each toss is independent of the outcomes of the other tosses.

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We can determine the total number of outcomes by considering the number of possibilities for each toss and using the Fundamental Principle of Counting (Multiplication Principle).

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For the first toss, there are 2 possible outcomes.

For the second toss, there are 2 possible outcomes.

For the third toss, there are 2 possible outcomes.

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The total number of possible outcomes when a coin is tossed 3 times is the product of the number of outcomes for each toss:

Total outcomes $= (\text{Outcomes for 1st toss}) \times (\text{Outcomes for 2nd toss}) \times (\text{Outcomes for 3rd toss})$

Total outcomes $= 2 \times 2 \times 2 = 2^3 = 8$

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The 8 possible outcomes can be listed as: HHH, HHT, HTH, THH, HTT, THT, TTH, TTT.

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Thus, the number of possible outcomes when a coin is tossed 3 times is 8.

Given:

We have 5 flags of different colours.

A signal requires the use of 2 flags placed one below the other.

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To Find:

The number of different signals that can be generated.

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Solution:

We need to select 2 flags out of 5 distinct flags and arrange them in a specific order (one above the other) to form a signal.

Let's consider the two positions for the flags in the signal:

Position 1 (Top flag)

Position 2 (Bottom flag)

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For the first position (top flag), there are 5 choices (any of the 5 different coloured flags).

Once a flag is chosen for the top position, there are 4 flags remaining.

For the second position (bottom flag), we have 4 choices (any of the remaining 4 flags).

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By the Fundamental Principle of Counting (Multiplication Principle), the total number of different signals is the product of the number of choices for each position.

Number of signals = (Number of choices for Top Position) $\times$ (Number of choices for Bottom Position)

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

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Alternatively, this problem is a permutation of 5 distinct objects taken 2 at a time, denoted as $P(5, 2)$.

The formula for permutation is $P(n, r) = \frac{n!}{(n-r)!}$.

Here, $n = 5$ (total number of distinct flags) and $r = 2$ (number of flags used to form a signal).

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

$P(5, 2) = \frac{5 \times 4 \times \cancel{3!}}{\cancel{3!}} = 5 \times 4 = 20$

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Thus, the number of different signals that can be generated using 2 flags out of 5 of different colours is 20.

Given:

We need to evaluate factorial expressions.

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To Find:

(i) The value of 5!

(ii) The value of 7!

(iii) The value of 7! - 5!

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Solution:

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

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(i) Evaluate 5!:

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

$5! = 20 \times 6$

$5! = 120$

Thus, the value of 5! is 120.

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(ii) Evaluate 7!:

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

We can also write 7! as $7 \times 6 \times (5 \times 4 \times 3 \times 2 \times 1)$, which is $7 \times 6 \times 5!$.

Using the value of 5! from part (i):

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

$7! = 42 \times 120$

$42 \times 120 = 42 \times 12 \times 10 = 504 \times 10 = 5040$

Thus, the value of 7! is 5040.

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(iii) Evaluate 7! - 5!:

Using the values calculated in parts (i) and (ii):

$7! - 5! = 5040 - 120$

$5040 - 120 = 4920$

Thus, the value of 7! - 5! is 4920.

Given:

We need to compute the values of given factorial expressions.

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To Compute:

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

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

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Solution:

The factorial of a non-negative integer $n$, denoted by $n!$, is the product of all positive integers less than or equal to $n$. $n! = n \times (n-1) \times (n-2) \times \dots \times 2 \times 1$. Also, $0! = 1$.

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(i) Compute $\frac{7\;!}{5\;!}$:

We know that $7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$.

We can write $7!$ as $7 \times 6 \times (5 \times 4 \times 3 \times 2 \times 1) = 7 \times 6 \times 5!$.

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 the denominator:

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

$\frac{7\;!}{5\;!} = 42$

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

We can write $12!$ in terms of $10!$ as $12! = 12 \times 11 \times 10!$.

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

Substitute these into the expression:

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

Cancel out the common term $10!$ from the numerator and the denominator:

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

Perform the multiplication in the numerator and division:

$\frac{12!}{(10!) (2!)} = \frac{132}{2}$

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

Alternatively, using cancellation format:

$\frac{12 \times 11}{2} = \frac{\cancel{12}^{6} \times 11}{\cancel{2}_{1}} = 6 \times 11 = 66$

Given:

The expression is $\frac{n!}{r! \;(n \;-\; r)!}$.

The given values are $n = 5$ and $r = 2$.

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To Evaluate:

Evaluate $\frac{n!}{r! \;(n \;-\; r)!}$ when $n=5$ and $r=2$.

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Solution:

Substitute the given values of $n$ and $r$ into the expression:

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

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First, calculate the term in the parenthesis in the denominator:

$(5 - 2)! = 3!$

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So the expression becomes:

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

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Now, calculate the factorial values:

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

$2! = 2 \times 1 = 2$

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

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Substitute the factorial values back into the expression:

$\frac{120}{2 \times 6}$

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Perform the multiplication in the denominator:

$\frac{120}{12}$

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Finally, perform the division:

$\frac{120}{12} = 10$

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Alternatively, we can expand the numerator until we reach the largest factorial in the denominator ($3!$) and cancel it out:

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

Cancel $3!$ from the numerator and denominator:

$\frac{5 \times 4}{2!} = \frac{20}{2 \times 1} = \frac{20}{2} = 10$

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Thus, the value of the expression $\frac{n!}{r! \;(n \;-\; r)!}$ when $n=5$ and $r=2$ is 10.

Given:

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

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To Find:

The value of $x$.

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Solution:

We are given the equation:

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

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We can express the factorials in the denominators in terms of the largest factorial, which is $10!$.

We know that $9! = 9 \times 8!$ and $10! = 10 \times 9! = 10 \times (9 \times 8!) = 90 \times 8!$.

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Rewrite the terms on the left side with a common denominator, $10!$.

For the first term, $\frac{1}{8!}$, multiply the numerator and denominator by $\frac{10 \times 9}{10 \times 9} = \frac{90}{90}$:

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

For the second term, $\frac{1}{9!}$, multiply the numerator and denominator by $\frac{10}{10}$:

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

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Substitute these back into the original equation:

$\frac{90}{10!} + \frac{10}{10!} = \frac{x}{10!}$

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Combine the terms on the left side:

$\frac{90 + 10}{10!} = \frac{x}{10!}$

$\frac{100}{10!} = \frac{x}{10!}$

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Since the denominators are equal and non-zero ($10! > 0$), the numerators must be equal.

$x = 100$

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Alternatively:

Multiply both sides of the original equation by $10!$:

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

Distribute $10!$ on the left side:

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

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Express $10!$ in terms of $8!$ and $9!$ for the respective fractions:

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

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Cancel the common factorial terms:

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

$90 + 10 = x$

$100 = x$

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Thus, the value of $x$ is 100.

Given:

We need to evaluate the given factorial expressions.

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To Evaluate:

(i) $8!$

(ii) $4! - 3!$

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Solution:

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

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(i) Evaluate $8!$:

The factorial of 8 is the product of all positive integers from 1 to 8.

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

Calculating the product:

$8 \times 7 = 56$

$56 \times 6 = 336$

$336 \times 5 = 1680$

$1680 \times 4 = 6720$

$6720 \times 3 = 20160$

$20160 \times 2 = 40320$

$40320 \times 1 = 40320$

$8! = 40320$

Thus, the value of 8! is 40320.

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(ii) Evaluate $4! - 3!$:

First, calculate the value of $4!$:

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

$4! = 12 \times 2$

$4! = 24$

Next, calculate the value of $3!$:

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

$3! = 6$

Now, perform the subtraction $4! - 3!$:

$4! - 3! = 24 - 6$

$4! - 3! = 18$

Thus, the value of $4! - 3!$ is 18.

Given:

The equation to check is $3! + 4! = 7!$.

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To Verify:

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

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Solution:

First, calculate the value of each factorial term separately.

Calculate $3!$:

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

Calculate $4!$:

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

Calculate $7!$:

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

We know that $4! = 24$, so $7! = 7 \times 6 \times 5 \times 4! = 42 \times 5 \times 24 = 210 \times 24 = 5040$.

Alternatively, calculate the full product:

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

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Now, evaluate the left side of the equation, $3! + 4!$:

Left Side $= 3! + 4! = 6 + 24 = 30$

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The right side of the equation is $7!$:

Right Side $= 7! = 5040$

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Compare the values of the left side and the right side:

Left Side = 30

Right Side = 5040

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

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The statement "Is $3! + 4! = 7!$?" is answered with No.

Given:

We need to compute the value of the expression $\frac{8!}{6! \; \times \; 2!}$.

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To Compute:

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

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Solution:

We have the expression $\frac{8!}{6! \; \times \; 2!}$.

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

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Expand the factorial in the numerator until we reach the largest factorial in the denominator ($6!$):

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

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

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

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Also, calculate the value of $2!$:

$2! = 2 \times 1 = 2$

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Substitute these expanded forms back into the expression:

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

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

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

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Perform the multiplication in the numerator and the division:

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

---

Alternatively, using cancellation format:

$\frac{8 \times 7 \times \cancel{6!}}{\cancel{6!} \times 2} = \frac{8 \times 7}{2} = \frac{\cancel{8}^{4} \times 7}{\cancel{2}_{1}} = 4 \times 7 = 28$

---

Thus, the value of $\frac{8!}{6! \; \times \; 2!}$ is 28.

Given:

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

---

To Find:

The value of $x$.

---

Solution:

We are given the equation:

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

---

We can rewrite the terms on the left side of the equation by expressing the larger factorials in terms of the smaller ones. We know that $7! = 7 \times 6!$ and $8! = 8 \times 7! = 8 \times (7 \times 6!) = 56 \times 6!$.

Rewrite the left side of the equation with a common denominator, which can be $7!$ or $8!$. Let's choose $7!$ initially:

$\frac{1}{6!} = \frac{1}{6!} \times \frac{7}{7} = \frac{7}{7 \times 6!} = \frac{7}{7!}$

So the equation becomes:

$\frac{7}{7!} + \frac{1}{7!} = \frac{x}{8!}$

---

Combine the terms on the left side:

$\frac{7 + 1}{7!} = \frac{x}{8!}$

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

---

Now, we can cross-multiply or multiply both sides by $8!$. Let's multiply both sides by $8!$:

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

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

---

Express $8!$ as $8 \times 7!$ in the fraction on the left side:

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

---

Cancel out the common term $7!$:

$8 \times 8 = x$

$64 = x$

---

Thus, the value of $x$ is 64.

---

Alternatively:

Start with the original equation:

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

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

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

Distribute $8!$ on the left side:

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

Express $8!$ in terms of $6!$ and $7!$ for the respective terms:

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

Cancel the common factorial terms:

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

$56 + 8 = x$

$64 = x$

The value of $x$ is 64.

Given:

The expression is $\frac{n!}{(n \;-\; r)!}$.

We need to evaluate this expression for two sets of values for $n$ and $r$.

---

To Evaluate:

(i) $\frac{n!}{(n \;-\; r)!}$ when $n = 6$, $r = 2$.

(ii) $\frac{n!}{(n \;-\; r)!}$ when $n = 9$, $r = 5$.

---

Solution:

The expression $\frac{n!}{(n \;-\; r)!}$ represents the number of permutations of $n$ distinct objects taken $r$ at a time, denoted as $P(n, r)$.

---

(i) Evaluate $\frac{n!}{(n \;-\; r)!}$ when $n = 6$, $r = 2$:

Substitute the given values into the expression:

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

Expand $6!$ until $4!$ and simplify:

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

Cancel out the common term $4!$:

$\frac{6 \times 5 \times \cancel{4!}}{\cancel{4!}} = 6 \times 5 = 30$

Thus, the value of $\frac{n!}{(n \;-\; r)!}$ when $n = 6, r = 2$ is 30.

---

(ii) Evaluate $\frac{n!}{(n \;-\; r)!}$ when $n = 9$, $r = 5$:

Substitute the given values into the expression:

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

Expand $9!$ until $4!$ and simplify:

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

Cancel out the common term $4!$:

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

Calculate the product:

$9 \times 8 = 72$

$72 \times 7 = 504$

$504 \times 6 = 3024$

$3024 \times 5 = 15120$

Thus, the value of $\frac{n!}{(n \;-\; r)!}$ when $n = 9, r = 5$ is 15120.

Given:

The word is ALLAHABAD.

---

To Find:

The number of different permutations of the letters of the word ALLAHABAD.

---

Solution:

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

The letters in the word are A, L, L, A, H, A, B, A, D.

Total number of letters ($n$) = 9.

---

Let's count the occurrences of each distinct letter:

The letter 'A' appears 4 times.

The letter 'L' appears 2 times.

The letter 'H' appears 1 time.

The letter 'B' appears 1 time.

The letter 'D' appears 1 time.

---

The total number of letters is $4 + 2 + 1 + 1 + 1 = 9$, which matches the total number of letters in the word.

---

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

Number of permutations $= \frac{n!}{p_1! \; p_2! \; \dots \; p_k!}$

---

In this case, $n = 9$. The distinct letters and their frequencies are:

$p_1$ (for 'A') = 4

$p_2$ (for 'L') = 2

The other letters ('H', 'B', 'D') appear only once, so their frequencies are $1! = 1$, and we can omit them in the denominator's product.

---

Using the formula, the number of permutations is:

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

---

Calculate the factorial values:

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

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

$2! = 2 \times 1 = 2$

---

Substitute the values into the expression:

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

Expand $9!$ until $4!$ to simplify the calculation:

Number of permutations $= \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4! \times 2}$

Cancel $4!$ from the numerator and denominator:

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

Perform the multiplication and division:

Number of permutations $= \frac{15120}{2} = 7560$

---

Alternatively, perform cancellation first:

$\frac{9 \times \cancel{8}^{4} \times 7 \times 6 \times 5}{\cancel{2}_{1}} = 9 \times 4 \times 7 \times 6 \times 5 = 36 \times 210 = 7560$

---

Thus, the number of different permutations of the letters of the word ALLAHABAD is 7560.

Given:

The available digits are 1, 2, 3, 4, 5, 6, 7, 8, 9.

Total number of distinct digits ($n$) = 9.

We need to form 4-digit numbers.

Number of digits to be used in the number ($r$) = 4.

Repetition of digits is not allowed.

---

To Find:

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

---

Solution:

A 4-digit number has four places: thousands, hundreds, tens, and units.

Available digits: {1, 2, 3, 4, 5, 6, 7, 8, 9}.

Since repetition of digits is not allowed, this is a permutation problem where we are selecting and arranging 4 digits out of 9 distinct digits.

We need to find the number of permutations of 9 distinct objects taken 4 at a time, which is denoted by $P(9, 4)$.

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

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

---

Substitute the given values $n=9$ and $r=4$ into the formula:

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

---

Now, calculate the factorial values and simplify by expanding the numerator until the $5!$ in the denominator:

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

Cancel out the common term $5!$:

$P(9, 4) = 9 \times 8 \times 7 \times 6 = 3024$

---

Thus, the number of 4-digit numbers that can be formed using the digits 1 to 9 without repetition is 3024.

Given:

The available digits are 0, 1, 2, 3, 4, 5.

Total number of distinct available digits = 6.

We need to form numbers lying between 100 and 1000.

This means we need to form 3-digit numbers.

Repetition of digits is not allowed.

---

To Find:

The number of 3-digit numbers that can be formed using the given digits without repetition, such that the number is between 100 and 1000.

---

Solution:

A number between 100 and 1000 is a 3-digit number.

A 3-digit number has three places: the hundreds place, the tens place, and the units place.

Available digits: {0, 1, 2, 3, 4, 5}.

