Permutation and Combination Questions in English

(C) We use the identity $^nC_r + ^nC_{r-1} = ^{n+1}C_r$.
Given expression: $^nC_r + 2^nC_{r-1} + ^nC_{r-2}$.
Split the middle term: $^nC_r + ^nC_{r-1} + ^nC_{r-1} + ^nC_{r-2}$.
Group the terms: $(^nC_r + ^nC_{r-1}) + (^nC_{r-1} + ^nC_{r-2})$.
Apply the identity: $^{n+1}C_r + ^{n+1}C_{r-1}$.
Apply the identity again: $^{n+2}C_r$.

(D) To find the total number of handshakes,we need to choose $2$ persons out of $8$ to perform one handshake.
This is a combination problem because the order of the two people shaking hands does not matter.
The number of ways to choose $2$ persons out of $8$ is given by the combination formula $^nC_r = \frac{n!}{r!(n-r)!}$.
Here,$n = 8$ and $r = 2$.
Total handshakes = $^8C_2 = \frac{8 \times 7}{2 \times 1} = 28$.
Therefore,the total number of handshakes is $28$.

(A) The expression $^nC_r + ^nC_{r-1}$ follows the Pascal's identity for combinations.
Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$,we have:
$^nC_r + ^nC_{r-1} = \frac{n!}{r!(n-r)!} + \frac{n!}{(r-1)!(n-r+1)!}$
Taking common factors out:
$= \frac{n!}{(r-1)!(n-r)!} \left[ \frac{1}{r} + \frac{1}{n-r+1} \right]$
$= \frac{n!}{(r-1)!(n-r)!} \left[ \frac{n-r+1+r}{r(n-r+1)} \right]$
$= \frac{n!}{(r-1)!(n-r)!} \left[ \frac{n+1}{r(n-r+1)} \right]$
$= \frac{(n+1)n!}{r(r-1)!(n-r+1)(n-r)!} = \frac{(n+1)!}{r!(n+1-r)!} = ^{n+1}C_r$.

(B) Given that $^nC_a = ^nC_b$,then either $a = b$ or $a + b = n$.
Here,$^8C_r = ^8C_{r+2}$.
Since $r \neq r + 2$,we must have $r + (r + 2) = 8$.
$2r + 2 = 8$
$2r = 6$
$r = 3$.
Now,we need to find the value of $^rC_2$,which is $^3C_2$.
$^3C_2 = \frac{3!}{2!(3-2)!} = \frac{3 \times 2 \times 1}{2 \times 1 \times 1} = 3$.

(D) The given equation is $^{20}C_{n+2} = ^nC_{16}$.
For a combination $^nC_r$ to be defined,the condition $n \ge r$ must be satisfied.
In the expression $^nC_{16}$,we must have $n \ge 16$.
In the expression $^{20}C_{n+2}$,we must have $20 \ge n+2$,which implies $n \le 18$.
Thus,$n$ must be in the range $16 \le n \le 18$.
If $n = 16$,then $^{20}C_{18} = ^{16}C_{16} \implies 190 = 1$,which is false.
If $n = 17$,then $^{20}C_{19} = ^{17}C_{16} \implies 20 = 17$,which is false.
If $n = 18$,then $^{20}C_{20} = ^{18}C_{16} \implies 1 = 153$,which is false.
Since no integer value of $n$ in the valid range satisfies the equation,there is no value for $n$.

(A) We use the property of combinations: $^{n}C_r = ^{n}C_{n-r}$.
First,simplify $^{15}C_{13}$ using this property:
$^{15}C_{13} = ^{15}C_{15-13} = ^{15}C_2$.
Now,use the Pascal's identity: $^{n}C_r + ^{n}C_{r-1} = ^{n+1}C_r$.
Substituting the values:
$^{15}C_3 + ^{15}C_2 = ^{15+1}C_3 = ^{16}C_3$.
Therefore,the correct option is $A$.

