Division into groups, Derangements Questions in English

(A) To divide $9$ persons into $3$ equal groups of $3$ persons each,we first select $3$ persons out of $9$,then $3$ out of the remaining $6$,and finally $3$ out of the remaining $3$.
Since the order of the groups does not matter,we divide by $3!$.
The number of ways is given by $\frac{\binom{9}{3} \binom{6}{3} \binom{3}{3}}{3!} = \frac{\frac{9!}{3!6!} \times \frac{6!}{3!3!} \times \frac{3!}{3!0!}}{3!} = \frac{9!}{(3!)^3 \times 3!}$.
Calculating the value: $\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4}{6 \times 6 \times 6} = \frac{60480}{216} = 280$.

(A) To divide $n$ distinct objects into $k$ groups of equal size $m$ (where $n = km$),the number of ways is given by the formula: $\frac{n!}{(m!)^k \times k!}$.
Here,$n = 15$,$m = 3$,and $k = 5$.
Substituting these values,the number of ways is $\frac{15!}{(3!)^5 \times 5!}$.
Thus,the correct option is $A$.

(A) The total number of cards is $52$,and they are to be divided equally among $4$ players,meaning each player receives $13$ cards.
The number of ways to choose $13$ cards for the first player is $^{52}C_{13}$.
The number of ways to choose $13$ cards for the second player from the remaining $39$ cards is $^{39}C_{13}$.
The number of ways to choose $13$ cards for the third player from the remaining $26$ cards is $^{26}C_{13}$.
The number of ways to choose $13$ cards for the fourth player from the remaining $13$ cards is $^{13}C_{13}$.
Therefore,the total number of ways is:
$^{52}C_{13} \times ^{39}C_{13} \times ^{26}C_{13} \times ^{13}C_{13}$
$= \frac{52!}{39! \times 13!} \times \frac{39!}{26! \times 13!} \times \frac{26!}{13! \times 13!} \times \frac{13!}{0! \times 13!}$
$= \frac{52!}{(13!)^4}$.

(C) This is a problem of derangements where $n = 4$ items are to be placed in $n$ boxes such that no item goes into its correct box.
The number of derangements $D_n$ is given by the formula:
$D_n = n! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \dots + \frac{(-1)^n}{n!} \right)$
For $n = 4$:
$D_4 = 4! \left( \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} \right)$
$D_4 = 24 \left( \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right)$
$D_4 = 12 - 4 + 1 = 9$.
Thus,there are $9$ ways.

(A) The total number of ways to distribute $52$ cards among four players such that three players receive $17$ cards each and the fourth player receives $1$ card is calculated as follows:
$1$. Choose $17$ cards for the first player from $52$ cards: $^{52}C_{17} = \frac{52!}{35!17!}$.
$2$. Choose $17$ cards for the second player from the remaining $35$ cards: $^{35}C_{17} = \frac{35!}{18!17!}$.
$3$. Choose $17$ cards for the third player from the remaining $18$ cards: $^{18}C_{17} = \frac{18!}{1!17!}$.
$4$. The remaining $1$ card goes to the fourth player: $^{1}C_{1} = 1$.
Multiplying these combinations together:
$\frac{52!}{35!17!} \times \frac{35!}{18!17!} \times \frac{18!}{1!17!} \times 1 = \frac{52!}{17!17!17!1!} = \frac{52!}{(17!)^3}$.

(B) The total number of ways to arrange $n$ letters in $n$ envelopes is $n!$.
There is only $1$ way in which all letters are placed in their correct envelopes.
Therefore,the probability that all letters are kept in the right envelope is $P(\text{correct}) = \frac{1}{n!}$.
The probability that all letters are not kept in the right envelope is $1 - P(\text{correct}) = 1 - \frac{1}{n!}$.

(D) The total number of ways to place $3$ distinct letters into $3$ distinct envelopes is $3! = 3 \times 2 \times 1 = 6$.
Out of these $6$ possible arrangements,only $1$ arrangement corresponds to all letters being placed in their correct envelopes.
Therefore,the required probability is $\frac{1}{3!} = \frac{1}{6}$.