---

We need to fill these three places using the available digits such that the number is between 100 and 1000, and no digit is repeated.

---

Consider the hundreds place (the first digit):

The hundreds digit cannot be 0, because if it were, the number would be a 2-digit number (or less), not a 3-digit number between 100 and 1000.

So, the possible digits for the hundreds place are {1, 2, 3, 4, 5}.

Number of choices for the hundreds place = 5.

---

Consider the tens place (the second digit):

One digit has been used for the hundreds place (it was one of 1, 2, 3, 4, 5). The remaining digits are the unused digits from the original set {0, 1, 2, 3, 4, 5}. There are 5 digits remaining (the 0 is now available). For example, if 1 was used for the hundreds place, the remaining available digits are {0, 2, 3, 4, 5}, which is 5 digits.

So, there are 5 choices for the tens place.

---

Consider the units place (the third digit):

Two distinct digits have been used for the hundreds and tens places. From the original 6 digits, $6 - 2 = 4$ digits remain.

So, there are 4 choices for the units place.

---

By the Fundamental Principle of Counting (Multiplication Principle), the total number of different 3-digit numbers that can be formed without repetition, starting with a non-zero digit, is the product of the number of choices for each place.

Number of numbers = (Choices for Hundreds Place) $\times$ (Choices for Tens Place) $\times$ (Choices for Units Place)

Number of numbers $= 5 \times 5 \times 4 = 25 \times 4 = 100$

---

Thus, the number of numbers lying between 100 and 1000 that can be formed with the digits 0, 1, 2, 3, 4, 5 without repetition is 100.

Given:

We need to find the value of $n$ satisfying the given equations involving permutations.

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

---

To Find:

The value of $n$ for each equation.

---

Solution:

---

(i) Solve $^nP_5 = 42\; ^nP_3$ for $n$, where $n > 4$.

Write the permutation terms using the formula:

$^nP_5 = \frac{n!}{(n-5)!}$

$^nP_3 = \frac{n!}{(n-3)!}$

---

Substitute these into the given equation:

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

---

Since $n > 4$, $n! \neq 0$. We can divide both sides by $n!$:

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

---

Cross-multiply:

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

---

Expand $(n-3)!$ until $(n-5)!$:

$(n-3)! = (n-3) \times (n-3-1)! = (n-3) \times (n-4)!$

$(n-3)! = (n-3) \times (n-4) \times (n-4-1)! = (n-3) \times (n-4) \times (n-5)!$

---

Substitute this expanded form back into the equation:

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

---

Since $n > 4$, $(n-5)! \neq 0$. We can divide both sides by $(n-5)!$:

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

---

Expand the left side:

$n^2 - 4n - 3n + 12 = 42$

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

Subtract 42 from both sides to form a quadratic equation:

$n^2 - 7n + 12 - 42 = 0$

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

---

Solve the quadratic equation by factoring. We need two numbers that multiply to -30 and add up to -7. These numbers are -10 and 3.

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

---

This gives two possible values for $n$:

$n - 10 = 0 \implies n = 10$

$n + 3 = 0 \implies n = -3$

---

The problem states that $n > 4$. Therefore, we must choose the value of $n$ that satisfies this condition.

$n = 10$ satisfies $10 > 4$.

$n = -3$ does not satisfy $n > 4$.

---

Thus, the value of $n$ is 10.

---

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

Write the permutation terms using the formula:

$^nP_4 = \frac{n!}{(n-4)!}$

$^{n-1}P_4 = \frac{(n-1)!}{((n-1)-4)!} = \frac{(n-1)!}{(n-5)!}$

---

Substitute these into the given equation:

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

---

Rewrite the left side by multiplying the numerator by the reciprocal of the denominator:

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

---

Rearrange the terms:

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

---

Simplify the factorial ratios. We know that $n! = n \times (n-1)!$ and $(n-4)! = (n-4) \times (n-5)!$.

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

---

Cancel out the common factorial terms:

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

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

---

Cross-multiply:

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

$3n = 5n - 20$

---

Subtract $5n$ from both sides:

$3n - 5n = -20$

$-2n = -20$

---

Divide both sides by -2:

$n = \frac{-20}{-2} = 10$

---

Check if the solution satisfies the condition $n > 4$. Since $10 > 4$, the solution is valid.

---

Thus, the value of $n$ is 10.

Given:

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

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

---

To Find:

The value of $r$.

---

Solution:

The given equation is:

$5 \; ^4P_r = 6 \; ^5P_{r-1}$

---

For the permutation $^nP_r$ to be defined, we must have $n \geq r \geq 0$.

From $^4P_r$, we have $4 \geq r \geq 0$.

From $^5P_{r-1}$, we have $5 \geq r-1 \geq 0$. This implies $5+1 \geq r \geq 0+1$, so $6 \geq r \geq 1$.

Combining these conditions, we need $4 \geq r \geq 1$. Possible integer values for $r$ are 1, 2, 3, 4.

---

Write the permutation terms using the formula:

$^4P_r = \frac{4!}{(4-r)!}$

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

---

Substitute these into the given equation:

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

---

Expand $5!$ as $5 \times 4!$ and $(6-r)!$ as $(6-r) \times (6-r-1)! = (6-r) \times (5-r)! = (6-r) \times (5-r) \times (5-r-1)! = (6-r) \times (5-r) \times (4-r)!$:

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

---

Cancel out the common terms $4!$ and $(4-r)!$ from both sides (since $4 \geq r \geq 1$, $(4-r)!$ is defined and non-zero):

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

---

Divide both sides by 5 (since $5 \neq 0$):

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

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

---

Multiply both sides by $(6-r)(5-r)$:

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

---

Expand the left side:

$30 - 6r - 5r + r^2 = 6$

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

Subtract 6 from both sides to form a quadratic equation:

$r^2 - 11r + 30 - 6 = 0$

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

---

Solve the quadratic equation by factoring. We need two numbers that multiply to 24 and add up to -11. These numbers are -3 and -8.

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

---

This gives two possible values for $r$:

$r - 3 = 0 \implies r = 3$

$r - 8 = 0 \implies r = 8$

---

We established that the possible integer values for $r$ are 1, 2, 3, 4. We need to check which of the obtained values satisfy this condition.

$r = 3$ satisfies $1 \leq 3 \leq 4$.

$r = 8$ does not satisfy $1 \leq r \leq 4$.

---

Thus, the value of $r$ is 3.

Given:

The word is DAUGHTER.

We need to form 8-letter arrangements using all the letters of the word.

---

To Find:

(i) The number of arrangements where all vowels occur together.

(ii) The number of arrangements where all vowels do not occur together.

---

Solution:

The word DAUGHTER has 8 distinct letters: D, A, U, G, H, T, E, R.

Number of letters ($n$) = 8.

The vowels in the word are A, U, E.

The consonants in the word are D, G, H, T, R.

Number of vowels = 3.

Number of consonants = 5.

---

(i) All vowels occur together:

Treat the group of all vowels (AUE) as a single unit or block. Let this unit be V.

The remaining letters are the consonants: D, G, H, T, R.

Now we need to arrange the consonants and the vowel block: D, G, H, T, R, V.

We have a total of 5 consonants + 1 vowel block = 6 units to arrange.

These 6 units are distinct (since the consonants are distinct and the vowel block is treated as a single distinct item). The number of ways to arrange these 6 distinct units is $6!$.

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

---

Within the vowel block (AUE), the 3 vowels can be arranged among themselves in $3!$ ways.

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

---

By the Fundamental Principle of Counting, the total number of arrangements where all vowels occur together is the product of the number of ways to arrange the units and the number of ways to arrange the vowels within their block.

Number of arrangements with vowels together = (Arrangements of 6 units) $\times$ (Arrangements of 3 vowels within the block)

Number of arrangements with vowels together $= 6! \times 3! = 720 \times 6 = 4320$

Thus, the number of different 8-letter arrangements where all vowels occur together is 4320.

---

(ii) All vowels do not occur together:

The total number of arrangements of the 8 distinct letters of the word DAUGHTER without any restrictions is the number of permutations of 8 distinct objects taken 8 at a time, which is $P(8, 8) = 8!$.

$8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320$

---

The number of arrangements where all vowels do not occur together is equal to the total number of arrangements minus the number of arrangements where all vowels do occur together.

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

Number of arrangements with vowels not together $= 8! - (6! \times 3!)$

Number of arrangements with vowels not together $= 40320 - 4320 = 36000$

Thus, the number of different 8-letter arrangements where all vowels do not occur together is 36000.

Given:

We have discs of three different colours: 4 red, 3 yellow, and 2 green.

The discs of the same colour are indistinguishable.

---

To Find:

The number of different ways the discs can be arranged in a row.

---

Solution:

First, find the total number of discs ($n$).

Total number of discs $= \text{Number of red discs} + \text{Number of yellow discs} + \text{Number of green discs}$

$n = 4 + 3 + 2 = 9$ discs.

---

We have a total of 9 items, but some of them are identical (indistinguishable) by colour.

Number of red discs ($p_1$) = 4.

Number of yellow discs ($p_2$) = 3.

Number of green discs ($p_3$) = 2.

---

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

Number of permutations $= \frac{n!}{p_1! \; p_2! \; \dots \; p_k!}$

---

Using this formula with $n=9$, $p_1=4$, $p_2=3$, and $p_3=2$:

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

---

Calculate the factorials in the denominator:

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

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

$2! = 2 \times 1 = 2$

The product of the denominators is $4! \times 3! \times 2! = 24 \times 6 \times 2 = 144 \times 2 = 288$.

---

Now, calculate the factorial in the numerator:

$9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 362880$

---

Substitute these values back into the expression:

Number of arrangements $= \frac{362880}{24 \times 6 \times 2} = \frac{362880}{288}$

---

Perform the division:

$\frac{362880}{288} = 1260$

---

Alternatively, we can simplify by expanding the numerator and canceling terms:

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

Cancel $4!$:

$\frac{9 \times 8 \times 7 \times 6 \times 5}{3 \times 2 \times 1 \times 2 \times 1} = \frac{9 \times 8 \times 7 \times 6 \times 5}{6 \times 2}$

Cancel $6$ and $2$ from the numerator with the denominator:

$\frac{9 \times \cancel{8}^{4} \times 7 \times \cancel{6}^{1} \times 5}{\cancel{6} \times \cancel{2}_{1}} = 9 \times 4 \times 7 \times 5 = 36 \times 35$

$36 \times 35 = 1260$

---

Thus, the number of different arrangements is 1260.

Given:

The word is INDEPENDENCE.

---

Analysis of the word INDEPENDENCE:

Let's count the total number of letters and the frequency of each distinct letter in the word INDEPENDENCE.

Total number of letters in the word = $12$.

The distinct letters and their counts are:

I appears $1$ time.

N appears $3$ times.

D appears $2$ times.

E appears $4$ times.

P appears $1$ time.

C appears $1$ time.

Total letters = $1 + 3 + 2 + 4 + 1 + 1 = 12$.

The vowels in the word are I, E, E, E, E. Total number of vowels = $1 + 4 = 5$.

The consonants in the word are N, N, N, D, D, P, C. Total number of consonants = $3 + 2 + 1 + 1 = 7$.

---

Calculation of Total Arrangements:

The total number of arrangements of $n$ objects where $p_1$ objects are of one type, $p_2$ are of a second type, ..., $p_k$ are of a $k^{th}$ type, and $p_1 + p_2 + \dots + p_k = n$, is given by the formula:

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

Here, $n=12$, and the frequencies are $1$ (I), $3$ (N), $2$ (D), $4$ (E), $1$ (P), $1$ (C).

Total number of arrangements of the letters of INDEPENDENCE is:

$ \frac{12!}{1! \; 3! \; 2! \; 4! \; 1! \; 1!} = \frac{12!}{3! \; 2! \; 4!} $

Calculate the factorials:

$12! = 47,90,01,600$

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

$2! = 2 \times 1 = 2$

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

Denominator $= 3! \times 2! \times 4! = 6 \times 2 \times 24 = 12 \times 24 = 288$.

Total number of arrangements $= \frac{479001600}{288}$.

$ \frac{479001600}{288} = 1663200 $

Expressing this in the Indian numbering system with commas:

Total number of arrangements $= 16,63,200$.

---

Solution to part (i): Words start with P

If the word must start with P, we fix the letter 'P' in the first position.

We need to arrange the remaining $12 - 1 = 11$ letters in the remaining $11$ positions.

The remaining letters are: I(1), N(3), D(2), E(4), C(1).

The number of arrangements of these 11 letters is:

$ \frac{11!}{1! \; 3! \; 2! \; 4! \; 1!} = \frac{11!}{3! \; 2! \; 4!} $

Calculate the factorials:

$11! = 3,99,16,800$

$3! = 6$, $2! = 2$, $4! = 24$. Denominator $= 6 \times 2 \times 24 = 288$.

Number of arrangements starting with P $= \frac{39916800}{288}$.

$ \frac{39916800}{288} = 138600 $

Expressing this in the Indian numbering system with commas:

Number of arrangements starting with P $= 1,38,600$.

---

Solution to part (ii): All the vowels always occur together

The vowels are I, E, E, E, E. Treat these $5$ vowels as a single block (unit).

The remaining letters are the $7$ consonants: N, N, N, D, D, P, C.

Now we consider the arrangement of the $7$ consonants and the $1$ vowel block. This makes a total of $7 + 1 = 8$ units.

The units are: (IEEEE), N, N, N, D, D, P, C.

Among these 8 units, the consonants have repetitions:

N appears $3$ times.

D appears $2$ times.

P appears $1$ time.

C appears $1$ time.

The vowel block (IEEEE) is a single distinct unit here.

The number of ways to arrange these 8 units is:

$ \frac{8!}{3! \; 2! \; 1! \; 1! \; 1!} = \frac{8!}{3! \; 2!} $

$ \frac{8!}{3! \; 2!} = \frac{40320}{6 \times 2} = \frac{40320}{12} = 3360 $

Number of arrangements of the 8 units $= 3,360$.

---

Next, consider the arrangements within the vowel block (IEEEE). The $5$ vowels have repetitions:

I appears $1$ time.

E appears $4$ times.

The number of ways to arrange these 5 vowels within the block is:

$ \frac{5!}{1! \; 4!} = \frac{5 \times 4!}{1 \times 4!} = 5 $

Number of arrangements within the vowel block $= 5$.

---

By the Fundamental Principle of Counting, the total number of arrangements where all vowels occur together is the product of the number of ways to arrange the units and the number of ways to arrange the vowels within their block.

Number of arrangements with vowels together = (Arrangements of 8 units) $\times$ (Arrangements of vowels within block)

Number of arrangements with vowels together $= 3,360 \times 5 = 16,800$.

Number of arrangements with vowels together $= 16,800$.

---

Solution to part (iii): Do the vowels never occur together

The number of arrangements where the vowels never occur together is the total number of arrangements minus the number of arrangements where all vowels occur together.

Total number of arrangements $= 16,63,200$ (from Total Arrangements calculation).

Number of arrangements with vowels together $= 16,800$ (from part ii calculation).

Number of arrangements with vowels never together = Total arrangements - Arrangements with vowels together

Number of arrangements with vowels never together $= 16,63,200 - 16,800$.

Expressing the result in the Indian numbering system with commas:

Number of arrangements with vowels never together $= 16,46,400$.

---

Solution to part (iv): Words begin with I and end in P

If the word must begin with I and end with P, we fix 'I' in the first position and 'P' in the last position.

We need to arrange the remaining $12 - 2 = 10$ letters in the middle $10$ positions.

The original letters are I, N, N, N, D, D, E, E, E, E, P, C.

With I and P fixed, the remaining letters are: N, N, N, D, D, E, E, E, E, C.