(B) Let the total number of persons in the room be $n$.
Since every person shakes hands with every other person exactly once,the total number of handshakes is given by the combination formula $^nC_2$,which represents choosing $2$ people out of $n$ to perform a handshake.
Given that the total number of handshakes is $66$,we have the equation: $^nC_2 = 66$.
Using the formula $^nC_2 = \frac{n(n-1)}{2}$,we get:
$\frac{n(n-1)}{2} = 66$
$n(n-1) = 132$
$n^2 - n - 132 = 0$
Factoring the quadratic equation: $(n - 12)(n + 11) = 0$.
Since the number of persons $n$ must be positive,we have $n = 12$.
Therefore,the total number of persons in the room is $12$.

(C) The given inequality is $^{10}C_{x-1} > 2 \cdot ^{10}C_x$.
Using the formula $^{n}C_r = \frac{n!}{r!(n-r)!}$, we expand the terms:
$\frac{10!}{(x-1)!(10-(x-1))!} > 2 \cdot \frac{10!}{x!(10-x)!}$
$\frac{10!}{(x-1)!(11-x)!} > 2 \cdot \frac{10!}{x!(10-x)!}$
Since $10!$ is common, we simplify:
$\frac{1}{(x-1)!(11-x)(10-x)!} > 2 \cdot \frac{1}{x(x-1)!(10-x)!}$
$\frac{1}{11-x} > \frac{2}{x}$
Since $x$ must be a positive integer and $1 \le x \le 10$, we cross-multiply:
$x > 2(11-x)$
$x > 22 - 2x$
$3x > 22$
$x > \frac{22}{3} \approx 7.33$
Since $x$ is an integer and $x \le 10$, the possible values for $x$ are ${8, 9, 10}$.

(A) We know the property of binomial coefficients: $^{n}{C_{r}} = ^{n}{C_{n-r}}$.
Thus,$^{n+r}{C_{n}} = ^{n+r}{C_{(n+r)-n}} = ^{n+r}{C_{r}}$.
We need to evaluate the sum $S = \sum_{r=0}^{m} {^{n+r}{C_{r}}} = ^{n}{C_{0}} + ^{n+1}{C_{1}} + ^{n+2}{C_{2}} + \dots + ^{n+m}{C_{m}}$.
Using the Hockey-stick identity,which states that $\sum_{i=r}^{n} {^{i}{C_{r}}} = ^{n+1}{C_{r+1}}$,we can rewrite our sum.
Here,the sum is $\sum_{r=0}^{m} {^{n+r}{C_{r}}}$.
By the identity $\sum_{i=0}^{m} {^{n+i}{C_{i}}} = ^{n+m+1}{C_{m}}$,or more standardly $\sum_{k=0}^{m} {^{n+k}{C_{k}}} = ^{n+m+1}{C_{m}}$.
Since $^{n+m+1}{C_{m}} = ^{n+m+1}{C_{(n+m+1)-m}} = ^{n+m+1}{C_{n+1}}$.
Therefore,the sum is equal to $^{n+m+1}{C_{n+1}}$.

(B) Let the number of teams be $n$.
Since every team plays one match with every other team,the total number of matches is given by the combination formula $^nC_2$,which represents choosing $2$ teams out of $n$ to play a match.
Given that the total number of matches is $153$,we have:
$^nC_2 = 153$
$\frac{n(n - 1)}{2} = 153$
$n(n - 1) = 306$
$n^2 - n - 306 = 0$
Solving the quadratic equation using the quadratic formula or factorization:
$(n - 18)(n + 17) = 0$
Since the number of teams cannot be negative,we have $n = 18$.
Therefore,the number of teams participating is $18$.

(D) Each question can be answered in $4$ ways.
Since there are $3$ questions,the total number of ways to answer all questions is $4 \times 4 \times 4 = 4^3 = 64$.
There is only $1$ way to answer all questions correctly (i.e.,choosing the correct option for every question).
Therefore,the number of ways a student can fail to get all answers correct is the total number of ways minus the way to get all correct: $64 - 1 = 63$.

(D) Given $\alpha = {^m}{C_2} = \frac{m(m-1)}{2}$.
We need to find ${^\alpha}{C_2} = \frac{\alpha(\alpha-1)}{2}$.
Substituting the value of $\alpha$:
${^\alpha}{C_2} = \frac{\frac{m(m-1)}{2} \left( \frac{m(m-1)}{2} - 1 \right)}{2}$
$= \frac{1}{4} \cdot \frac{m(m-1)}{2} \cdot (m^2 - m - 2)$
$= \frac{1}{8} m(m-1)(m-2)(m+1)$
$= \frac{3}{1} \cdot \frac{(m+1)m(m-1)(m-2)}{4 \cdot 3 \cdot 2 \cdot 1}$
$= 3 \cdot {^{m+1}}{C_4}$.