(C) The total number of ways to place $4$ letters into $4$ envelopes is $4! = 24$.
This is a problem of derangements. The number of ways in which no letter goes into its correct envelope is given by $D_4 = 4! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} \right) = 24 \left( \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right) = 12 - 4 + 1 = 9$.
The probability is $\frac{9}{24} = \frac{3}{8}$.
Wait,the standard interpretation of 'not going into the proper envelope' usually implies at least one is wrong,but the phrasing 'no letter goes into its correct envelope' refers to derangements.
However,if the question implies the complement of 'all letters go into their correct envelopes',the probability is $1 - \frac{1}{4!} = 1 - \frac{1}{24} = \frac{23}{24}$.

(A) Let $E_i$ denote the event that the $i^{th}$ object goes to the $i^{th}$ place.
We have $P(E_i) = \frac{(n-1)!}{n!} = \frac{1}{n}$ for all $i$.
The probability that exactly $3$ specific objects occupy their correct places is $P(E_i \cap E_j \cap E_k) = \frac{(n-3)!}{n!}$ for $i < j < k$.
Since we can choose $3$ places out of $n$ in $\binom{n}{3}$ ways,the probability that at least three objects occupy their corresponding places is given by $\binom{n}{3} \times \frac{(n-3)!}{n!} = \frac{n!}{3!(n-3)!} \times \frac{(n-3)!}{n!} = \frac{1}{3!} = \frac{1}{6}$.

(A) We need to select $2$ balls out of $4$ and place them in the boxes corresponding to the other $2$ balls such that no ball goes into its own box.
Number of ways to select $2$ balls out of $4$ is $^4C_2 = \frac{4 \times 3}{2 \times 1} = 6$.
For the $2$ selected balls,the number of ways to place them such that neither goes into its own box is given by the derangement $D_2 = 2! \times (1 - 1 + \frac{1}{2!}) = 1$.
Thus,the total number of ways is $6 \times 1 = 6$.

(B) The set $S$ contains $12$ elements. Since it is partitioned into three sets $A, B, C$ of equal size,each set must contain $12 / 3 = 4$ elements.
The number of ways to choose $4$ elements for set $A$ is $\binom{12}{4}$.
The number of ways to choose $4$ elements for set $B$ from the remaining $8$ is $\binom{8}{4}$.
The number of ways to choose $4$ elements for set $C$ from the remaining $4$ is $\binom{4}{4}$.
Since the sets $A, B, C$ are distinct (labeled),the total number of ways is:
$\binom{12}{4} \times \binom{8}{4} \times \binom{4}{4} = \frac{12!}{4! \times 8!} \times \frac{8!}{4! \times 4!} \times \frac{4!}{4! \times 0!} = \frac{12!}{(4!)^3}$.

(A) The number of ways to distribute $52$ cards equally among four players is given by the multinomial coefficient:
$\frac{52!}{13! \times 13! \times 13! \times 13!} = \frac{52!}{(13!)^4}$
This can also be calculated as:
$^{52}C_{13} \times ^{39}C_{13} \times ^{26}C_{13} \times ^{13}C_{13} = \frac{52!}{39!13!} \times \frac{39!}{26!13!} \times \frac{26!}{13!13!} \times \frac{13!}{0!13!} = \frac{52!}{(13!)^4}$

(C) This is a problem of derangements of $4$ objects,denoted as $D_4$.
The formula for derangements of $n$ objects is $D_n = n! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \dots + \frac{(-1)^n}{n!} \right)$.
For $n = 4$:
$D_4 = 4! \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right)$
$D_4 = 24 \left( \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right)$
$D_4 = 24 \left( \frac{12 - 4 + 1}{24} \right)$
$D_4 = 9$.
Thus,there are $9$ ways to place the balls such that no ball is in its own color box.

(A) The number of ways to place $n$ letters in wrong envelopes is given by the derangement formula $D_n = n! \left[ 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \dots + (-1)^n \frac{1}{n!} \right]$.
For $n = 6$,the number of ways is $D_6 = 6! \left[ 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{6!} \right]$.
$D_6 = 6! \left[ \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{6!} \right]$.
$D_6 = \frac{720}{2} - \frac{720}{6} + \frac{720}{24} - \frac{720}{120} + \frac{720}{720}$.
$D_6 = 360 - 120 + 30 - 6 + 1 = 265$.