These 10 letters have the following repetitions:

N appears $3$ times.

D appears $2$ times.

E appears $4$ times.

C appears $1$ time.

The number of arrangements of these 10 letters is:

$ \frac{10!}{3! \; 2! \; 4! \; 1!} = \frac{10!}{3! \; 2! \; 4!} $

Calculate the factorials:

$10! = 36,28,800$

$3! = 6$, $2! = 2$, $4! = 24$. Denominator $= 6 \times 2 \times 24 = 288$.

Number of arrangements beginning with I and ending in P $= \frac{3628800}{288}$.

$ \frac{3628800}{288} = 12600 $

Expressing this in the Indian numbering system with commas:

Number of arrangements beginning with I and ending in P $= 12,600$.

---

Answer Summary:

Total number of arrangements of the letters of INDEPENDENCE is $16,63,200$.

(i) The number of arrangements that start with P is $1,38,600$.

(ii) The number of arrangements where all the vowels always occur together is $16,800$.

(iii) The number of arrangements where the vowels never occur together is $16,46,400$.

(iv) The number of arrangements that begin with I and end in P is $12,600$.

Given:

The available digits are 1, 2, 3, 4, 5, 6, 7, 8, 9.

Total number of distinct digits ($n$) = 9.

We need to form 3-digit numbers.

Number of digits to be used in the number ($r$) = 3.

No digit is repeated.

---

To Find:

The number of different 3-digit numbers that can be formed using the given digits if no digit is repeated.

---

Solution:

A 3-digit number has three places: the hundreds place, the tens place, and the units place.

Available digits: {1, 2, 3, 4, 5, 6, 7, 8, 9}.

Since no digit is repeated, the digits used for each place must be distinct.

---

Consider the hundreds place (the first digit):

We can use any of the 9 available digits.

Number of choices for the hundreds place = 9.

---

Consider the tens place (the second digit):

One digit has been used for the hundreds place, and it cannot be repeated. So, there are $9 - 1 = 8$ distinct digits remaining for the tens place.

Number of choices for the tens place = 8.

---

Consider the units place (the third digit):

Two distinct digits have been used for the hundreds and tens places. So, there are $9 - 2 = 7$ distinct digits remaining for the units place.

Number of choices for the units place = 7.

---

By the Fundamental Principle of Counting (Multiplication Principle), the total number of different 3-digit numbers that can be formed without repetition is the product of the number of choices for each place.

Number of 3-digit numbers = (Choices for Hundreds Place) $\times$ (Choices for Tens Place) $\times$ (Choices for Units Place)

Number of 3-digit numbers $= 9 \times 8 \times 7 = 72 \times 7 = 504$

---

Alternatively, this problem is a permutation of 9 distinct objects taken 3 at a time, which is denoted by $P(9, 3)$.

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

Here, $n = 9$ (total number of distinct digits) and $r = 3$ (number of digits to be used to form the number).

$P(9, 3) = \frac{9!}{(9-3)!} = \frac{9!}{6!}$

$P(9, 3) = \frac{9 \times 8 \times 7 \times 6!}{6!} = 9 \times 8 \times 7 = 504$

---

Thus, the number of 3-digit numbers that can be formed by using the digits 1 to 9 without repetition is 504.

Given:

We need to form 4-digit numbers.

The available digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

Total number of distinct available digits = 10.

No digit is repeated.

---

To Find:

The number of different 4-digit numbers that can be formed with no digit repeated.

---

Solution:

A 4-digit number has four places: thousands, hundreds, tens, and units.

Available digits: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.

Since no digit is repeated, the digits used for each place must be distinct.

---

Consider the thousands place (the first digit):

The thousands digit cannot be 0, because if it were, the number would be a 3-digit number (or less). The possible digits for the thousands place are {1, 2, 3, 4, 5, 6, 7, 8, 9}.

Number of choices for the thousands place = 9.

---

Consider the hundreds place (the second digit):

One non-zero digit has been used for the thousands place. The remaining available digits are the 9 digits from the original set {0, 1, ..., 9} that were not used for the thousands place. This set of 9 digits includes 0.

So, there are 9 distinct digits remaining for the hundreds place.

Number of choices for the hundreds place = 9.

---

Consider the tens place (the third digit):

Two distinct digits have been used for the thousands and hundreds places. From the original 10 digits, $10 - 2 = 8$ distinct digits remain.

Number of choices for the tens place = 8.

---

Consider the units place (the fourth digit):

Three distinct digits have been used for the thousands, hundreds, and tens places. From the original 10 digits, $10 - 3 = 7$ distinct digits remain.

Number of choices for the units place = 7.

---

By the Fundamental Principle of Counting (Multiplication Principle), the total number of different 4-digit numbers with no digit repeated is the product of the number of choices for each place.

Number of 4-digit numbers = (Thousands Place Choices) $\times$ (Hundreds Place Choices) $\times$ (Tens Place Choices) $\times$ (Units Place Choices)

Number of 4-digit numbers $= 9 \times 9 \times 8 \times 7$

Number of 4-digit numbers $= 4536$

---

Thus, the number of 4-digit numbers with no digit repeated is 4536.

Given:

The available digits are 1, 2, 3, 4, 6, 7.

Total number of distinct available digits = 6.

We need to form 3-digit even numbers.

No digit is repeated.

---

To Find:

The number of possible 3-digit even numbers using the given digits without repetition.

---

Solution:

A 3-digit number has three places: the hundreds place, the tens place, and the units place.

Available digits: {1, 2, 3, 4, 6, 7}.

For the number to be even, the units place must be filled with an even digit from the available set.

The even digits from the available set {1, 2, 3, 4, 6, 7} are 2, 4, and 6.

Number of choices for the units place = 3 (2, 4, or 6).

---

Consider the hundreds place (the first digit):

One digit (an even digit) has been used for the units place. From the original 6 distinct digits, $6 - 1 = 5$ digits remain. These 5 remaining digits can be used for the hundreds place (none of the original digits is 0, so the hundreds place will not be 0).

Number of choices for the hundreds place = 5.

---

Consider the tens place (the second digit):

One digit has been used for the units place, and another distinct digit has been used for the hundreds place. From the original 6 distinct digits, $6 - 2 = 4$ distinct digits remain for the tens place.

Number of choices for the tens place = 4.

---

By the Fundamental Principle of Counting (Multiplication Principle), the total number of 3-digit even numbers that can be formed without repetition is the product of the number of choices for each place.

Number of 3-digit even numbers = (Choices for Hundreds Place) $\times$ (Choices for Tens Place) $\times$ (Choices for Units Place)

Number of 3-digit even numbers $= 5 \times 4 \times 3 = 20 \times 3 = 60$

---

Thus, the number of 3-digit even numbers that can be made using the digits 1, 2, 3, 4, 6, 7 without repetition is 60.

Given:

The available digits are 1, 2, 3, 4, 5.

Total number of distinct available digits ($n$) = 5.

We need to form 4-digit numbers.

No digit is repeated.

---

To Find:

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

(ii) The number of these 4-digit numbers that will be even.

---

Solution:

---

(i) Total number of 4-digit numbers:

We need to select and arrange 4 digits out of the 5 distinct available digits {1, 2, 3, 4, 5} without repetition.

This is a permutation of 5 distinct objects taken 4 at a time, denoted as $P(5, 4)$.

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

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

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

Alternatively, using the Multiplication Principle:

Hundreds place: 5 choices

Tens place: 4 choices (after using one for hundreds)

Thousands place: 3 choices (after using two)

Units place: 2 choices (after using three)

Number of 4-digit numbers $= 5 \times 4 \times 3 \times 2 = 120$.

Thus, the total number of 4-digit numbers that can be formed without repetition is 120.

---

(ii) Number of even 4-digit numbers:

For a number to be even, the units place must be filled with an even digit from the available set {1, 2, 3, 4, 5}.

The even digits are 2 and 4.

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

---

Now, consider the remaining three places (thousands, hundreds, and tens) using the remaining digits. We have 4 digits left from the original set of 5 (as one even digit was used for the units place).

We need to arrange 3 of these 4 remaining digits in the thousands, hundreds, and tens places without repetition.

This is a permutation of 4 distinct objects taken 3 at a time, denoted as $P(4, 3)$.

$P(4, 3) = \frac{4!}{(4-3)!} = \frac{4!}{1!} = 4!$

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

Number of ways to arrange the remaining 4 digits in the other 3 places $= 4 \times 3 \times 2 = 24$.

---

By the Fundamental Principle of Counting, the total number of 4-digit even numbers formed without repetition is the product of the number of choices for the units place and the number of ways to arrange the remaining digits in the other places.

Number of even 4-digit numbers = (Choices for Units Place) $\times$ (Arrangements of remaining 4 digits in 3 places)

Number of even 4-digit numbers $= 2 \times P(4, 3) = 2 \times 24 = 48$.

---

Thus, among the 120 possible 4-digit numbers, 48 will be even.

Given:

Total number of persons in the committee ($n$) = 8.

We need to choose a chairman and a vice chairman.

The two positions (chairman and vice chairman) are distinct.

One person cannot hold more than one position (no repetition of persons).

---

To Find:

The number of ways to choose a chairman and a vice chairman from 8 persons, with no repetition.

---

Solution:

We need to select 2 persons from the 8 available persons and assign them to specific roles (Chairman and Vice Chairman). The order in which we select and assign matters, because choosing person A as Chairman and person B as Vice Chairman is different from choosing person B as Chairman and person A as Vice Chairman.

This is a permutation problem where we are selecting and arranging 2 persons out of 8 distinct persons.

We need to find the number of permutations of 8 distinct objects taken 2 at a time, which is denoted by $P(8, 2)$.

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

Here, $n = 8$ (total number of persons) and $r = 2$ (number of positions to fill).

$P(8, 2) = \frac{8!}{(8-2)!} = \frac{8!}{6!}$

---

Calculate the factorial values and simplify:

$P(8, 2) = \frac{8 \times 7 \times 6!}{6!}$

Cancel out the common term $6!$:

$P(8, 2) = 8 \times 7 = 56$

---

Alternatively, using the Fundamental Principle of Counting:

Number of choices for the Chairman position = 8 (any of the 8 persons).

Once the Chairman is chosen, there are $8 - 1 = 7$ persons remaining.

Number of choices for the Vice Chairman position = 7 (any of the remaining 7 persons).

By the Multiplication Principle, the total number of ways to choose a Chairman and a Vice Chairman is:

Number of ways $= (\text{Choices for Chairman}) \times (\text{Choices for Vice Chairman})$

Number of ways $= 8 \times 7 = 56$

---

Thus, there are 56 ways to choose a chairman and a vice chairman from a committee of 8 persons, assuming one person cannot hold more than one position.

Given:

The ratio of permutations is given as $^{n – 1}P_3 : ^nP_4 = 1 : 9$. This can be written as the equation $\frac{^{n – 1}P_3}{^nP_4} = \frac{1}{9}$.

---

To Find:

The value of $n$.

---

Solution:

The given equation is:

The formula for permutations is $^mP_r = \frac{m!}{(m-r)!}$.

For $^{n-1}P_3$, we have $m = n-1$ and $r = 3$. For this permutation to be defined, we require $n-1 \geq 3$, which implies $n \geq 4$. Thus, $^{n-1}P_3 = \frac{(n-1)!}{((n-1)-3)!} = \frac{(n-1)!}{(n-4)!}$.

For $^nP_4$, we have $m = n$ and $r = 4$. For this permutation to be defined, we require $n \geq 4$. Thus, $^nP_4 = \frac{n!}{(n-4)!}$.

Substitute these expressions into equation (1):

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

Rewrite the left side by multiplying the numerator by the reciprocal of the denominator:

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

Assuming $n \geq 4$, $(n-4)!$ is defined and non-zero, so we can cancel it from the numerator and the denominator on the left side:

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

We know that $n! = n \times (n-1)!$. Substitute this into the denominator on the left side:

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

Assuming $n \geq 4$, $(n-1)!$ is defined and non-zero, so we can cancel it from the numerator and the denominator on the left side:

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

Taking the reciprocal of both sides (or cross-multiplying):

$n = 9$

We check if the value $n=9$ satisfies the condition $n \geq 4$. Since $9 \geq 4$, the value $n=9$ is valid.

---

Answer:

The value of $n$ is $9$.

Given:

We need to find the value of $r$ satisfying the given equations involving permutations: (i) $^{5}P_r = 2 \; ^{6}P_{r-1}$ and (ii) $^{5}P_r = ^{6}P_{r-1}$.

---

To Find:

The value of $r$ for each equation.

---

Solution:

The formula for permutations is $^nP_r = \frac{n!}{(n-r)!}$. For a permutation $^nP_r$ to be defined, we must have $n \geq r \geq 0$.

For $^{5}P_r$, we need $5 \geq r \geq 0$.

For $^{6}P_{r-1}$, we need $6 \geq r-1 \geq 0$, which implies $6+1 \geq r \geq 0+1$, so $7 \geq r \geq 1$.

Combining these conditions, the possible integer values for $r$ that make both permutations defined are those where $5 \geq r \geq 1$, i.e., $r \in \{1, 2, 3, 4, 5\}$. We must check our solutions for $r$ against this set.

---

(i) Solve $^{5}P_r = 2 \; ^{6}P_{r-1}$ for $r$.

Using the permutation formula, we write the terms:

$^{5}P_r = \frac{5!}{(5-r)!}$

$^{6}P_{r-1} = \frac{6!}{(6-(r-1))!} = \frac{6!}{(6-r+1)!} = \frac{6!}{(7-r)!}$

Substitute these into the given equation:

Expand $6!$ as $6 \times 5!$ and $(7-r)!$ as $(7-r) \times (6-r) \times (5-r)!$ (since $r \leq 5$, $7-r \geq 2$ and $6-r \geq 1$). Substitute these expansions into equation (1):

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

Assuming $r \leq 5$, $5!$ and $(5-r)!$ are defined and non-zero. Cancel $5!$ from both sides:

$\frac{1}{(5-r)!} = \frac{12}{(7-r)(6-r)(5-r)!}$

Multiply both sides by $(5-r)!$:

$1 = \frac{12}{(7-r)(6-r)}$

Multiply both sides by $(7-r)(6-r)$:

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

Expand the left side:

$42 - 7r - 6r + r^2 = 12$

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

Subtract 12 from both sides to form a quadratic equation:

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

Solve the quadratic equation by factoring. We look for two numbers that multiply to $30$ and add up to $-13$. These numbers are $-3$ and $-10$.

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

This gives two possible values for $r$: $r - 3 = 0 \implies r = 3$ or $r - 10 = 0 \implies r = 10$.

We check these values against the valid range for $r$, which is $r \in \{1, 2, 3, 4, 5\}$.

$r = 3$ falls within the valid range.

$r = 10$ does not fall within the valid range.

Thus, the value of $r$ for the first equation is $3$.

---

(ii) Solve $^{5}P_r = ^{6}P_{r-1}$ for $r$.

Using the permutation formula, we write the terms as in part (i):

$^{5}P_r = \frac{5!}{(5-r)!}$

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

Substitute these into the given equation:

Expand $6!$ as $6 \times 5!$ and $(7-r)!$ as $(7-r)(6-r)(5-r)!$ (since $r \leq 5$, $7-r \geq 2$ and $6-r \geq 1$). Substitute these expansions into equation (2):

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

Assuming $r \leq 5$, $5!$ and $(5-r)!$ are defined and non-zero. Cancel $5!$ from both sides:

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

Multiply both sides by $(5-r)!$:

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

Multiply both sides by $(7-r)(6-r)$:

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

Expand the left side:

$42 - 7r - 6r + r^2 = 6$

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

Subtract 6 from both sides to form a quadratic equation:

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

Solve the quadratic equation by factoring. We look for two numbers that multiply to $36$ and add up to $-13$. These numbers are $-4$ and $-9$.