(B) In a class of $20$ students,if each student sends a card to every other student,we first consider the number of pairs that can be formed. The number of ways to select $2$ students out of $20$ is given by $^{20}C_2$.
Since each pair of students involves two cards being exchanged (student $A$ sends to $B$,and student $B$ sends to $A$),the total number of cards exchanged is $2 \times ^{20}C_2$.
Calculating this: $2 \times \frac{20 \times 19}{2 \times 1} = 380$ cards.

(C) Each person has $32$ potential positions for teeth.
For each position,there are $2$ possibilities: either a tooth is present or it is absent.
Since there are $32$ such positions,the total number of possible combinations of teeth is $2 \times 2 \times \dots \times 2$ ($32$ times),which equals $2^{32}$.
However,the problem states that there is no person without a tooth,meaning the case where all $32$ positions are empty must be excluded.
Therefore,the maximum population of the city is $2^{32} - 1$.

(B) Given the ratio $^{2n}C_2 : ^nC_2 = 9:2$.
Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$,we have:
$\frac{2n(2n-1)}{2} : \frac{n(n-1)}{2} = 9:2$
$\frac{2n(2n-1)}{n(n-1)} = \frac{9}{2}$
$4(2n-1) = 9(n-1)$
$8n - 4 = 9n - 9$
$n = 5$
Now,substitute $n = 5$ into $^nC_r = 10$:
$^5C_r = 10$
Since $^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$,we get $r = 2$.

(D) Given the property of combinations,if $^nC_x = ^nC_y$,then either $x = y$ or $x + y = n$.
Here,$^{10}C_r = ^{10}C_{r+2}$.
Since $r \neq r + 2$,we must have $r + (r + 2) = 10$.
$2r + 2 = 10$
$2r = 8$
$r = 4$.
Now,we need to find the value of $^5C_r$,which is $^5C_4$.
Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$,we get $^5C_4 = \frac{5!}{4!(5-4)!} = \frac{5 \times 4!}{4! \times 1!} = 5$.

(B) We are given the following combinations:
$1$) $\frac{^nC_r}{^nC_{r-1}} = \frac{84}{36} = \frac{7}{3}$
Using the formula $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$,we get $\frac{n-r+1}{r} = \frac{7}{3} \implies 3n - 3r + 3 = 7r \implies 3n - 10r = -3$ (Equation $1$)
$2$) $\frac{^nC_{r+1}}{^nC_r} = \frac{126}{84} = \frac{3}{2}$
Using the formula $\frac{^nC_{r+1}}{^nC_r} = \frac{n-r}{r+1}$,we get $\frac{n-r}{r+1} = \frac{3}{2} \implies 2n - 2r = 3r + 3 \implies 2n - 5r = 3$ (Equation $2$)
Multiply Equation $2$ by $2$: $4n - 10r = 6$ (Equation $3$)
Subtract Equation $1$ from Equation $3$: $(4n - 10r) - (3n - 10r) = 6 - (-3) \implies n = 9$.
Substituting $n=9$ into Equation $2$: $2(9) - 5r = 3 \implies 18 - 5r = 3 \implies 5r = 15 \implies r = 3$.

(A) We use the Pascal's identity: $^nC_r + ^nC_{r-1} = ^{n+1}C_r$.
Applying this to the given inequality: $^nC_3 + ^nC_4 = ^{n+1}C_4$.
So,the inequality becomes: $^{n+1}C_4 > ^{n+1}C_3$.
Expanding the combinations: $\frac{(n+1)!}{4!(n-3)!} > \frac{(n+1)!}{3!(n-2)!}$.
Dividing both sides by $(n+1)!$ and simplifying the factorials:
$\frac{1}{4 \times 3!(n-3)!} > \frac{1}{3!(n-2)(n-3)!}$.
$\frac{1}{4} > \frac{1}{n-2}$.
Since $n > 3$ for the combinations to be defined,$n-2$ is positive,so we can cross-multiply:
$n - 2 > 4$.
$n > 6$.