(C) This is a problem of derangements,denoted by $D_n$ or $!n$.
For $n = 4$ objects,the number of derangements is given by the formula:
$D_n = n! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} \right)$
Substituting $n = 4$:
$D_4 = 4! \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right)$
$D_4 = 24 \left( \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right)$
$D_4 = 12 - 4 + 1 = 9$
Thus,there are $9$ ways to place the balls such that no ball goes into a box of its own color.

(B) The number of ways to divide $52$ distinct items into $n$ groups of equal size $k$ is given by the formula $\frac{n!}{(k!)^n \times n!}$.
Here,$n = 4$ and $k = 13$.
Therefore,the number of ways is $\frac{52!}{(13!)^4 \times 4!}$.

(B) The total number of students is $150$. We need to divide them into $3$ groups of $50$ students each.
Since the sections $A, B,$ and $C$ are distinct (labeled),the number of ways to distribute $150$ students into $3$ groups of $50$ is given by the multinomial coefficient:
$\frac{150!}{50! 50! 50!} = \frac{150!}{(50!)^3}$.
Therefore,the correct option is $B$.

(B) This is a problem of derangements of $n$ objects,denoted by $D_n$.
For $n = 4$,the number of derangements is given by the formula:
$D_n = n! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} \right)$
Substituting $n = 4$:
$D_4 = 4! \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right)$
$D_4 = 24 \left( \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right)$
$D_4 = 24 \left( \frac{12 - 4 + 1}{24} \right)$
$D_4 = 12 - 4 + 1 = 9$
Thus,there are $9$ ways to place all letters in the wrong envelopes.

(D) The number of ways to place $n$ letters into $n$ envelopes such that no letter is in its correct envelope is given by the derangement formula $D_n = n! \sum_{k=0}^{n} \frac{(-1)^k}{k!}$.
The total number of ways to arrange $4$ letters in $4$ envelopes is $4! = 24$.
The number of derangements for $n=4$ is $D_4 = 4! \left( \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} \right) = 24 \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right) = 12 - 4 + 1 = 9$.
The probability that all letters are in the wrong envelopes is $\frac{D_4}{4!} = \frac{9}{24} = \frac{3}{8}$.

(D) Let $E_i$ be the event that the $i^{th}$ item is in the $i^{th}$ position.
The probability that a specific set of $3$ items are in their correct positions is given by $\frac{(n-3)!}{n!}$.
There are $\binom{n}{3}$ ways to choose $3$ positions out of $n$.
For large $n$,the probability that exactly $k$ items are in their correct positions follows a Poisson distribution with parameter $\lambda = 1$,given by $P(X=k) = \frac{e^{-1}}{k!}$.
The probability that at least $3$ items are in their correct positions is $1 - [P(X=0) + P(X=1) + P(X=2)]$.
$P(X=0) = \frac{e^{-1}}{0!} = e^{-1}$,$P(X=1) = \frac{e^{-1}}{1!} = e^{-1}$,$P(X=2) = \frac{e^{-1}}{2!} = \frac{e^{-1}}{2}$.
Sum $= e^{-1}(1 + 1 + 0.5) = 2.5 e^{-1} = \frac{2.5}{e} \approx \frac{2.5}{2.718} \approx 0.919$.
Probability $= 1 - 0.919 = 0.081$,which is not equal to $\frac{1}{6} \approx 0.166$.
Thus,the correct answer is $\text{None of these}$.

(C) The total number of ways to place $4$ letters into $4$ envelopes is $4! = 24$.
There is only $1$ way for all letters to be in their correct envelopes.
The probability that all letters are in their correct envelopes is $\frac{1}{24}$.
The probability that not all letters are in their correct envelopes is $1 - \frac{1}{24} = \frac{23}{24}$.

(D) The total number of ways to place $3$ letters into $3$ envelopes is $n = 3! = 6$.
Out of these,there is only $r = 1$ way to place all letters into their correct envelopes.
The probability of the described event is $\frac{r}{n} = \frac{1}{3!} = \frac{1}{6}$.