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

This gives two possible values for $r$: $r - 4 = 0 \implies r = 4$ or $r - 9 = 0 \implies r = 9$.

We check these values against the valid range for $r$, which is $r \in \{1, 2, 3, 4, 5\}$.

$r = 4$ falls within the valid range.

$r = 9$ does not fall within the valid range.

Thus, the value of $r$ for the second equation is $4$.

---

Answer:

(i) The value of $r$ is $3$.

(ii) The value of $r$ is $4$.

Given:

The word is EQUATION.

We need to form words using all the letters of the word.

Each letter is used exactly once.

---

To Find:

The number of different words that can be formed.

---

Solution:

First, let's identify the letters in the word EQUATION and check if there are any repetitions.

The letters are E, Q, U, A, T, I, O, N.

Counting the letters, we have 8 distinct letters.

Total number of letters ($n$) = 8.

---

We need to form words using all 8 letters, and each letter is used exactly once. This means we are finding the number of ways to arrange 8 distinct letters in 8 positions.

This is a permutation of 8 distinct objects taken 8 at a time, denoted as $P(8, 8)$ or simply $8!$.

The formula for the number of permutations of $n$ distinct objects is $n!$.

Number of arrangements $= 8!$

---

Calculate the factorial value:

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

$8! = 40320$

---

Thus, the number of words, with or without meaning, that can be formed using all the letters of the word EQUATION, using each letter exactly once, is 40320.

Given:

The word is MONDAY.

No letter is repeated (all letters in MONDAY are distinct).

---

Analysis of the word MONDAY:

The letters in the word MONDAY are M, O, N, D, A, Y.

Total number of distinct letters ($n$) = 6.

The vowels in the word are O, A.

Number of vowels = 2.

The consonants in the word are M, N, D, Y.

Number of consonants = 4.

---

To Find:

The number of words formed under different conditions.

---

Solution:

Since no letter is repeated, the arrangements are permutations of distinct objects.

---

(i) 4 letters are used at a time:

We need to form 4-letter words using 4 distinct letters out of the 6 available distinct letters (M, O, N, D, A, Y).

This is a permutation of 6 distinct objects taken 4 at a time, denoted by $P(6, 4)$.

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

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

$P(6, 4) = \frac{6 \times 5 \times 4 \times 3 \times 2!}{2!} = 6 \times 5 \times 4 \times 3 = 30 \times 12 = 360$

Thus, the number of 4-letter words that can be formed is 360.

---

(ii) All letters are used at a time:

We need to form words using all 6 letters of MONDAY, with no repetition.

This is a permutation of 6 distinct objects taken 6 at a time, denoted by $P(6, 6)$ or $6!$.

$P(6, 6) = \frac{6!}{(6-6)!} = \frac{6!}{0!} = \frac{6!}{1} = 6!$

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

Thus, the number of words formed using all letters at a time is 720.

---

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

We need to form 6-letter words using all letters of MONDAY, with no repetition, and the first letter must be a vowel.

The vowels in MONDAY are O and A (2 vowels).

Consider the first position (the first letter):

Number of choices for the first position = 2 (either O or A).

---

Consider the remaining 5 positions:

One vowel has been used for the first position. There are $6 - 1 = 5$ letters remaining from the original set of 6 distinct letters.

We need to arrange these 5 remaining distinct letters in the remaining 5 positions without repetition.

This is a permutation of 5 distinct objects taken 5 at a time, which is $P(5, 5) = 5!$.

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

Number of ways to arrange the remaining 5 letters in the remaining 5 positions = 120.

---

By the Fundamental Principle of Counting, the total number of words formed with the first letter being a vowel is the product of the number of choices for the first position and the number of ways to arrange the remaining letters.

Number of words with first letter a vowel = (Choices for First Position) $\times$ (Arrangements of remaining 5 letters)

Number of words with first letter a vowel $= 2 \times 5! = 2 \times 120 = 240$

Thus, the number of words formed using all letters with the first letter being a vowel is 240.

Given:

The word is MISSISSIPPI.

---

Analysis of the word MISSISSIPPI:

Let's count the total number of letters and the frequency of each distinct letter.

Total number of letters ($n$) = 11.

---

Letters and their frequencies:

M: 1

I: 4

S: 4

P: 2

Sum of frequencies: $1 + 4 + 4 + 2 = 11$.

---

To Find:

The number of distinct permutations of the letters in MISSISSIPPI where the four I's do not come together.

---

Solution:

The number of distinct permutations where the four I's do not come together is equal to the total number of distinct permutations minus the number of distinct permutations where the four I's come together.

---

Step 1: Calculate the total number of distinct permutations.

Using the formula for permutations with repetitions, where $n=11$, with 4 I's, 4 S's, and 2 P's:

Total number of distinct permutations $= \frac{11!}{4! \; 4! \; 2!}$

Calculate the factorials:

$11! = 39916800$

$4! = 24$

$4! = 24$

$2! = 2$

Denominator $= 4! \times 4! \times 2! = 24 \times 24 \times 2 = 576 \times 2 = 1152$.

Total number of distinct permutations $= \frac{39916800}{1152} = 34650$.

---

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

Treat the group of four I's (IIII) as a single unit or block. Let this block be IIII.

The remaining letters are M, S, S, S, S, P, P.

Now we need to arrange these 7 letters and the block of I's: M, S, S, S, S, P, P, (IIII).

We have a total of 7 letters + 1 block = 8 units to arrange.

These 8 units have repetitions among the letters:

M: 1 time

S: 4 times

P: 2 times

The block (IIII) is treated as a single distinct item for this arrangement step.

The number of distinct arrangements of these 8 units is $\frac{8!}{1! \; 4! \; 2! \; 1!} = \frac{8!}{4! \; 2!}$.

Calculate the factorials:

$8! = 40320$

$4! = 24$

$2! = 2$

Denominator $= 4! \times 2! = 24 \times 2 = 48$.

Number of distinct arrangements with I's together $= \frac{40320}{48} = 840$.

---

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

Number of arrangements with I's not together = (Total number of distinct permutations) - (Number of distinct arrangements with I's together)

Number of arrangements with I's not together $= 34650 - 840 = 33810$.

---

Thus, the number of distinct permutations of the letters in MISSISSIPPI where the four I's do not come together is 33810.

Given:

The word is PERMUTATIONS.

---

Analysis of the word PERMUTATIONS:

Let's count the total number of letters and the frequency of each distinct letter.

Total number of letters ($n$) = 12.

---

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

Sum of frequencies: $1 \times 10 + 2 = 12$.

All letters are distinct except for the letter 'T', which appears 2 times.

---

The vowels in the word are E, U, A, I, O.

Number of vowels = 5.

The consonants in the word are P, R, M, T, T, N, S.

Number of consonants = 7.

Total letters = 5 (vowels) + 7 (consonants) = 12.

---

The total number of distinct arrangements of the letters in PERMUTATIONS $= \frac{12!}{2!}$.

$12! = 479001600$

$2! = 2$

Total arrangements $= \frac{479001600}{2} = 239500800$.

---

Now, let's address each part of the question.

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(i) Words start with P and end with S:

Fix the letter 'P' in the first position and the letter 'S' in the last position. The remaining 10 letters need to be arranged in the middle 10 positions.

The original letters are P, E, R, M, U, T, T, A, I, O, N, S.

The letters P and S are fixed. The remaining 10 letters are: E, R, M, U, T, T, A, I, O, N.

These 10 letters contain one repetition: 'T' appears 2 times.

The number of distinct arrangements of these 10 letters in the 10 middle positions is $\frac{10!}{2!}$.

$10! = 3628800$

$2! = 2$

Number of arrangements starting with P and ending with S $= \frac{10!}{2!} = \frac{3628800}{2} = 1814400$.

Thus, the number of arrangements that start with P and end with S is 1814400.

---

(ii) Vowels are all together:

The vowels are E, U, A, I, O. These are 5 distinct vowels.

Treat these 5 vowels as a single unit or block. Let this block be VVVVV.

The remaining letters are the 7 consonants: P, R, M, T, T, N, S.

Now we need to arrange the 7 consonants and the vowel block: P, R, M, T, T, N, S, (EU A I O).

We have a total of 7 consonants + 1 vowel block = 8 units to arrange.

These 8 units have one repetition: 'T' appears 2 times among the consonants. The vowel block (EU A I O) is treated as a single distinct item for this arrangement step.

The number of distinct arrangements of these 8 units is $\frac{8!}{2! \; 1! \; 1! \; 1! \; 1! \; 1! \; 1!} = \frac{8!}{2!}$.

$8! = 40320$

$2! = 2$

Number of arrangements of units $= \frac{40320}{2} = 20160$.

---

Within the vowel block (EU A I O), the 5 distinct vowels can be arranged among themselves in $5!$ ways.

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

---

By the Fundamental Principle of Counting, the total number of arrangements where all vowels occur together is the product of the number of ways to arrange the units and the number of ways to arrange the vowels within their block.

Number of arrangements with vowels together = (Arrangements of 8 units) $\times$ (Arrangements of 5 distinct vowels within the block)

Number of arrangements with vowels together $= \frac{8!}{2!} \times 5! = 20160 \times 120 = 2419200$.

Thus, the number of arrangements where all the vowels always occur together is 2419200.

---

(iii) There are always 4 letters between P and S:

The word has 12 positions.

We need to place P and S such that there are exactly 4 letters between them.

Let the positions be numbered 1 to 12.

Possible positions for (P, S) with 4 letters in between:

(1, 6), (2, 7), (3, 8), (4, 9), (5, 10), (6, 11), (7, 12).

There are 7 such pairs of positions where P comes before S.

Positions: P _ _ _ _ S

The letters P and S are distinct. Once the positions are chosen, there are 2 ways to arrange P and S in those positions (P in the first chosen position, S in the second, OR S in the first chosen position, P in the second).

So, there are $7 \times 2 = 14$ ways to choose and place P and S such that there are 4 letters between them.

---

After placing P and S, the remaining $12 - 2 = 10$ letters need to be arranged in the remaining 10 positions.

The remaining letters are the letters of PERMUTATIONS excluding one P and one S: E, R, M, U, T, T, A, I, O, N.

These 10 remaining letters have one repetition: 'T' appears 2 times.

The number of distinct arrangements of these 10 letters in the remaining 10 positions is $\frac{10!}{2!}$.

$\frac{10!}{2!} = \frac{3628800}{2} = 1814400$.

---

By the Fundamental Principle of Counting, 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.

Number of arrangements = (Ways to place P and S with 4 letters between) $\times$ (Arrangements of remaining 10 letters)

Number of arrangements $= 14 \times \frac{10!}{2!} = 14 \times 1814400 = 25401600$.

Thus, the number of arrangements where there are always 4 letters between P and S is 25401600.

Given:

The equation is $^nC_9 = ^nC_8$.

---

To Find:

The value of $^nC_{17}$.

---

Solution:

We are given the equation $^nC_9 = ^nC_8$.

---

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 $b = 8$.

Case 1: $a = b$

$9 = 8$ (This is false).

Case 2: $a + b = n$

$9 + 8 = n$

$17 = n$

So, the value of $n$ is 17.

---

The combination formula $^nC_r$ is defined for non-negative integers $n$ and $r$ such that $0 \leq r \leq n$. With $n=17$, the given combinations $^nC_9 = ^{17}C_9$ and $^nC_8 = ^{17}C_8$ are well-defined since $0 \leq 9 \leq 17$ and $0 \leq 8 \leq 17$.

---

Now we need to find the value of $^nC_{17}$ with $n=17$.

We need to evaluate $^{17}C_{17}$.

---

The formula for $^nC_r$ is $^nC_r = \frac{n!}{r! \; (n-r)!}$.

Using this formula for $^{17}C_{17}$:

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

We know that $0! = 1$.

$^{17}C_{17} = \frac{17!}{17! \times 1} = \frac{17!}{17!} = 1$

---

Alternatively, the combination $^nC_n$ represents the number of ways to choose $n$ objects from a set of $n$ distinct objects, which is always 1 (choose all of them).

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

---

Thus, the value of $^nC_{17}$ is 1.

Given:

A group consists of 2 men and 3 women.

A committee of 3 persons is to be constituted from this group.

---

To Find:

(i) The total number of ways to constitute a committee of 3 persons.

(ii) The number of these committees that would consist of exactly 1 man and 2 women.

---

Solution:

The total number of persons in the group is $2 + 3 = 5$.

We need to choose a committee of 3 persons from these 5 persons. The order in which the persons are chosen does not matter in forming a committee, so this is a combination problem.

---

(i) Total number of ways to constitute a committee of 3 persons:

We need to choose 3 persons from a total of 5 distinct persons.

The number of ways to do this is given by the combination formula $^nC_r = \frac{n!}{r! \; (n-r)!}$, where $n$ is the total number of objects and $r$ is the number of objects to choose.

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

Total number of committees $= ^5C_3 = \frac{5!}{3! \; (5-3)!} = \frac{5!}{3! \; 2!}$

Calculate the factorial values:

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

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

$2! = 2 \times 1 = 2$

Total number of committees $= \frac{120}{6 \times 2} = \frac{120}{12} = 10$.

Alternatively, using cancellation:

$^5C_3 = \frac{5 \times 4 \times \cancel{3!}}{\cancel{3!} \times 2!} = \frac{5 \times 4}{2 \times 1} = \frac{20}{2} = 10$.

Thus, the total number of ways to constitute a committee of 3 persons is 10.

---

(ii) Number of committees with 1 man and 2 women:

We need to choose 1 man from the 2 available men, AND 2 women from the 3 available women.

The number of ways to choose 1 man from 2 men is $^2C_1$.

$^2C_1 = \frac{2!}{1! \; (2-1)!} = \frac{2!}{1! \; 1!} = \frac{2}{1 \times 1} = 2$. (Alternatively, choosing 1 item from 2 is simply 2 ways).

---

The number of ways to choose 2 women from 3 women is $^3C_2$.

$^3C_2 = \frac{3!}{2! \; (3-2)!} = \frac{3!}{2! \; 1!} = \frac{3 \times 2!}{2! \times 1} = 3$. (Alternatively, $^3C_2$ = $^3C_{3-2}$ = $^3C_1 = 3$).

---

Since the choice of men and the choice of women are independent events, by the Fundamental Principle of Counting (Multiplication Principle), the total number of committees consisting of 1 man and 2 women is the product of the number of ways to choose the men and the number of ways to choose the women.

Number of committees with 1 man and 2 women = (Ways to choose 1 man) $\times$ (Ways to choose 2 women)

Number of committees with 1 man and 2 women = $^2C_1 \times$ $^3C_2 = 2 \times 3 = 6$.

Thus, the number of committees that would consist of 1 man and 2 women is 6.

Given:

A standard deck of $52$ playing cards.

---

Analysis of a Standard Deck:

A standard deck has $52$ cards divided into $4$ suits:

Hearts (Red), Diamonds (Red), Clubs (Black), Spades (Black).

Each suit has $13$ cards: Ace (A), $2, 3, 4, 5, 6, 7, 8, 9, 10$, Jack (J), Queen (Q), King (K).

Total red cards = $13$ (Hearts) + $13$ (Diamonds) = $26$.

Total black cards = $13$ (Clubs) + $13$ (Spades) = $26$.

Face cards are J, Q, K. There are $3$ face cards in each of the $4$ suits. Total face cards = $3 \times 4 = 12$.

---

To Find:

The number of ways to choose $4$ cards such that:

(i) four cards are of the same suit,

(ii) four cards belong to four different suits,

(iii) are face cards,

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

(v) cards are of the same colour.