(C) We know that if ${}^nC_x = {}^nC_y$,then either $x = y$ or $x + y = n$.
Given equation: ${}^{15}C_{r + 3} = {}^{15}C_{2r - 6}$.
Case $1$: $r + 3 = 2r - 6$
$r - 2r = -6 - 3$
$-r = -9$
$r = 9$.
Case $2$: $(r + 3) + (2r - 6) = 15$
$3r - 3 = 15$
$3r = 18$
$r = 6$.
Since $r$ must satisfy the condition $0 \le r+3 \le 15$ and $0 \le 2r-6 \le 15$,both $r=9$ and $r=6$ are valid solutions. However,looking at the options provided,$6$ is the correct choice.

(C) Given equation: $^{n + 1}{C_3} = 2{\,^n}{C_2}$
Using the formula $^{n}{C_r} = \frac{n!}{r!(n-r)!}$,we expand both sides:
$\frac{(n + 1)!}{3!(n + 1 - 3)!} = 2 \times \frac{n!}{2!(n - 2)!}$
$\frac{(n + 1) \times n!}{6 \times (n - 2)!} = 2 \times \frac{n!}{2 \times (n - 2)!}$
Cancel $n!$ and $(n - 2)!$ from both sides:
$\frac{n + 1}{6} = \frac{2}{2}$
$\frac{n + 1}{6} = 1$
$n + 1 = 6$
$n = 5$

(D) We know that $\binom{n}{k} = \binom{n}{n-k}$.
Therefore,$\binom{n}{n-r} = \binom{n}{r}$.
Substituting this into the expression,we get:
$\binom{n}{n-r} + \binom{n}{r+1} = \binom{n}{r} + \binom{n}{r+1}$.
Using the Pascal's identity formula $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$,we obtain:
$\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$.

(A) Using the Pascal's identity,$^nC_r + ^nC_{r-1} = ^{n+1}C_r$,we have $^nC_6 + ^nC_5 = ^{n+1}C_6$.
Given the inequality: $^{n+1}C_6 > ^{n+1}C_5$.
Expanding the combinations: $\frac{(n+1)!}{6!(n-5)!} > \frac{(n+1)!}{5!(n-4)!}$.
Dividing both sides by $(n+1)!$ and simplifying the factorials:
$\frac{1}{6 \times 5!(n-5)!} > \frac{1}{5!(n-4)(n-5)!}$.
$\frac{1}{6} > \frac{1}{n-4}$.
Since $n$ is a natural number,$n-4 > 6$,which implies $n > 10$.
The least natural number $n$ satisfying this condition is $11$.

(B) In a party of $n$ persons,if every person shakes hands with every other person exactly once,the total number of handshakes is equivalent to choosing $2$ persons out of $n$ to form a pair.
This is calculated using the combination formula: $^nC_r = \frac{n!}{r!(n-r)!}$.
Here,$n = 15$ and $r = 2$.
Total handshakes = $^{15}C_2 = \frac{15 \times 14}{2 \times 1} = 105$.

(D) The given expression is a standard identity in combinations known as Pascal's Identity.
According to Pascal's Identity,for any positive integers $n$ and $r$ where $n \ge r$,the sum of two consecutive binomial coefficients is given by:
$^nC_{r-1} + ^nC_r = ^{n+1}C_r$
Therefore,the correct option is $D$.

(A) We know that if $^nC_a = ^nC_b$,then either $a = b$ or $a + b = n$.
Given the equation: $^{43}C_{r-6} = ^{43}C_{3r+1}$.
Case $1$: $r - 6 = 3r + 1$
$-7 = 2r$
$r = -3.5$ (This is not possible as $r$ must be a non-negative integer).
Case $2$: $(r - 6) + (3r + 1) = 43$
$4r - 5 = 43$
$4r = 48$
$r = 12$.
Thus,the value of $r$ is $12$.