(C) The set $S$ contains $12$ elements.
We need to partition $S$ into three disjoint sets $A, B, C$ of equal size.
Since $|S| = 12$,each set must contain $12 / 3 = 4$ elements.
The number of ways to distribute $12$ distinct items into $3$ unlabeled groups of size $4$ is given by the formula $\frac{1}{3!} \binom{12}{4, 4, 4}$.
This is calculated as:
$\frac{1}{3!} \times \binom{12}{4} \times \binom{8}{4} \times \binom{4}{4} = \frac{1}{3!} \times \frac{12!}{4!8!} \times \frac{8!}{4!4!} \times \frac{4!}{4!0!} = \frac{12!}{3!(4!)^3}$.

(B) This problem is equivalent to finding the number of derangements of $4$ distinct elements,denoted as $D_4$.
The formula for derangements of $n$ elements is given by $D_n = n! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \dots + \frac{(-1)^n}{n!} \right)$.
For $n = 4$:
$D_4 = 4! \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right)$
$D_4 = 24 \left( \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right)$
$D_4 = 12 - 4 + 1 = 9$.
Thus,the number of such one-one functions is $9$.

(B) Let $S = \{1, 2, 3, 4, 5\}$. We want to find the number of permutations $\sigma$ of $S$ such that at least $3$ elements are not in their original positions.
This is equivalent to saying that at most $2$ elements are in their original positions.
Let $A_i$ be the set of permutations where the $i$-th digit is in its original position.
We want to find the number of permutations where at most $2$ digits are fixed.
Total permutations = $5! = 120$.
Let $f(k)$ be the number of permutations with exactly $k$ fixed points. $f(k) = \binom{5}{k} D_{5-k}$,where $D_n$ is the number of derangements of $n$ objects.
$D_0 = 1, D_1 = 0, D_2 = 1, D_3 = 2, D_4 = 9, D_5 = 44$.
The number of permutations with at least $3$ digits not in their original positions is the total permutations minus those with $4$ or $5$ fixed points.
Permutations with $5$ fixed points: $\binom{5}{5} D_0 = 1 \times 1 = 1$.
Permutations with $4$ fixed points: $\binom{5}{4} D_1 = 5 \times 0 = 0$.
Permutations with $3$ fixed points: $\binom{5}{3} D_2 = 10 \times 1 = 10$.
Number of permutations with at most $2$ fixed points = $120 - (1 + 0 + 10) = 120 - 11 = 109$.

(A) square matrix of order $n=5$ where each row and each column has exactly one non-zero element (which is $1$) is a permutation matrix.
There are $n!$ such matrices in total.
For $n=5$,the total number of such matrices is $5! = 120$.
However,we are given the condition $a_{ij} = 0$ whenever $i + j = n + 1 = 6$.
This means the non-zero element cannot be placed on the anti-diagonal of the matrix.
Let $S_n$ be the set of all permutation matrices of order $n$. We want to find the number of matrices in $S_n$ such that no $1$ is on the anti-diagonal.
This is equivalent to finding the number of permutations $\sigma$ of $\{1, 2, 3, 4, 5\}$ such that $\sigma(i) \neq 6 - i$ for all $i \in \{1, 2, 3, 4, 5\}$.
This is the definition of the number of derangements of a specific type,often related to the ménage problem or restricted permutations.
For $n=5$,the number of such permutations is given by the formula for the number of permutations with no fixed points on the anti-diagonal,which is $D_n$ for even $n$ and a related value for odd $n$.
Specifically,for $n=5$,the number of such permutations is $44$.

(C) Let the letters be $L_1, L_2, L_3$ and their corresponding envelopes be $E_1, E_2, E_3$. The total number of ways to arrange $3$ letters in $3$ envelopes is $3! = 6$.
These arrangements are:
$(L_1E_1, L_2E_2, L_3E_3)$ - All $3$ correct
$(L_1E_1, L_2E_3, L_3E_2)$ - $L_1$ correct
$(L_2E_2, L_1E_1, L_3E_3)$ - $L_3$ correct
$(L_3E_3, L_2E_2, L_1E_1)$ - $L_2$ correct
$(L_1E_2, L_2E_3, L_3E_1)$ - None correct
$(L_1E_3, L_2E_1, L_3E_2)$ - None correct
There are $4$ arrangements where at least one letter is in its proper envelope.
Thus,the required probability is $\frac{4}{6} = \frac{2}{3}$.