---

Solution - Total Number of Ways to Choose 4 Cards:

The total number of ways to choose $4$ cards from a pack of $52$ distinct cards is given by the combination formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$, where $n=52$ (total cards) and $r=4$ (cards to choose).

Total number of ways $= \binom{52}{4}$

$ \binom{52}{4} = \frac{52!}{4!(52-4)!} = \frac{52!}{4!48!} $

Expanding the factorials:

$ \binom{52}{4} = \frac{52 \times 51 \times 50 \times 49 \times \cancel{48!}}{ (4 \times 3 \times 2 \times 1) \times \cancel{48!} } $

Simplify the denominator: $4 \times 3 \times 2 \times 1 = 24$.

$ \binom{52}{4} = \frac{52 \times 51 \times 50 \times 49}{24} $

Cancel common factors:

$ \binom{52}{4} = \frac{\cancel{52}^{13} \times \cancel{51}^{17} \times \cancel{50}^{25} \times 49}{\cancel{24}_1} = 13 \times 17 \times 25 \times 49 $

Calculate the product:

$13 \times 17 = 221$

$25 \times 49 = 1225$

$221 \times 1225 = 270725$

Total number of ways to choose $4$ cards from $52$ is $2,70,725$.

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Solution to part (i): Four cards are of the same suit

For the four cards to be of the same suit, all four must come from one specific suit (Hearts, Diamonds, Clubs, or Spades).

Each suit has $13$ cards.

Number of ways to choose 4 cards from the 13 cards of Hearts $= \binom{13}{4}$.

Number of ways to choose 4 cards from the 13 cards of Diamonds $= \binom{13}{4}$.

Number of ways to choose 4 cards from the 13 cards of Clubs $= \binom{13}{4}$.

Number of ways to choose 4 cards from the 13 cards of Spades $= \binom{13}{4}$.

The number of ways to choose 4 cards of the same suit is the sum of the ways for each suit:

Number of ways $= \binom{13}{4} + \binom{13}{4} + \binom{13}{4} + \binom{13}{4} = 4 \times \binom{13}{4}$.

Calculate $\binom{13}{4}$:

$ \binom{13}{4} = \frac{13!}{4!9!} = \frac{13 \times 12 \times 11 \times 10 \times \cancel{9!}}{ (4 \times 3 \times 2 \times 1) \times \cancel{9!} } = \frac{13 \times 12 \times 11 \times 10}{24} $

Cancel common factors:

$ \binom{13}{4} = \frac{13 \times \cancel{12}^1 \times 11 \times \cancel{10}^5}{\cancel{24}_2} = \frac{13 \times 1 \times 11 \times 5}{\cancel{2}_1} = 13 \times 11 \times 5 = 715 $

Number of ways to choose 4 cards of the same suit $= 4 \times 715 = 2860$.

The number of ways to choose 4 cards of the same suit is $2,860$.

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Solution to part (ii): Four cards belong to four different suits

To choose four cards belonging to four different suits, we must select exactly one card from each of the four suits (Hearts, Diamonds, Clubs, Spades).

Number of ways to choose 1 card from the 13 Hearts $= \binom{13}{1} = 13$.

Number of ways to choose 1 card from the 13 Diamonds $= \binom{13}{1} = 13$.

Number of ways to choose 1 card from the 13 Clubs $= \binom{13}{1} = 13$.

Number of ways to choose 1 card from the 13 Spades $= \binom{13}{1} = 13$.

By the Fundamental Principle of Counting, the number of ways to make these selections is the product of the number of ways for each selection.

Number of ways $= \binom{13}{1} \times \binom{13}{1} \times \binom{13}{1} \times \binom{13}{1} = 13 \times 13 \times 13 \times 13 = 13^4$.

$13^2 = 169$

$13^4 = (13^2)^2 = 169^2 = 28561$.

The number of ways to choose four cards belonging to four different suits is $28,561$.

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Solution to part (iii): Are face cards

There are $12$ face cards in total in a standard deck (J, Q, K from each of the 4 suits).

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

Number of ways $= \binom{12}{4}$.

$ \binom{12}{4} = \frac{12!}{4!(12-4)!} = \frac{12!}{4!8!} = \frac{12 \times 11 \times 10 \times 9 \times \cancel{8!}}{ (4 \times 3 \times 2 \times 1) \times \cancel{8!} } = \frac{12 \times 11 \times 10 \times 9}{24} $

Cancel common factors:

$ \binom{12}{4} = \frac{\cancel{12}^1 \times 11 \times \cancel{10}^5 \times 9}{\cancel{24}_2} = \frac{1 \times 11 \times 5 \times 9}{\cancel{2}_1} = 11 \times 5 \times 9 = 495 $

The number of ways to choose 4 face cards is $495$.

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Solution to part (iv): Two are red cards and two are black cards

There are $26$ red cards and $26$ black cards in a deck.

We need to choose $2$ red cards from the $26$ red cards AND $2$ black cards from the $26$ black cards.

Number of ways to choose 2 red cards $= \binom{26}{2}$.

Number of ways to choose 2 black cards $= \binom{26}{2}$.

By the Fundamental Principle of Counting, the number of ways to choose 2 red cards and 2 black cards is the product of the number of ways for each selection.

Number of ways $= \binom{26}{2} \times \binom{26}{2}$.

Calculate $\binom{26}{2}$:

$ \binom{26}{2} = \frac{26!}{2!(26-2)!} = \frac{26!}{2!24!} = \frac{26 \times 25 \times \cancel{24!}}{ (2 \times 1) \times \cancel{24!} } = \frac{26 \times 25}{2} $

Cancel common factors:

$ \binom{26}{2} = \frac{\cancel{26}^{13} \times 25}{\cancel{2}_1} = 13 \times 25 = 325 $

Number of ways to choose 2 red and 2 black cards $= 325 \times 325$.

$325 \times 325 = 1,05,625$.

The number of ways to choose two red cards and two black cards is $1,05,625$.

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Solution to part (v): Cards are of the same colour

For the four cards to be of the same colour, they must either be all red or all black.

There are $26$ red cards.

Number of ways to choose 4 red cards from 26 red cards $= \binom{26}{4}$.

Calculate $\binom{26}{4}$:

$ \binom{26}{4} = \frac{26!}{4!(26-4)!} = \frac{26!}{4!22!} = \frac{26 \times 25 \times 24 \times 23 \times \cancel{22!}}{ (4 \times 3 \times 2 \times 1) \times \cancel{22!} } = \frac{26 \times 25 \times 24 \times 23}{24} $

Cancel common factors:

$ \binom{26}{4} = \frac{26 \times 25 \times \cancel{24} \times 23}{\cancel{24}} = 26 \times 25 \times 23 $

$26 \times 25 = 650$

$650 \times 23 = 14950$

Number of ways to choose 4 red cards $= 14,950$.

There are $26$ black cards.

Number of ways to choose 4 black cards from 26 black cards $= \binom{26}{4} = 14,950$ (same calculation as for red cards).

Number of ways to choose 4 cards of the same colour = (Ways to choose 4 Red cards) + (Ways to choose 4 Black cards)

Number of ways $= 14,950 + 14,950 = 29,900$.

The number of ways to choose 4 cards of the same colour is $29,900$.

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Answer Summary:

The total number of ways to choose 4 cards from a pack of 52 playing cards is $2,70,725$.

(i) The number of ways where four cards are of the same suit is $2,860$.

(ii) The number of ways where four cards belong to four different suits is $28,561$.

(iii) The number of ways where the four cards are face cards is $495$.

(iv) The number of ways where two cards are red and two are black is $1,05,625$.

(v) The number of ways where the four cards are of the same colour is $29,900$.

Solution:

Given that $\binom{n}{8} = \binom{n}{2}$.

We know the property of combinations that if $\binom{n}{a} = \binom{n}{b}$, then either $a=b$ or $a+b=n$.

In this case, $a=8$ and $b=2$. Since $8 \neq 2$, we must have $a+b=n$.

---

Now we need to find the value of $\binom{n}{2}$, with $n=10$.

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

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

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

$\binom{10}{2} = \frac{10 \times 9}{2 \times 1}$

$\binom{10}{2} = \frac{\cancel{10}^{5} \times 9}{\cancel{2} \times 1}$

$\binom{10}{2} = 5 \times 9$

Thus, the value of $\binom{n}{2}$ is $45$.

We are given ratios involving combinations. The formula for combination is $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.

First, let's write down the general expressions for $\binom{2n}{3}$ and $\binom{n}{3}$:

$\binom{2n}{3} = \frac{(2n)!}{3!(2n-3)!} = \frac{2n(2n-1)(2n-2)(2n-3)!}{3 \times 2 \times 1 \times (2n-3)!} = \frac{2n(2n-1)2(n-1)}{6} = \frac{2n(2n-1)(n-1)}{3}$

$\binom{n}{3} = \frac{n!}{3!(n-3)!} = \frac{n(n-1)(n-2)(n-3)!}{3 \times 2 \times 1 \times (n-3)!} = \frac{n(n-1)(n-2)}{6}$

For these combinations to be defined, we must have $2n \ge 3$ and $n \ge 3$. This implies $n \ge 3$. Also, the denominators cannot be zero, which means $n \neq 0$ and $n \neq 1$. The condition $n \ge 3$ satisfies these.

---

(i) $\binom{2n}{3} : \binom{n}{3} = 12 : 1$

This can be written as $\frac{\binom{2n}{3}}{\binom{n}{3}} = \frac{12}{1}$.

Substitute the expressions for the combinations:

$\frac{\frac{2n(2n-1)(n-1)}{3}}{\frac{n(n-1)(n-2)}{6}} = 12$

Multiply the numerator by the reciprocal of the denominator:

$\frac{2n(2n-1)(n-1)}{3} \times \frac{6}{n(n-1)(n-2)} = 12$

Assuming $n \ge 3$, we have $n \neq 0$ and $n-1 \neq 0$. We can cancel $n$ and $(n-1)$ from the numerator and denominator:

$\frac{2(2n-1) \times \cancel{6}^2}{\cancel{3} \times (n-2)} = 12$

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

Divide both sides by 4:

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

Multiply both sides by $(n-2)$:

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

$2n-1 = 3n - 6$

Rearrange the terms to solve for $n$:

$3n - 2n = 6 - 1$

Since $n=5$ satisfies the condition $n \ge 3$, this is a valid solution.

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(ii) $\binom{2n}{3} : \binom{n}{3} = 11 : 1$

This can be written as $\frac{\binom{2n}{3}}{\binom{n}{3}} = \frac{11}{1}$.

Substitute the expressions for the combinations as derived before:

$\frac{2n(2n-1)(n-1)}{3} \times \frac{6}{n(n-1)(n-2)} = 11$

Assuming $n \ge 3$, cancel common terms:

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

Multiply both sides by $(n-2)$:

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

$8n - 4 = 11n - 22$

Rearrange the terms to solve for $n$:

$11n - 8n = 22 - 4$

$3n = 18$

Divide by 3:

Since $n=6$ satisfies the condition $n \ge 3$, this is a valid solution.

Solution:

A chord is a line segment joining any two distinct points on a circle.

To draw a chord, we need to select 2 points from the given 21 points on the circle.

The order in which we choose the two points does not matter (choosing point A and then point B results in the same chord as choosing point B and then point A). This is a combination problem.

The number of ways to choose 2 points out of 21 is given by $\binom{21}{2}$.

---

Number of chords = $\binom{21}{2}$

Using the combination formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$, with $n=21$ and $r=2$:

$\binom{21}{2} = \frac{21!}{2!(21-2)!} = \frac{21!}{2!19!}$

$\binom{21}{2} = \frac{21 \times 20 \times 19!}{2 \times 1 \times 19!}$

Cancel out the $19!$ from the numerator and denominator:

$\binom{21}{2} = \frac{21 \times 20}{2}$

$\binom{21}{2} = \frac{21 \times \cancel{20}^{10}}{\cancel{2}}$

$\binom{21}{2} = 21 \times 10$

Therefore, $210$ chords can be drawn through 21 points on a circle.

Solution:

We need to select a team consisting of 3 boys and 3 girls from a group of 5 boys and 4 girls.

The selection of boys and the selection of girls are independent events.

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Number of ways to select 3 boys from 5 boys is given by the combination $\binom{5}{3}$.

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

So, there are $10$ ways to select 3 boys from 5 boys.

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Number of ways to select 3 girls from 4 girls is given by the combination $\binom{4}{3}$.

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

So, there are $4$ ways to select 3 girls from 4 girls.

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To find the total number of ways to select a team of 3 boys and 3 girls, we multiply the number of ways to select the boys by the number of ways to select the girls.

Total number of ways = (Ways to select boys) $\times$ (Ways to select girls)

Total number of ways = $\binom{5}{3} \times \binom{4}{3} = 10 \times 4 = 40$

Therefore, a team of 3 boys and 3 girls can be selected in $40$ ways from 5 boys and 4 girls.

Solution:

We need to select a total of 9 balls with the condition that there are exactly 3 balls of each colour (red, white, and blue).

We have: - 6 red balls - 5 white balls - 5 blue balls

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Number of ways to select 3 red balls from 6 red balls = $\binom{6}{3}$.

$\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4 \times 3!}{3 \times 2 \times 1 \times 3!} = \frac{\cancel{6}^1 \times 5 \times 4}{\cancel{6}} = 20$

There are $20$ ways to select 3 red balls.

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Number of ways to select 3 white balls from 5 white balls = $\binom{5}{3}$.

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

There are $10$ ways to select 3 white balls.

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Number of ways to select 3 blue balls from 5 blue balls = $\binom{5}{3}$.

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

There are $10$ ways to select 3 blue balls.

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Since the selection of balls of each colour is independent, the total number of ways to make the required selection is the product of the number of ways for each colour.

Total number of ways = (Ways to select red balls) $\times$ (Ways to select white balls) $\times$ (Ways to select blue balls)

Total number of ways = $\binom{6}{3} \times \binom{5}{3} \times \binom{5}{3} = 20 \times 10 \times 10 = 2000$

Therefore, there are $2000$ ways to select 9 balls with 3 balls of each colour.

Solution:

We need to form a 5-card combination from a standard deck of 52 cards such that each combination contains exactly one ace.

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

To form a combination with exactly one ace, we need to make two independent selections:

1. Select exactly one ace from the 4 available aces.

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

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Number of ways to select 1 ace from 4 aces = $\binom{4}{1}$.

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

There are $4$ ways to select one ace.

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Number of ways to select the remaining 4 cards from the 48 non-ace cards = $\binom{48}{4}$.

$\binom{48}{4} = \frac{48!}{4!(48-4)!} = \frac{48!}{4!44!}$

$\binom{48}{4} = \frac{48 \times 47 \times 46 \times 45 \times 44!}{4 \times 3 \times 2 \times 1 \times 44!}$

Cancel out the $44!$ from the numerator and denominator:

$\binom{48}{4} = \frac{48 \times 47 \times 46 \times 45}{4 \times 3 \times 2 \times 1}$

$\binom{48}{4} = \frac{\cancel{48}^{2}}{\cancel{4 \times 3 \times 2 \times 1}} \times 47 \times 46 \times 45$ (Since $4 \times 3 \times 2 \times 1 = 24$, and $48/24 = 2$)

$\binom{48}{4} = 2 \times 47 \times 46 \times 45$

$\binom{48}{4} = 2 \times 47 \times (46 \times 45)$

$46 \times 45 = 2070$

$\binom{48}{4} = 2 \times 47 \times 2070$

$\binom{48}{4} = 94 \times 2070$

$\binom{48}{4} = 194580$

There are $194580$ ways to select 4 non-ace cards.