(C) The given number is $112233$,which consists of $6$ digits in total.
The digits are $1, 1, 2, 2, 3, 3$.
Here,the digit $1$ repeats $2$ times,the digit $2$ repeats $2$ times,and the digit $3$ repeats $2$ times.
The number of ways to arrange these $6$ digits is given by the formula for permutations of a multiset:
$\text{Number of ways} = \frac{n!}{n_1! \cdot n_2! \cdot n_3!}$
Substituting the values: $n = 6$,$n_1 = 2$,$n_2 = 2$,$n_3 = 2$.
$\text{Number of ways} = \frac{6!}{2! \cdot 2! \cdot 2!} = \frac{720}{2 \cdot 2 \cdot 2} = \frac{720}{8} = 90$.
Thus,$90$ such numbers can be formed.

(C) voter can vote for $1, 2, 3, 4,$ or $5$ candidates.
Since the order of selection does not matter,we use combinations.
The total number of ways is given by the sum of combinations:
Total ways $= ^8C_1 + ^8C_2 + ^8C_3 + ^8C_4 + ^8C_5$
$= 8 + 28 + 56 + 70 + 56$
$= 218$
Thus,there are $218$ ways for a voter to cast their vote.

(C) Let the number of candidates be $n$. The number of persons to be elected is $n-1$.
A voter can vote for any number of candidates from $1$ to $n-1$.
The total number of ways a voter can vote is:
$^nC_1 + {}^nC_2 + \cdots + {}^nC_{n-1} = 254$
We know that:
$\sum_{r=0}^{n} {}^nC_r = 2^n$
Therefore,
$^nC_0 + {}^nC_1 + {}^nC_2 + \cdots + {}^nC_{n-1} + {}^nC_n = 2^n$
Substituting the known values:
$1 + 254 + 1 = 2^n$
$256 = 2^n$
$2^8 = 2^n$
Thus, $n = 8$

(A) To ensure that no two Hindi books are together,we first arrange the $21$ English books in a row.
These $21$ English books create $22$ possible gaps (including the ends) where the Hindi books can be placed: $\bullet E_1 \bullet E_2 \bullet E_3 \bullet ... \bullet E_{21} \bullet$.
Since there are $19$ Hindi books,we need to choose $19$ gaps out of the $22$ available gaps.
The number of ways to choose these gaps is given by the combination formula $^{n}C_{r} = \frac{n!}{r!(n-r)!}$.
Here,$n = 22$ and $r = 19$.
Total ways = $^{22}C_{19} = ^{22}C_{22-19} = ^{22}C_{3}$.
Calculating the value: $^{22}C_{3} = \frac{22 \times 21 \times 20}{3 \times 2 \times 1} = 22 \times 7 \times 10 = 1540$.
Thus,there are $1540$ ways to arrange the books.

(B) The given expression is a sum of combinations: $S = ^rC_r + ^{r+1}C_r + ^{r+2}C_r + ...... + ^{n-1}C_r + ^nC_r$.
We use the Hockey-Stick Identity (also known as the Christmas Stocking Identity),which states that $\sum_{i=r}^{n} {^iC_r} = ^{n+1}C_{r+1}$.
Step-by-step application:
$1$. Since $^rC_r = 1$ and $^{r+1}C_{r+1} = 1$,we can write $^rC_r$ as $^{r+1}C_{r+1}$.
$2$. Using the Pascal's Identity: $^nC_r + ^nC_{r-1} = ^{n+1}C_r$.
$3$. Applying this iteratively:
$(^{r+1}C_{r+1} + ^{r+1}C_r) + ^{r+2}C_r + ...... + ^nC_r$
$= ^{r+2}C_{r+1} + ^{r+2}C_r + ...... + ^nC_r$
$= ^{r+3}C_{r+1} + ...... + ^nC_r$
$= ^{n+1}C_{r+1}$.
Thus,the correct option is $B$.

(D) Step $1$: Select $3$ consonants from $5$ available consonants,which can be done in $^5C_3$ ways.
Step $2$: Select $2$ vowels from $4$ available vowels,which can be done in $^4C_2$ ways.
Step $3$: The total number of ways to select the letters is $^5C_3 \times ^4C_2$.
Step $4$: Since we have selected $5$ letters in total ($3$ consonants + $2$ vowels),these $5$ letters can be arranged among themselves in $5!$ ways.
Step $5$: Therefore,the total number of words that can be formed is $(^5C_3 \times ^4C_2) \times 5!$.