(B) To transport $8$ persons in $3$ cars with a maximum capacity of $3$ persons each,the distribution of persons in the cars must be $(3, 3, 2)$.
First,we divide the $8$ persons into groups of $3, 3,$ and $2$:
$\text{Number of ways to group} = \frac{8!}{3!3!2!} \times \frac{1}{2!}$
Since the cars are of different makes,the order of the groups matters,so we multiply by $3!$:
$\text{Total ways} = \left( \frac{8!}{3!3!2! \times 2!} \right) \times 3!$
$= \frac{40320}{6 \times 6 \times 2 \times 2} \times 6$
$= \frac{40320}{144} \times 6 = 280 \times 6 = 1680$.

(B) The number of ways in which none of the $n$ students sits on their allotted seat is given by the derangement formula $D_n = n! \left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \dots + \frac{(-1)^n}{n!}\right)$.
For $n = 5$:
$D_5 = 5! \left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!}\right)$
$D_5 = 120 \left(1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}\right)$
$D_5 = 120 \left(\frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120}\right)$
$D_5 = 60 - 20 + 5 - 1$
$D_5 = 44$.

(C) Given $\alpha = \frac{(4!)!}{(4!)^{3!}} = \frac{24!}{(24)^6}$ and $\beta = \frac{(5!)!}{(5!)^{4!}} = \frac{120!}{(120)^{24}}$.
The number of ways to divide $n$ distinct objects into $k$ groups of size $m$ each (where $n = km$) is given by $\frac{n!}{(m!)^k \cdot k!}$.
For $\alpha$,we have $n=24, m=4, k=6$. The number of ways is $\frac{24!}{(4!)^6 \cdot 6!} = K_1$,where $K_1 \in N$.
Thus,$\alpha = K_1 \cdot 6!$. Since $K_1$ and $6!$ are integers,$\alpha \in N$.
For $\beta$,we have $n=120, m=5, k=24$. The number of ways is $\frac{120!}{(5!)^{24} \cdot 24!} = K_2$,where $K_2 \in N$.
Thus,$\beta = K_2 \cdot 24!$. Since $K_2$ and $24!$ are integers,$\beta \in N$.
Therefore,both $\alpha$ and $\beta$ are natural numbers.

(A) The set $S = \{2^1, 2^2, 2^3, \ldots, 2^9\}$ contains $9$ elements.
We need to partition these $9$ elements into $3$ sets $A, B, C$ each containing $3$ elements.
The number of ways to divide $9$ distinct objects into $3$ groups of $3$ is given by the multinomial coefficient:
$\frac{9!}{3! 3! 3! 3!}$
Since the sets $A, B, C$ are distinct (labeled),we multiply by $3!$ to assign the groups to the sets $A, B, C$:
$\text{Number of ways} = \frac{9!}{3! 3! 3! 3!} \times 3! = \frac{9!}{3! 3! 3!} = \frac{362880}{6 \times 6 \times 6} = \frac{362880}{216} = 1680$.

(A, C) $(1)$ Total ways to arrange $5$ students in $5$ seats is $n(S) = 5! = 120$.
Let $A$ be the event that $S_1$ gets seat $R_1$ and none of the remaining $4$ students $(S_2, S_3, S_4, S_5)$ get their original seats $(R_2, R_3, R_4, R_5)$.
This is a derangement of $4$ items,denoted by $D_4$.
$n(A) = D_4 = 4! \left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}\right) = 24 \left(\frac{1}{2} - \frac{1}{6} + \frac{1}{24}\right) = 12 - 4 + 1 = 9$.
$P(A) = \frac{9}{120} = \frac{3}{40}$.
$(2)$ Let $E_i$ be the event that $S_i$ and $S_{i+1}$ sit adjacent. We want $P(T_1 \cap T_2 \cap T_3 \cap T_4) = 1 - P(E_1 \cup E_2 \cup E_3 \cup E_4)$.
Using the principle of inclusion-exclusion,the number of permutations where at least one pair is adjacent is calculated. For $n=5$,the number of permutations where no two adjacent students $S_i, S_{i+1}$ are together is $14$.
Thus,$P(T_1 \cap T_2 \cap T_3 \cap T_4) = \frac{14}{120} = \frac{7}{60}$.