---

To find the total number of 5-card combinations with exactly one ace, we multiply the number of ways to select the ace by the number of ways to select the remaining 4 cards.

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

Total number of ways = $\binom{4}{1} \times \binom{48}{4} = 4 \times 194580$

$4 \times 194580 = 778320$

Therefore, the number of 5-card combinations with exactly one ace is $778320$.

Given:

Total number of players available = $17$.

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

The team to be selected must have exactly $11$ players.

The team must include exactly $4$ bowlers.

---

Analysis of Player Groups:

The 17 players available can be divided into two groups:

1. Bowlers: $5$ players.

2. Non-bowlers: Total players - Number of bowlers = $17 - 5 = 12$ players.

---

Team Composition Required:

The team of $11$ players must have exactly $4$ bowlers.

If the team has exactly $4$ bowlers, the remaining players must be non-bowlers.

Number of non-bowlers in the team = Total team size - Number of bowlers in the team

Number of non-bowlers in the team = $11 - 4 = 7$.

So, to form the team, we need to select:

a) Exactly $4$ bowlers from the $5$ available bowlers.

b) Exactly $7$ non-bowlers from the $12$ available non-bowlers.

---

Solution:

The number of ways to select $k$ items from a set of $n$ items is given by the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

Step 1: Select the bowlers.

We need to select $4$ bowlers from the $5$ available bowlers.

Number of ways to select bowlers $= \binom{5}{4}$.

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

There are $5$ ways to select the $4$ bowlers.

---

Step 2: Select the non-bowlers.

We need to select $7$ non-bowlers from the $12$ available non-bowlers.

Number of ways to select non-bowlers $= \binom{12}{7}$.

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

Expand the factorials in the numerator and the larger factorial in the denominator to cancel terms:

$ \binom{12}{7} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times \cancel{7!}}{ \cancel{7!} \times (5 \times 4 \times 3 \times 2 \times 1) } = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} $

Simplify the denominator: $5 \times 4 \times 3 \times 2 \times 1 = 120$.

$ \binom{12}{7} = \frac{12 \times 11 \times 10 \times 9 \times 8}{120} $

Perform cancellations:

$ \binom{12}{7} = \frac{\cancel{12}^{1} \times 11 \times \cancel{10}^{1} \times 9 \times 8}{\cancel{120}_{1}} = 11 \times 9 \times 8 $

$11 \times 9 = 99$

$99 \times 8 = 792$

There are $792$ ways to select the $7$ non-bowlers.

---

Step 3: Combine the selections.

To get the total number of ways to form the team, we multiply the number of ways to select the bowlers by the number of ways to select the non-bowlers (by the Fundamental Principle of Counting).

Total number of ways to select the team = (Ways to select 4 bowlers) $\times$ (Ways to select 7 non-bowlers)

Total number of ways $= \binom{5}{4} \times \binom{12}{7} = 5 \times 792$.

$5 \times 792 = 3960$.

The total number of ways to select a cricket team of eleven with exactly 4 bowlers is $3,960$.

---

Answer:

A cricket team of eleven players with exactly 4 bowlers can be selected in $3,960$ ways.

Solution:

We are given a bag containing 5 black balls and 6 red balls.

We need to select a total of $2 + 3 = 5$ balls such that exactly 2 are black and exactly 3 are red.

The selection of black balls and the selection of red balls are independent events.

---

Number of ways to select 2 black balls from 5 black balls is given by the combination $\binom{5}{2}$.

Using the combination formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$, with $n=5$ and $r=2$:

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

$\binom{5}{2} = \frac{5 \times 4 \times 3!}{2 \times 1 \times 3!}$

Cancel out the $3!$ from the numerator and denominator:

$\binom{5}{2} = \frac{5 \times 4}{2}$

$\binom{5}{2} = \frac{\cancel{20}^{10}}{\cancel{2}^1}$

$\binom{5}{2} = 10$

There are $10$ ways to select 2 black balls.

---

Number of ways to select 3 red balls from 6 red balls is given by the combination $\binom{6}{3}$.

Using the combination formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$, with $n=6$ and $r=3$:

$\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!}$

$\binom{6}{3} = \frac{6 \times 5 \times 4 \times 3!}{3 \times 2 \times 1 \times 3!}$

Cancel out the $3!$ from the numerator and denominator:

$\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1}$

$\binom{6}{3} = \frac{\cancel{120}}{\cancel{6}}$

$\binom{6}{3} = 20$

There are $20$ ways to select 3 red balls.

---

To find the total number of ways to select 2 black balls and 3 red balls, we multiply the number of ways to select the black balls by the number of ways to select the red balls.

Total number of ways = (Ways to select black balls) $\times$ (Ways to select red balls)

Total number of ways = $\binom{5}{2} \times \binom{6}{3} = 10 \times 20 = 200$

Therefore, there are $200$ ways to select 2 black balls and 3 red balls.

Solution:

Total number of courses available = 9.

Number of courses to be chosen for the programme = 5.

Number of specific compulsory courses = 2.

---

Since 2 specific courses are compulsory, the student must select these 2 courses.

Number of courses already selected (compulsory) = 2.

Number of courses remaining to be selected = Total courses needed - Compulsory courses = $5 - 2 = 3$ courses.

---

These remaining 3 courses must be selected from the non-compulsory courses.

Number of non-compulsory courses available = Total available courses - Compulsory courses = $9 - 2 = 7$ courses.

---

The problem now reduces to finding the number of ways to select 3 courses from the 7 available non-compulsory courses. This is a combination problem.

The number of ways to select 3 courses from 7 is given by $\binom{7}{3}$.

Using the combination formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$, with $n=7$ and $r=3$:

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

$\binom{7}{3} = \frac{7 \times 6 \times 5 \times 4!}{3 \times 2 \times 1 \times 4!}$

Cancel out the $4!$ from the numerator and denominator:

$\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1}$

$\binom{7}{3} = \frac{7 \times \cancel{6}}{\cancel{6}}$ (Since $3 \times 2 \times 1 = 6$)

$\binom{7}{3} = 7 \times 5$

Thus, there are $35$ ways to select the remaining 3 courses.

Since the 2 compulsory courses are always included, the total number of ways to choose the programme of 5 courses is the number of ways to choose the remaining 3 courses.

Therefore, the number of ways a student can choose the programme is $35$.

Solution:

The given word is INVOLUTE.

Let's identify the vowels and consonants in the word:

Vowels: I, O, U, E

Consonants: N, V, L, T

Total number of vowels = 4

Total number of consonants = 4

---

We need to form words each containing exactly 3 vowels and 2 consonants.

This process involves two steps:

1. Selecting the required number of vowels and consonants from the available letters.

2. Arranging the selected letters to form words.

---

Step 1: Selection of letters

We need to select 3 vowels from the 4 available vowels. The number of ways to do this is given by the combination $\binom{4}{3}$.

Number of ways to select 3 vowels = $\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4!}{3!1!} = \frac{4 \times 3!}{3! \times 1} = 4$

We need to select 2 consonants from the 4 available consonants. The number of ways to do this is given by the combination $\binom{4}{2}$.

Number of ways to select 2 consonants = $\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3 \times 2!}{2 \times 1 \times 2!} = \frac{12}{2} = 6$

The total number of ways to select 3 vowels and 2 consonants is the product of the number of ways to select the vowels and the number of ways to select the consonants.

Total number of ways to select the letters = (Ways to select 3 vowels) $\times$ (Ways to select 2 consonants)

Total number of ways to select the letters = $\binom{4}{3} \times \binom{4}{2} = 4 \times 6 = 24$

There are 24 different combinations of 5 letters (3 vowels and 2 consonants) that can be selected from the word INVOLUTE.

---

Step 2: Arrangement of selected letters

For each of the 24 selected combinations of 5 letters, we can arrange these 5 distinct letters in $5!$ ways to form a word.

Number of ways to arrange 5 distinct letters = $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

---

Total number of words formed

The total number of words is the product of the number of ways to select the 5 letters and the number of ways to arrange them.

Total number of words = (Number of ways to select letters) $\times$ (Number of ways to arrange letters)

Total number of words = $24 \times 120$

$24 \times 120 = 2880$

Therefore, $2880$ words, with or without meaning, each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE.

Solution:

Total number of members in the group = Number of girls + Number of boys = $4 + 7 = 11$.

We need to select a team of 5 members from this group of 11 members.

---

(i) The team has no girl

If the team has no girl, then all 5 members must be boys.

We need to select 5 boys from the 7 available boys.

Number of ways to select 5 boys from 7 boys = $\binom{7}{5}$.

$\binom{7}{5} = \frac{7!}{5!(7-5)!} = \frac{7!}{5!2!} = \frac{7 \times 6 \times 5!}{5! \times 2 \times 1} = \frac{7 \times 6}{2} = \frac{42}{2} = 21$

There are $21$ ways to select a team with no girl.

---

(ii) The team has at least one boy and one girl

The total number of ways to select a team of 5 members from the 11 members is $\binom{11}{5}$.

$\binom{11}{5} = \frac{11!}{5!(11-5)!} = \frac{11!}{5!6!} = \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6!}{5 \times 4 \times 3 \times 2 \times 1 \times 6!} = \frac{11 \times 10 \times 9 \times 8 \times 7}{120}$

$\binom{11}{5} = \frac{11 \times \cancel{10}^1 \times \cancel{9}^3 \times \cancel{8}^2 \times 7}{\cancel{5} \times \cancel{4} \times \cancel{3} \times \cancel{2} \times 1}$

$\binom{11}{5} = 11 \times 1 \times 3 \times 2 \times 7 = 11 \times 42 = 462$

Total number of ways to select the team is $462$.

A team with "at least one boy and one girl" means the team is not composed of only boys or only girls.

A team of 5 cannot be composed of only girls, as there are only 4 girls available.

So, "at least one boy and one girl" is the complement of having "no girl" (which means all 5 members are boys).

Number of ways with at least one boy and one girl = (Total number of ways) - (Number of ways with no girl)

Number of ways = $462 - 21 = 441$

Alternatively, we can sum the possibilities for (boys, girls) combinations that satisfy the condition (at least one boy and one girl): (1, 4), (2, 3), (3, 2), (4, 1).

Ways (1 B, 4 G) = $\binom{7}{1} \times \binom{4}{4} = 7 \times 1 = 7$

Ways (2 B, 3 G) = $\binom{7}{2} \times \binom{4}{3} = \frac{7 \times 6}{2 \times 1} \times 4 = 21 \times 4 = 84$

Ways (3 B, 2 G) = $\binom{7}{3} \times \binom{4}{2} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times \frac{4 \times 3}{2 \times 1} = 35 \times 6 = 210$

Ways (4 B, 1 G) = $\binom{7}{4} \times \binom{4}{1} = \frac{7!}{4!3!} \times 4 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} \times 4 = 35 \times 4 = 140$

Total ways = $7 + 84 + 210 + 140 = 441$. This matches the previous result.

There are $441$ ways to select a team with at least one boy and one girl.

---

(iii) The team has at least 3 girls

The team size is 5, and there are only 4 girls available. "At least 3 girls" means the number of girls in the team can be 3 or 4.

Case 1: Exactly 3 girls and $5-3=2$ boys.

Number of ways to select 3 girls from 4 = $\binom{4}{3} = 4$.

Number of ways to select 2 boys from 7 = $\binom{7}{2} = 21$.

Number of ways for Case 1 = $\binom{4}{3} \times \binom{7}{2} = 4 \times 21 = 84$.

Case 2: Exactly 4 girls and $5-4=1$ boy.

Number of ways to select 4 girls from 4 = $\binom{4}{4} = 1$.

Number of ways to select 1 boy from 7 = $\binom{7}{1} = 7$.

Number of ways for Case 2 = $\binom{4}{4} \times \binom{7}{1} = 1 \times 7 = 7$.

The total number of ways to select a team with at least 3 girls is the sum of the ways in Case 1 and Case 2.

Total number of ways = (Ways for 3 girls and 2 boys) + (Ways for 4 girls and 1 boy)

Total number of ways = $84 + 7 = 91$

There are $91$ ways to select a team with at least 3 girls.

Solution:

The given word is AGAIN.

The letters in the word are A, G, A, I, N.

There are 5 letters in total, with the letter 'A' repeating 2 times.

---

Number of words formed using all the letters:

The number of permutations of $n$ objects where $p$ objects are of one type, $q$ objects are of another type, and so on, is given by $\frac{n!}{p!q!...}$.

Here, $n=5$ (total letters), and the letter 'A' repeats $p=2$ times.

Number of words = $\frac{5!}{2!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = \frac{120}{2} = 60$

There are $60$ words that can be formed using all the letters of the word AGAIN.

---

Finding the 50^th word in dictionary order:

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

We list the words in dictionary order based on the first letter.

Words starting with A:

The first letter is A. The remaining letters are A, G, I, N. These 4 letters can be arranged in $4!$ ways, but since 'A' is one of the remaining letters and is treated as distinct for now in terms of position, let's consider the available letters for the remaining positions. The set of available letters is {A, G, I, N}. One 'A' is fixed at the start. The remaining letters are A, G, I, N. There are 4 letters, with 'A' repeating 1 time in this set of 4 if we were to pick from the original set. However, it's easier to think of the letters as {A, G, I, N}. We are arranging these 4 letters. The number of arrangements is $4! = 24$.

Let's rephrase: If the first letter is A, the remaining letters are {A, G, I, N}. We arrange these 4 letters. The number of permutations of these 4 letters is $4! = 24$.

Number of words starting with A = Permutations of {A, G, I, N} = $4! = 24$.

The words starting with A are the first 24 words.

Words starting with G:

The first letter is G. The remaining letters are A, A, I, N. These 4 letters can be arranged in $\frac{4!}{2!}$ ways because 'A' repeats 2 times.

Number of words starting with G = Permutations of {A, A, I, N} = $\frac{4!}{2!} = \frac{24}{2} = 12$.

The words starting with G are from rank $24 + 1 = 25$ to $24 + 12 = 36$.

Words starting with I:

The first letter is I. The remaining letters are A, A, G, N. These 4 letters can be arranged in $\frac{4!}{2!}$ ways because 'A' repeats 2 times.

Number of words starting with I = Permutations of {A, A, G, N} = $\frac{4!}{2!} = \frac{24}{2} = 12$.

The words starting with I are from rank $36 + 1 = 37$ to $36 + 12 = 48$.

The 48^th word is the last word starting with I.

The 49^th word will be the first word starting with the next letter in alphabetical order, which is N.

Words starting with N:

The first letter is N. The remaining letters are A, A, G, I. We need the first word starting with N in dictionary order.

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

The words starting with N, in alphabetical order of the remaining letters, will be:

NAAGI (Arrangement of AAGI in order) - 49^th word

NAAIG (Next arrangement of AAGI) - 50^th word

To be precise, let's list the arrangements of AAGI in alphabetical order:

AA G I

AA I G

AG A I

AG I A

AI A G

AI G A

... and so on.

The permutations of A, A, G, I in dictionary order are:

AAGI

AAIG

AGAI

AGIA

AIAG

AIGA

GAAI

GAIA

GIAA

IAAG

IAGA

IGAA

Number of permutations = $\frac{4!}{2!} = 12$.

The words starting with N are formed by prefixing 'N' to these 12 arrangements of AAGI in dictionary order.

The first word starting with N is obtained by arranging A, A, G, I in alphabetical order and prefixing N:

N + AAGI $\implies$ NAAGI (49^th word)

The second word starting with N is obtained by the next arrangement of AAGI:

N + AAIG $\implies$ NAAIG (50^th word)

Therefore, the 50^th word in dictionary order is NAAIG.

Solution:

The given digits are 1, 2, 0, 2, 4, 2, 4.

There are a total of 7 digits.

The digits are: one 1, three 2s, one 0, and two 4s.

We need to form numbers greater than 1000000. A number greater than 1000000 must have at least 7 digits. Since we only have 7 digits available, we must form 7-digit numbers using all the given digits.

---

The total number of 7-digit numbers that can be formed using these digits (including those starting with 0) is the number of permutations of these 7 digits with repetitions.

Number of permutations = $\frac{n!}{p_1! p_2! ... p_k!}$, where $n$ is the total number of items and $p_i$ is the frequency of each distinct item.

Here, $n=7$ (total digits).

The repetitions are: digit 2 appears 3 times ($p_1=3$), digit 4 appears 2 times ($p_2=2$). Other digits (1, 0) appear once.

Total number of 7-digit arrangements = $\frac{7!}{3!2!1!1!} = \frac{7!}{3!2!}$

$\frac{7!}{3!2!} = \frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1) \times (2 \times 1)} = \frac{5040}{6 \times 2} = \frac{5040}{12} = 420$

So, there are 420 total 7-digit numbers that can be formed using these digits.

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A number is greater than 1000000 if and only if its first digit is not 0.

The numbers that are not greater than 1000000 are the 7-digit numbers that start with 0.

Let's find the number of 7-digit numbers that start with 0.

If the first digit is 0, the remaining 6 digits are 1, 2, 2, 2, 4, 4.

We need to arrange these 6 digits in the remaining 6 positions.

The digits to be arranged are: one 1, three 2s, and two 4s.

Number of arrangements of these 6 digits = $\frac{6!}{3!2!1!} = \frac{6!}{3!2!}$

$\frac{6!}{3!2!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1) \times (2 \times 1)} = \frac{720}{6 \times 2} = \frac{720}{12} = 60$

So, there are 60 seven-digit numbers that start with 0.

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The number of numbers greater than 1000000 is the total number of 7-digit numbers minus the number of 7-digit numbers starting with 0.

Number of numbers greater than 1000000 = (Total 7-digit numbers) - (7-digit numbers starting with 0)

Number of numbers greater than 1000000 = $420 - 60 = 360$

Therefore, $360$ numbers greater than 1000000 can be formed using the digits 1, 2, 0, 2, 4, 2, 4.

Solution:

We have 5 girls and 3 boys to be seated in a row such that no two boys are together.

To ensure that no two boys are together, we can arrange the girls first and then place the boys in the spaces between the girls or at the ends.

---

First, arrange the 5 girls. The number of ways to arrange 5 distinct girls in a row is $5!$.

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

These arrangements create spaces where the boys can be placed. If we represent the girls by 'G', the arrangement looks like:

$\_$ G $\_$ G $\_$ G $\_$ G $\_$ G $\_$

There are $5+1 = 6$ possible spaces where the 3 boys can be seated so that no two boys are adjacent.

Spaces are denoted by $\_$.

---

Now, we need to place the 3 boys in these 6 spaces such that each boy occupies a different space. This is a permutation problem because the boys are distinct (unless stated otherwise, individuals are treated as distinct) and the order of placing them in the spaces matters.

We need to choose 3 spaces out of the 6 available spaces and arrange the 3 boys in these 3 chosen spaces. The number of ways to do this is the number of permutations of 6 items taken 3 at a time, denoted by $P(6, 3)$ or $^6P_3$.

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

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

There are $120$ ways to place the 3 boys in the available spaces.

---

Since the arrangement of girls and the placement of boys are independent processes, the total number of ways to seat 5 girls and 3 boys such that no two boys are together is the product of the number of ways for each step.

Total number of ways = (Number of ways to arrange girls) $\times$ (Number of ways to place boys)

Total number of ways = $5! \times P(6, 3) = 120 \times 120 = 14400$

Therefore, there are $14400$ ways to seat 5 girls and 3 boys in a row so that no two boys are together.

Solution:

The given word is DAUGHTER.

The letters in the word are D, A, U, G, H, T, E, R.

Let's identify the vowels and consonants in the word:

Vowels: A, U, E (3 vowels)

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

Total number of letters in the word = $3 + 5 = 8$. All letters are distinct.

---

We need to form words each containing exactly 2 vowels and 3 consonants.

This process involves two steps:

1. Selecting the required number of vowels and consonants from the available letters.

2. Arranging the selected letters to form words.

---

Step 1: Selection of letters

We need to select 2 vowels from the 3 available vowels. The number of ways to do this is given by the combination $\binom{3}{2}$.

Number of ways to select 2 vowels = $\binom{3}{2} = \frac{3!}{2!(3-2)!} = \frac{3!}{2!1!} = \frac{3 \times 2 \times 1}{(2 \times 1) \times 1} = \frac{6}{2} = 3$

There are $3$ ways to select 2 vowels.

---

We need to select 3 consonants from the 5 available consonants. The number of ways to do this is given by the combination $\binom{5}{3}$.

Number of ways to select 3 consonants = $\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5 \times 4 \times 3!}{3! \times 2 \times 1} = \frac{5 \times 4}{2} = \frac{20}{2} = 10$

There are $10$ ways to select 3 consonants.

---

The total number of ways to select a group of 2 vowels and 3 consonants is the product of the number of ways to select the vowels and the number of ways to select the consonants.

Total number of ways to select the letters = (Ways to select 2 vowels) $\times$ (Ways to select 3 consonants)

Total number of ways to select the letters = $\binom{3}{2} \times \binom{5}{3} = 3 \times 10 = 30$

There are 30 different combinations of 5 letters (2 vowels and 3 consonants) that can be selected from the word DAUGHTER.

---

Step 2: Arrangement of selected letters

For each of the 30 selected combinations of 5 letters, all the selected letters are distinct (since all letters in DAUGHTER are distinct).

The number of ways to arrange 5 distinct letters is the permutation of 5 items, which is $5!$.

Number of ways to arrange 5 distinct letters = $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$

---

Total number of words formed

The total number of words is the product of the number of ways to select the 5 letters and the number of ways to arrange them for each selection.

Total number of words = (Number of ways to select letters) $\times$ (Number of ways to arrange letters)

Total number of words = $30 \times 120 = 3600$

Therefore, $3600$ words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER.

Solution:

The given word is EQUATION.

The letters in the word are E, Q, U, A, T, I, O, N.

Let's identify the vowels and consonants in the word:

Vowels: E, U, A, I, O (5 vowels)

Consonants: Q, T, N (3 consonants)

Total number of letters = $5 + 3 = 8$. All the letters are distinct.

---

We need to form words using all 8 letters such that all vowels occur together and all consonants occur together.

This means the word will be formed by arranging a block of vowels and a block of consonants.

We can treat the set of all vowels as a single unit (V) and the set of all consonants as a single unit (C).

The two possible arrangements of these units are (Vowel block) (Consonant block) or (Consonant block) (Vowel block).

---

Arrangements within the vowel block:

There are 5 distinct vowels (E, U, A, I, O). These 5 vowels can be arranged among themselves within the vowel block in $5!$ ways.

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

There are $120$ ways to arrange the vowels within their block.

---

Arrangements within the consonant block:

There are 3 distinct consonants (Q, T, N). These 3 consonants can be arranged among themselves within the consonant block in $3!$ ways.

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

There are $6$ ways to arrange the consonants within their block.

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Arrangement of the blocks:

We have two blocks: the vowel block (V) and the consonant block (C).

These two blocks can be arranged in $2!$ ways.

The possible arrangements are (Vowel block) (Consonant block) or (Consonant block) (Vowel block).

$2! = 2 \times 1 = 2$

There are $2$ ways to arrange the blocks.

---

The total number of words is the product of the number of ways to arrange the letters within the vowel block, the number of ways to arrange the letters within the consonant block, and the number of ways to arrange the blocks themselves.

Total number of words = (Arrangements of vowels) $\times$ (Arrangements of consonants) $\times$ (Arrangement of blocks)

Total number of words = $5! \times 3! \times 2! = 120 \times 6 \times 2$

$120 \times 6 = 720$

$720 \times 2 = 1440$

Therefore, $1440$ words can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together.

Solution:

We need to form a committee of 7 members from a group of 9 boys and 4 girls.

Total number of people = $9 + 4 = 13$.

Committee size = 7.

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(i) The committee consists of exactly 3 girls

If the committee has exactly 3 girls, then the number of boys must be $7 - 3 = 4$.

We need to select 3 girls from the 4 available girls AND 4 boys from the 9 available boys.

Number of ways to select 3 girls from 4 = $\binom{4}{3}$.

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

Number of ways to select 4 boys from 9 = $\binom{9}{4}$.

$\binom{9}{4} = \frac{9!}{4!(9-4)!} = \frac{9!}{4!5!} = \frac{9 \times 8 \times 7 \times 6 \times 5!}{4 \times 3 \times 2 \times 1 \times 5!} = \frac{9 \times 8 \times 7 \times 6}{24}$

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

Total number of ways to form the committee with exactly 3 girls = (Ways to select girls) $\times$ (Ways to select boys)

Total number of ways = $\binom{4}{3} \times \binom{9}{4} = 4 \times 126 = 504$

There are $504$ ways to form a committee with exactly 3 girls.

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(ii) The committee consists of at least 3 girls

"At least 3 girls" means the number of girls in the committee can be 3 or 4 (since there are only 4 girls available). The remaining members will be boys to make the total committee size 7.

Case 1: Exactly 3 girls and $7-3=4$ boys.

Number of ways = $\binom{4}{3} \times \binom{9}{4} = 4 \times 126 = 504$ (calculated in part (i)).

Case 2: Exactly 4 girls and $7-4=3$ boys.

Number of ways to select 4 girls from 4 = $\binom{4}{4}$.

$\binom{4}{4} = \frac{4!}{4!(4-4)!} = \frac{4!}{4!0!} = 1$

Number of ways to select 3 boys from 9 = $\binom{9}{3}$.

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

$\binom{9}{3} = \frac{\cancel{9}^3 \times \cancel{8}^4 \times 7}{\cancel{6}^1} = 3 \times 4 \times 7 = 84$

Number of ways for Case 2 = $\binom{4}{4} \times \binom{9}{3} = 1 \times 84 = 84$.

The total number of ways for "at least 3 girls" is the sum of the ways in Case 1 and Case 2.

Total number of ways = (Ways for 3 girls and 4 boys) + (Ways for 4 girls and 3 boys)

Total number of ways = $504 + 84 = 588$

There are $588$ ways to form a committee with at least 3 girls.

---

(iii) The committee consists of at most 3 girls

"At most 3 girls" means the number of girls in the committee can be 0, 1, 2, or 3 (since there are only 4 girls available). The remaining members will be boys to make the total committee size 7.

Case 1: Exactly 0 girls and $7-0=7$ boys.

Number of ways to select 0 girls from 4 = $\binom{4}{0} = 1$.

Number of ways to select 7 boys from 9 = $\binom{9}{7}$.

$\binom{9}{7} = \binom{9}{9-7} = \binom{9}{2} = \frac{9!}{2!(9-2)!} = \frac{9!}{2!7!} = \frac{9 \times 8 \times 7!}{2 \times 1 \times 7!} = \frac{9 \times 8}{2} = 36$

Number of ways for Case 1 = $\binom{4}{0} \times \binom{9}{7} = 1 \times 36 = 36$.

Case 2: Exactly 1 girl and $7-1=6$ boys.

Number of ways to select 1 girl from 4 = $\binom{4}{1} = 4$.

Number of ways to select 6 boys from 9 = $\binom{9}{6}$.

$\binom{9}{6} = \binom{9}{9-6} = \binom{9}{3} = 84$ (calculated in part (ii)).

Number of ways for Case 2 = $\binom{4}{1} \times \binom{9}{6} = 4 \times 84 = 336$.

Case 3: Exactly 2 girls and $7-2=5$ boys.

Number of ways to select 2 girls from 4 = $\binom{4}{2}$.

$\binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3 \times 2!}{2 \times 1 \times 2!} = \frac{12}{2} = 6$

Number of ways to select 5 boys from 9 = $\binom{9}{5}$.

$\binom{9}{5} = \binom{9}{9-5} = \binom{9}{4} = 126$ (calculated in part (i)).

Number of ways for Case 3 = $\binom{4}{2} \times \binom{9}{5} = 6 \times 126 = 756$.

Case 4: Exactly 3 girls and $7-3=4$ boys.

Number of ways for Case 4 = 504 (calculated in part (i)).

The total number of ways for "at most 3 girls" is the sum of the ways in Case 1, Case 2, Case 3, and Case 4.

Total number of ways = $36 + 336 + 756 + 504 = 1632$

Therefore, there are $1632$ ways to form a committee with at most 3 girls.

Alternate approach for (iii) atmost 3 girls:

The opposite of "at most 3 girls" is having "more than 3 girls". Since there are only 4 girls, "more than 3 girls" means exactly 4 girls.

The total number of ways to form a committee of 7 from 13 people is $\binom{13}{7}$.

$\binom{13}{7} = \frac{13!}{7!6!} = \frac{13 \times \cancel{12}^2 \times 11 \times \cancel{10}^2 \times \cancel{9}^3 \times \cancel{8} \times \cancel{7!}}{\cancel{7!} \times \cancel{6} \times \cancel{5} \times \cancel{4} \times \cancel{3} \times \cancel{2} \times 1} = 1716$.

Total number of ways to form the committee is $1716$.

Number of ways with exactly 4 girls and 3 boys is $\binom{4}{4} \times \binom{9}{3} = 1 \times 84 = 84$ (calculated in part (ii)).

Number of ways with at most 3 girls = (Total number of ways) - (Number of ways with exactly 4 girls)

Number of ways = $1716 - 84 = 1632$.

Both methods give the same result.

There are $1632$ ways to form a committee with at most 3 girls.

Solution:

The given word is EXAMINATION.

The letters in the word are E, X, A, M, I, N, A, T, I, O, N.

Let's count the total number of letters and the frequency of each letter:

Total number of letters = 11.

The distinct letters are A, E, I, M, N, O, T, X.

Frequencies:

A: 2 times

I: 2 times

N: 2 times

E, X, M, T, O: 1 time each

---

We need to find the number of words in the dictionary list that come before the first word starting with E.

The words in the dictionary list are arranged in alphabetical order of the first letter.

The distinct letters in alphabetical order are A, E, I, M, N, O, T, X.

The words before the first word starting with E are those that start with a letter that comes alphabetically before E.

The only letter before E in the set of distinct letters is A.

Therefore, we need to count the number of words that start with A.

---

Words starting with A:

If the first letter is fixed as A, the remaining $11 - 1 = 10$ letters are X, A, M, I, N, A, T, I, O, N.

The set of remaining letters is {X, A, M, I, N, A, T, I, O, N}. Let's list their frequencies:

A: 1 time (since one A is used as the first letter, and there were two A's initially)

I: 2 times

N: 2 times

E, X, M, T, O: 1 time each (these letters are not among the remaining ones? This is wrong. The set of remaining letters is what's left from the original word after taking one A.)

The original letters are: A, A, I, I, N, N, E, M, O, T, X.

If the first letter is A, the remaining 10 letters are: A, I, I, N, N, E, M, O, T, X.

The frequencies of the remaining letters are:

A: 1 time

I: 2 times

N: 2 times

E: 1 time

M: 1 time

O: 1 time

T: 1 time

X: 1 time

The number of ways to arrange these 10 letters is given by the permutation formula with repetitions.

Number of words starting with A = $\frac{10!}{1!2!2!1!1!1!1!1!} = \frac{10!}{2!2!}$

$\frac{10!}{2!2!} = \frac{3628800}{(2 \times 1) \times (2 \times 1)} = \frac{3628800}{4} = 907200$

There are $907200$ words that start with the letter A.

---

The words before the first word starting with E are precisely the words starting with A.