(C) Let the cards be $C_1, C_2, C_3, C_4, C_5, C_6$ and envelopes be $E_1, E_2, E_3, E_4, E_5, E_6$.
Given $C_1$ is in $E_2$.
We need to place $C_2, C_3, C_4, C_5, C_6$ in $E_1, E_3, E_4, E_5, E_6$ such that $C_i$ is not in $E_i$ for $i \in \{2, 3, 4, 5, 6\}$.
Case $1$: $C_2$ is in $E_1$.
Then we need to place $C_3, C_4, C_5, C_6$ in $E_3, E_4, E_5, E_6$ such that no card $C_i$ is in $E_i$. This is a derangement of $4$ objects,$D_4 = 4!(\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}) = 24(\frac{1}{2} - \frac{1}{6} + \frac{1}{24}) = 12 - 4 + 1 = 9$.
Case $2$: $C_2$ is not in $E_1$.
We have $5$ cards $C_2, C_3, C_4, C_5, C_6$ to be placed in $E_1, E_3, E_4, E_5, E_6$ such that $C_2 \neq E_1, C_3 \neq E_3, C_4 \neq E_4, C_5 \neq E_5, C_6 \neq E_6$. This is a derangement of $5$ objects,$D_5 = 5!(\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!}) = 120(\frac{60-20+5-1}{120}) = 44$.
Total ways = $9 + 44 = 53$.

(C) The total number of ways to arrange $5$ letters in $5$ envelopes is $5! = 120$.
There is only $1$ way in which all letters go into their respective correct envelopes.
Therefore,the probability that all letters go into the correct envelopes is $P(E) = \frac{1}{5!} = \frac{1}{120}$.
The probability that all letters are not dispatched in the respective right envelopes is $P(E') = 1 - P(E)$.
$P(E') = 1 - \frac{1}{120} = \frac{119}{120}$.

(B) To divide $52$ cards into groups of sizes $17, 17, 17,$ and $1$,we first consider the number of ways to partition the set of $52$ cards into these groups.
Since $3$ groups have the same size $(17)$,we must account for the indistinguishability of these groups if we were just partitioning,but here the players are distinct.
The number of ways to distribute $52$ distinct items into $4$ distinct groups of sizes $n_1, n_2, n_3, n_4$ is given by the multinomial coefficient: $\frac{52!}{n_1! n_2! n_3! n_4!}$.
Here,$n_1=17, n_2=17, n_3=17, n_4=1$.
Thus,the number of ways is $\frac{52!}{17! 17! 17! 1!} = \frac{52!}{(17!)^3}$.

(B) The total number of ways to place $3$ letters in $3$ envelopes is $3! = 6$.
There is only $1$ way to place all letters in their correct envelopes.
The probability that all letters are placed in the correct envelopes is $\frac{1}{3!} = \frac{1}{6}$.
The probability that all letters are not placed in the right envelopes is $1 - \frac{1}{6} = \frac{5}{6}$.

(B) Total number of fruits = $3 \times 12 = 36$.
Number of persons = $9$.
Each person receives an equal number of fruits,so each person gets $\frac{36}{9} = 4$ fruits.
The number of ways to distribute $36$ distinct fruits into $9$ groups of $4$ each is given by the multinomial coefficient:
$\frac{36!}{4! \times 4! \times 4! \times 4! \times 4! \times 4! \times 4! \times 4! \times 4!} = \frac{36!}{(4!)^9}$.
Thus,the correct option is $B$.

(D) Total persons = $15$. Groups required are of sizes $3, 5, 7$.
Let the two particular persons be $P_1$ and $P_2$.
We need to form groups such that $P_1$ and $P_2$ are not in the group of $5$.
Case $1$: Both $P_1$ and $P_2$ are in the group of $3$.
Remaining $13$ persons are to be divided into groups of $1, 5, 7$.
Number of ways = $\binom{13}{1} \times \binom{12}{5} \times \binom{7}{7} = 13 \times \frac{12!}{5!7!} = 13 \times \frac{12 \times 11!}{5!7!} = \frac{156 \times 11!}{5!7!}$.
Case $2$: Both $P_1$ and $P_2$ are in the group of $7$.
Remaining $13$ persons are to be divided into groups of $3, 5, 5$.
Number of ways = $\binom{13}{3} \times \binom{10}{5} \times \binom{5}{5} = \frac{13!}{3!10!} \times \frac{10!}{5!5!} = \frac{13!}{3!5!5!}$.
Case $3$: One is in the group of $3$ and one is in the group of $7$.
Number of ways = $\binom{2}{1} \times \binom{13}{2} \times \binom{11}{4} \times \binom{7}{6} = 2 \times \frac{13!}{2!11!} \times \frac{11!}{4!7!} \times 7 = \frac{13!}{4!6!}$.
Summing these cases or using the complement method: Total ways to divide $15$ into $3, 5, 7$ is $\frac{15!}{3!5!7!}$.
Ways where $P_1, P_2$ are in the group of $5$ is $\binom{13}{3} \times \binom{10}{7} = \frac{13!}{3!10!} \times \frac{10!}{7!3!} = \frac{13!}{3!3!7!}$.
Subtracting this from total: $\frac{15!}{3!5!7!} - \frac{13!}{3!3!7!} = \frac{13!}{3!7!} [\frac{15 \times 14}{5 \times 4} - \frac{1}{3}] = \frac{13!}{3!7!} [4.2 - 0.33] = 13 \times \frac{11!}{3!5!}$.

(C) The number of ways to divide $mn$ distinct objects into $m$ groups of equal size $n$ is given by the formula $\frac{(mn)!}{(n!)^m \cdot m!}$.
Here,$mn = 200$ and $n = 20$,which implies $m = 10$.
Substituting these values into the formula,we get the number of ways as $\frac{200!}{(20!)^{10} \cdot 10!}$.

(C) The total number of ways to place $6$ letters in $6$ envelopes is $6!$.
Let $S$ be the total number of ways,which is $6!$.
Let $E_0$ be the case where all letters are in the correct envelopes ($1$ way).
Let $E_1$ be the case where exactly one letter is in the wrong envelope. This is impossible,as if $5$ are correct,the $6^{th}$ must also be correct.
Therefore,the number of ways where at least two letters are in the wrong envelopes is the total ways minus the ways where $0$ or $1$ letter is in the wrong envelope.
This is equivalent to $6! - (1 + 0) = 6! - 1$.
Alternatively,using the derangement formula $D_r$ (where $r$ is the number of letters in wrong envelopes),we sum the cases where $r$ letters are in wrong envelopes for $r = 2, 3, 4, 5, 6$.
The number of ways is $\sum_{r=2}^6 { }^6 C_{6-r} D_r = { }^6 C_4 D_2 + { }^6 C_3 D_3 + { }^6 C_2 D_4 + { }^6 C_1 D_5 + { }^6 C_0 D_6$.

(D) The total number of ways to place $5$ letters in $5$ addressed envelopes is $5! = 120$.
The number of ways in which all letters are placed in the correct envelopes is $1$.
Thus,the probability that all letters are placed correctly is $\frac{1}{120}$.
The probability that at least one letter is placed in the wrongly addressed envelope is $1 - P(\text{all correct}) = 1 - \frac{1}{120} = \frac{119}{120}$.

(D) The total number of ways to place $8$ balls in $8$ bags is $8!$.
We want to find the number of ways such that exactly $5$ balls are in their respective coloured bags.
First,we choose $5$ balls out of $8$ to be in their correct bags,which can be done in $\binom{8}{5}$ ways.
The remaining $3$ balls must be placed in the remaining $3$ bags such that none of them is in its respective coloured bag. This is a derangement of $3$ items,denoted by $D_3$.
$D_3 = 3! \times (1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!}) = 6 \times (1 - 1 + 0.5 - 0.1667) = 2$.
So,the number of favorable outcomes is $\binom{8}{5} \times D_3 = \binom{8}{3} \times 2 = 56 \times 2 = 112$.
The probability is $\frac{112}{8!} = \frac{112}{40320} = \frac{1}{360}$.