Therefore, the number of words before the first word starting with E is $907200$.

Solution:

The given digits are 0, 1, 3, 5, 7, and 9. There are 6 distinct digits available.

We need to form a 6-digit number using these digits without repetition.

The condition is that the 6-digit number must be divisible by 10.

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For a number to be divisible by 10, its units digit must be 0.

Since the digit 0 is available among the given digits and no digit is to be repeated, we must place 0 in the units place (the last digit) of the 6-digit number.

There is only one choice for the units digit, which is 0.

---

After placing 0 in the units place, we are left with the remaining $6 - 1 = 5$ digits: 1, 3, 5, 7, and 9.

We need to arrange these 5 distinct digits in the remaining $6 - 1 = 5$ places (the ten thousands, thousands, hundreds, tens, and units place).

Since the digits are distinct and we are arranging them in different positions, this is a permutation problem.

The number of ways to arrange 5 distinct digits in 5 places is the number of permutations of 5 items taken 5 at a time, which is $P(5, 5)$ or $5!$.

---

Number of ways to arrange the remaining 5 digits = $5!$

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

$5! = 120$

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The total number of 6-digit numbers that can be formed under the given conditions is the number of ways to fix the units digit multiplied by the number of ways to arrange the remaining digits.

Total number of ways = (Ways to choose units digit) $\times$ (Ways to arrange remaining 5 digits)

Total number of ways = $1 \times 120$

Total number of ways = $120$

Therefore, $120$ six-digit numbers can be formed from the digits 0, 1, 3, 5, 7, and 9 which are divisible by 10 and no digit is repeated.

Solution:

We are given the number of vowels and consonants in the English alphabet:

- Number of vowels = 5

- Number of consonants = 21

We need to form words consisting of two different vowels and two different consonants.

The formation of such a word involves two steps:

1. Selecting the required number of vowels and consonants.

2. Arranging the selected letters to form words.

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Step 1: Selection of letters

We need to select 2 different vowels from the 5 available vowels. The number of ways to do this is given by the combination $\binom{5}{2}$.

Number of ways to select 2 vowels = $\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5!}{2!3!} = \frac{5 \times 4 \times 3!}{2 \times 1 \times 3!} = \frac{20}{2} = 10$

There are $10$ ways to select 2 different vowels.

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We need to select 2 different consonants from the 21 available consonants. The number of ways to do this is given by the combination $\binom{21}{2}$.

Number of ways to select 2 consonants = $\binom{21}{2} = \frac{21!}{2!(21-2)!} = \frac{21!}{2!19!} = \frac{21 \times 20 \times 19!}{2 \times 1 \times 19!} = \frac{21 \times 20}{2} = \frac{420}{2} = 210$

There are $210$ ways to select 2 different consonants.

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The total number of ways to select a group of 2 vowels and 2 consonants is the product of the number of ways to select the vowels and the number of ways to select the consonants.

Total number of ways to select the letters = (Ways to select 2 vowels) $\times$ (Ways to select 2 consonants)

Total number of ways to select the letters = $\binom{5}{2} \times \binom{21}{2} = 10 \times 210 = 2100$

There are 2100 different combinations of 4 letters (2 different vowels and 2 different consonants) that can be selected.

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Step 2: Arrangement of selected letters

For each of the 2100 selected combinations, we have a set of $2 + 2 = 4$ distinct letters (since the vowels are different from each other and the consonants are different from each other, and vowels are distinct from consonants).

The number of ways to arrange 4 distinct letters is the permutation of 4 items, which is $4!$.

Number of ways to arrange 4 distinct letters = $4! = 4 \times 3 \times 2 \times 1 = 24$

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Total number of words formed

The total number of words is the product of the number of ways to select the 4 letters and the number of ways to arrange them for each selection.

Total number of words = (Number of ways to select letters) $\times$ (Number of ways to arrange letters)

Total number of words = $2100 \times 24$

$2100 \times 24 = 50400$

Therefore, $50400$ words with two different vowels and 2 different consonants can be formed from the English alphabet.

Solution:

The question paper has a total of 12 questions, divided into two parts:

- Part I: 5 questions

- Part II: 7 questions

The student must attempt a total of 8 questions.

The student must select at least 3 questions from each part.

---

Let $x$ be the number of questions selected from Part I, and $y$ be the number of questions selected from Part II.

We have the following conditions:

Also, the number of questions selected from each part cannot exceed the total number of questions in that part:

We need to find the possible integer values for $x$ and $y$ that satisfy all these conditions.

From (i), $y = 8 - x$. Substitute this into (iii) and (v):

$8 - x \ge 3 \implies x \le 8 - 3 \implies x \le 5$

$8 - x \le 7 \implies 8 - 7 \le x \implies 1 \le x$

So, from (i), (iii), and (v), we have $1 \le x \le 5$.

Combining this with condition (ii) $x \ge 3$, the possible values for $x$ are $3, 4, 5$.

For each value of $x$, the corresponding value of $y$ is determined by $y = 8 - x$. Let's check if these pairs $(x, y)$ satisfy all conditions:

Case 1: $x = 3$. Then $y = 8 - 3 = 5$. (3 from Part I, 5 from Part II).

Check conditions: $x=3 \ge 3$ (True), $x=3 \le 5$ (True), $y=5 \ge 3$ (True), $y=5 \le 7$ (True). This case is valid.

Case 2: $x = 4$. Then $y = 8 - 4 = 4$. (4 from Part I, 4 from Part II).

Check conditions: $x=4 \ge 3$ (True), $x=4 \le 5$ (True), $y=4 \ge 3$ (True), $y=4 \le 7$ (True). This case is valid.

Case 3: $x = 5$. Then $y = 8 - 5 = 3$. (5 from Part I, 3 from Part II).

Check conditions: $x=5 \ge 3$ (True), $x=5 \le 5$ (True), $y=3 \ge 3$ (True), $y=3 \le 7$ (True). This case is valid.

So, there are three possible scenarios for the number of questions to be selected from each part:

Scenario A: 3 questions from Part I and 5 questions from Part II.

Scenario B: 4 questions from Part I and 4 questions from Part II.

Scenario C: 5 questions from Part I and 3 questions from Part II.

---

Now, let's calculate the number of ways for each scenario.

Scenario A: Select 3 from Part I (5 questions) and 5 from Part II (7 questions)

Number of ways to select 3 questions from Part I = $\binom{5}{3} = \frac{5!}{3!2!} = 10$.

Number of ways to select 5 questions from Part II = $\binom{7}{5} = \frac{7!}{5!2!} = \frac{7 \times 6}{2 \times 1} = 21$.

Number of ways for Scenario A = $\binom{5}{3} \times \binom{7}{5} = 10 \times 21 = 210$.

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Scenario B: Select 4 from Part I (5 questions) and 4 from Part II (7 questions)

Number of ways to select 4 questions from Part I = $\binom{5}{4} = \frac{5!}{4!1!} = 5$.

Number of ways to select 4 questions from Part II = $\binom{7}{4} = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.

Number of ways for Scenario B = $\binom{5}{4} \times \binom{7}{4} = 5 \times 35 = 175$.

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Scenario C: Select 5 from Part I (5 questions) and 3 from Part II (7 questions)

Number of ways to select 5 questions from Part I = $\binom{5}{5} = \frac{5!}{5!0!} = 1$.

Number of ways to select 3 questions from Part II = $\binom{7}{3} = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.

Number of ways for Scenario C = $\binom{5}{5} \times \binom{7}{3} = 1 \times 35 = 35$.

---

The total number of ways a student can select the questions is the sum of the number of ways in each valid scenario.

Total number of ways = (Ways for Scenario A) + (Ways for Scenario B) + (Ways for Scenario C)

Total number of ways = $210 + 175 + 35$

$210 + 175 = 385$

$385 + 35 = 420$

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

Solution:

We need to form a 5-card combination from a standard deck of 52 cards such that each combination contains exactly one king.

A standard deck has 52 cards, which include 4 kings and $52 - 4 = 48$ non-king cards.

To form a combination with exactly one king, we need to make two independent selections:

1. Select exactly one king from the 4 available kings.

2. Select the remaining $5 - 1 = 4$ cards from the 48 non-king cards.

---

Number of ways to select 1 king from 4 kings is given by the combination $\binom{4}{1}$.

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

There are $4$ ways to select one king.

---

Number of ways to select the remaining 4 cards from the 48 non-king cards is given by the combination $\binom{48}{4}$.

$\binom{48}{4} = \frac{48!}{4!(48-4)!} = \frac{48!}{4!44!}$

$\binom{48}{4} = \frac{48 \times 47 \times 46 \times 45 \times 44!}{4 \times 3 \times 2 \times 1 \times 44!}$

Cancel out the $44!$ from the numerator and denominator:

$\binom{48}{4} = \frac{48 \times 47 \times 46 \times 45}{4 \times 3 \times 2 \times 1}$

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

$\binom{48}{4} = \frac{\cancel{48}^{2} \times 47 \times 46 \times 45}{\cancel{24}}$

$\binom{48}{4} = 2 \times 47 \times 46 \times 45$

$\binom{48}{4} = 94 \times (46 \times 45)$

$46 \times 45 = 2070$

$\binom{48}{4} = 94 \times 2070 = 194580$

There are $194580$ ways to select 4 non-king cards.

---

To find the total number of 5-card combinations with exactly one king, we multiply the number of ways to select the king by the number of ways to select the remaining 4 cards.

Total number of ways = (Ways to select 1 king) $\times$ (Ways to select 4 non-kings)

Total number of ways = $\binom{4}{1} \times \binom{48}{4} = 4 \times 194580$

$4 \times 194580 = 778320$

Therefore, the number of 5-card combinations out of a deck of 52 cards with exactly one king is $778320$.

Solution:

We need to seat 5 men and 4 women in a row, so there are a total of $5 + 4 = 9$ positions in the row.

The condition is that the women must occupy the even places.

The positions in the row are 1, 2, 3, 4, 5, 6, 7, 8, 9.

The even places are the 2nd, 4th, 6th, and 8th positions.

There are 4 even places, and there are 4 women.

---

First, let's consider the seating of the women.

There are 4 women to be seated in the 4 even places (positions 2, 4, 6, 8).

Since the women are distinct and the positions are distinct, the number of ways to arrange the 4 women in the 4 even places is the number of permutations of 4 items taken 4 at a time, which is $P(4, 4)$ or $4!$.

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

There are $24$ ways to seat the women in the even places.

---

Next, let's consider the seating of the men.

There are 5 men to be seated in the remaining $9 - 4 = 5$ places.

The remaining places are the odd places (positions 1, 3, 5, 7, 9).

Since the men are distinct and the positions are distinct, the number of ways to arrange the 5 men in the 5 odd places is the number of permutations of 5 items taken 5 at a time, which is $P(5, 5)$ or $5!$.

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

There are $120$ ways to seat the men in the odd places.

---

Since the seating of the women and the seating of the men are independent events, the total number of arrangements is the product of the number of ways for each group.

Total number of arrangements = (Number of ways to seat women) $\times$ (Number of ways to seat men)

Total number of arrangements = $4! \times 5! = 24 \times 120$

$24 \times 120 = 2880$

Therefore, there are $2880$ possible arrangements where 5 men and 4 women are seated in a row so that the women occupy the even places.

Given:

Total number of students in the class = 25

Number of students to be chosen for the excursion party = 10

Condition regarding 3 specific students: Either all 3 join the party or none of the 3 join the party.

---

To Find:

The total number of ways the excursion party can be chosen under the given condition.

---

Solution:

The problem can be divided into two mutually exclusive cases based on the condition about the 3 specific students.

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Case 1: The 3 specific students are included in the excursion party.

If these 3 students are included, we need to select the remaining students for the party from the rest of the class.

Number of students already included = 3

Number of students still needed for the party = $10 - 3 = 7$

Number of students remaining in the class (excluding the 3 specific ones) = $25 - 3 = 22$

So, we need to choose 7 students from these 22 remaining students.

The number of ways to do this is given by the combination formula $^{n}C_r = \frac{n!}{r!(n-r)!}$.

Number of ways for Case 1 = $^{22}C_7$

$^{22}C_7 = \frac{22!}{7!(22-7)!} = \frac{22!}{7!15!}$

$^{22}C_7 = \frac{22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}$

Calculating the value:

$^{22}C_7 = 170544$

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Case 2: The 3 specific students are NOT included in the excursion party.

If these 3 students are not included, the entire party of 10 students must be chosen from the remaining students in the class.

Number of students to be chosen = 10

Number of students available for selection (excluding the 3 specific ones) = $25 - 3 = 22$

So, we need to choose 10 students from these 22 available students.

The number of ways to do this is given by the combination formula $^{n}C_r$.

Number of ways for Case 2 = $^{22}C_{10}$

$^{22}C_{10} = \frac{22!}{10!(22-10)!} = \frac{22!}{10!12!}$

$^{22}C_{10} = \frac{22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13}{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}$

Calculating the value:

$^{22}C_{10} = 646646$

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Since the excursion party can be formed either by including the 3 specific students (Case 1) or by excluding them (Case 2), the total number of ways to choose the party 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 = $^{22}C_7 + ^{22}C_{10}$

Total number of ways = $170544 + 646646$

Total number of ways = $817190$

---

Therefore, the excursion party can be chosen in 817190 ways.

Solution:

The given word is ASSASSINATION.

Let's count the total number of letters and the frequency of each letter:

Total number of letters = 13.

The distinct letters are A, S, I, N, T, O.

Frequencies:

A: 3 times

S: 4 times

I: 2 times

N: 2 times

T: 1 time

O: 1 time

Sum of frequencies = $3 + 4 + 2 + 2 + 1 + 1 = 13$.

---

We need to arrange the letters such that all the S's are together.

To handle the condition "all the S's are together", we treat all four S's as a single block or a single unit.

Let this block of S's be denoted by (SSSS).

Now, we need to arrange this single unit (SSSS) along with the remaining letters from the word ASSASSINATION.

The remaining letters are A, A, A, I, I, N, N, T, O.

So, we are arranging the following units:

(SSSS), A, A, A, I, I, N, N, T, O

There are $1 + 3 + 2 + 2 + 1 + 1 = 10$ units in total to arrange.

The units to be arranged are (SSSS), A, A, A, I, I, N, N, T, O. Note that the distinct items among these units are (SSSS), A, I, N, T, O, with repetitions for A, I, and N.

The number of items to arrange is 10.

The repetitions among these items are:

A: 3 times

I: 2 times

N: 2 times

(SSSS): 1 time (as a single block)

T: 1 time

O: 1 time

---

The number of ways to arrange these 10 units is given by the permutation formula with repetitions:

Number of arrangements = $\frac{(\text{Total number of units})!}{(\text{Frequency of A})! \times (\text{Frequency of I})! \times (\text{Frequency of N})! \times ...}$

Number of arrangements = $\frac{10!}{3!2!2!1!1!1!} = \frac{10!}{3!2!2!}$

$\frac{10!}{3!2!2!} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1) \times (2 \times 1) \times (2 \times 1)} = \frac{3628800}{6 \times 2 \times 2} = \frac{3628800}{24}$

$\frac{3628800}{24} = 151200$

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Since the letters within the (SSSS) block are identical, there is only $\frac{4!}{4!} = 1$ way to arrange the letters within the block of S's. If the S's were distinct (e.g., $S_1, S_2, S_3, S_4$), we would multiply by $4!$ ways to arrange them within the block. However, since they are identical, this factor is 1.

The number of ways to arrange the letters such that all S's are together is the number of arrangements of the 10 units.

Total number of arrangements = $151200$.

Therefore, the letters of the word ASSASSINATION can be arranged in $151200$ ways so that all the S's are together.