(C) Given that $5$ letters are written to $5$ different people and $5$ envelopes are addressed to them.
The number of ways in which these letters can be arranged so that no letter goes into its corresponding envelope is equal to the number of derangements of $5$ objects,denoted by $D_n$ or $!n$.
The formula for the derangement of $n$ objects is $D_n = n! \left[ 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \ldots + (-1)^n \frac{1}{n!} \right]$.
Here,$n = 5$.
$D_5 = 5! \left[ 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} \right]$.
$D_5 = 120 \left[ 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} \right]$.
$D_5 = 120 \left[ \frac{60 - 20 + 5 - 1}{120} \right]$.
$D_5 = 60 - 20 + 5 - 1 = 44$.
Hence,the number of ways to put $5$ letters in $5$ addressed envelopes so that all are in wrong envelopes is $44$.

(D) This is a problem of derangements,denoted by $D_n$,where $n$ is the number of items.
For $n=4$,the number of derangements is given by the formula $D_n = n! \times \left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}\right)$.
Substituting $n=4$:
$D_4 = 24 \times \left(1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24}\right)$
$D_4 = 24 \times \left(\frac{12 - 4 + 1}{24}\right)$
$D_4 = 24 \times \frac{9}{24} = 9$.
Alternatively,using the inclusion-exclusion principle:
Total ways $= 4! = 24$.
Ways where at least one letter is in the correct envelope $= \binom{4}{1} \times 3! - \binom{4}{2} \times 2! + \binom{4}{3} \times 1! - \binom{4}{4} \times 0! = 24 - 12 + 4 - 1 = 15$.
Required ways $= 24 - 15 = 9$.

(C) The total number of ways to match $5$ men with $5$ wives is given by $5! = 120$.
The number of ways in which no man is matched with his own wife is a derangement of $5$ objects,denoted by $D_5$.
The formula for derangement is $D_n = n! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \dots + \frac{(-1)^n}{n!} \right)$.
For $n = 5$:
$D_5 = 5! \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} \right) = 120 \left( \frac{60 - 20 + 5 - 1}{120} \right) = 44$.
Therefore,the probability that no man is matched with his wife is $\frac{D_5}{5!} = \frac{44}{120} = \frac{11}{30}$.

(B) To find the number of ways such that exactly $4$ cards go into their respective envelopes:
$1$. First,select $4$ cards out of $7$ to be placed in their correct envelopes. This can be done in ${ }^{7} C_{4}$ ways.
$2$. The remaining $3$ cards must be placed in the remaining $3$ envelopes such that none of them go into their respective envelopes (this is a derangement of $3$ objects,denoted by $D(3)$).
$3$. The number of derangements of $n$ objects is given by $D(n) = n! \left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \dots + \frac{(-1)^n}{n!}\right)$.
$4$. For $n = 3$,$D(3) = 3! \left(1 - 1 + \frac{1}{2} - \frac{1}{6}\right) = 6 \times \left(\frac{1}{3}\right) = 2$.
$5$. Total ways = ${ }^{7} C_{4} \times D(3) = { }^{7} C_{3} \times 2 = 2 \times { }^{7} C_{3}$.

(B) We need to distribute $4$ distinct books into $3$ distinct bags such that no bag is empty.
This is equivalent to finding the number of onto functions from a set of $4$ elements to a set of $3$ elements.
The formula for the number of onto functions from a set of $n$ elements to a set of $m$ elements is given by $\sum_{k=0}^{m} (-1)^k \binom{m}{k} (m-k)^n$.
Here,$n = 4$ and $m = 3$.
Number of ways = $\binom{3}{0} 3^4 - \binom{3}{1} 2^4 + \binom{3}{2} 1^4 = 1 \times 81 - 3 \times 16 + 3 \times 1 = 81 - 48 + 3 = 36$.