(B) Total players available = $25$.
Players to be included = $6$.
Players to be excluded = $5$.
Remaining players to choose from = $25 - 6 - 5 = 14$.
Players already selected = $6$.
Players still needed to complete the team of $11$ = $11 - 6 = 5$.
Therefore,the number of ways to choose $5$ players from the remaining $14$ is given by the combination formula $^{n}C_{r} = \frac{n!}{r!(n-r)!}$.
$^{14}C_{5} = \frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1} = 14 \times 13 \times 11 = 2002$.

(A) The total number of ways to form a committee of any number of members from a group of $n$ members is given by the sum of combinations: $\sum_{r=1}^{n} {^nC_r} = 2^n - 1$.
Here,$n = 12$.
Therefore,the number of ways to form a committee with one or more members is $2^{12} - 1$.
Calculating the value: $2^{12} = 4096$.
So,$4096 - 1 = 4095$.

(C) To select one or more balls from a collection of identical items of different types,we use the formula $(n_1 + 1)(n_2 + 1)(n_3 + 1) - 1$,where $n_1, n_2, n_3$ are the number of balls of each color.
Here,$n_1 = 10$ (white),$n_2 = 9$ (black),and $n_3 = 7$ (red).
The total number of ways to select one or more balls is given by $(10 + 1)(9 + 1)(7 + 1) - 1$.
$= 11 \times 10 \times 8 - 1$.
$= 880 - 1 = 879$.
Thus,the total number of ways is $879$.

(C) To find the number of permutations of $n$ things taken $r$ at a time such that $p$ specific things are always included:
$1$. First,we select the $r$ objects. Since $p$ objects must always be included,we need to choose the remaining $r-p$ objects from the remaining $n-p$ objects. This can be done in $^{n-p}C_{r-p}$ ways.
$2$. Once the $r$ objects are selected,they can be arranged among themselves in $r!$ ways.
$3$. Therefore,the total number of permutations is $^{n-p}C_{r-p} \times r!$.

(A) There are $52$ distinct types of cards (suit and denomination combination). Since there are two packs,each of these $52$ types appears twice.
To select $26$ cards such that no two cards have the same suit and same denomination,we must first choose $26$ distinct types out of the $52$ available types. This can be done in $^{52}C_{26}$ ways.
For each of the $26$ chosen cards,we have $2$ choices (either from the first pack or the second pack).
Since there are $26$ such cards,the total number of ways is $^{52}C_{26} \times 2^{26}$.

(B) Total players = $16$.
Number of bowlers = $5$.
Number of wicket-keepers = $2$.
Remaining players (batsmen/all-rounders) = $16 - (5 + 2) = 9$.
We need to select a team of $11$ players including $3$ bowlers,$1$ wicket-keeper,and the remaining $11 - (3 + 1) = 7$ players from the remaining $9$ players.
The number of ways to select the team is given by:
$= ^5C_3 \times ^2C_1 \times ^9C_7$
$= \frac{5 \times 4}{2 \times 1} \times 2 \times \frac{9 \times 8}{2 \times 1}$
$= 10 \times 2 \times 36 = 720$.

(B) The total number of ways to choose any number of books from $n$ distinct books is given by $2^n$.
This includes the case where no books are chosen (the empty set).
Since we need to choose one or more books,we must exclude the case where zero books are chosen.
Therefore,the required number of ways is $2^n - 1$.
For $n = 6$,the number of ways is $2^6 - 1 = 64 - 1 = 63$.

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

(A) To divide $52$ cards equally among four players,each player receives $13$ cards.
The number of ways to choose $13$ cards for the first player from $52$ is $^{52}C_{13}$.
The number of ways to choose $13$ cards for the second player from the remaining $39$ is $^{39}C_{13}$.
The number of ways to choose $13$ cards for the third player from the remaining $26$ is $^{26}C_{13}$.
The number of ways to choose $13$ cards for the fourth player from the remaining $13$ is $^{13}C_{13}$.
Total number of ways = $^{52}C_{13} \times ^{39}C_{13} \times ^{26}C_{13} \times ^{13}C_{13}$
$= \frac{52!}{39! \cdot 13!} \times \frac{39!}{26! \cdot 13!} \times \frac{26!}{13! \cdot 13!} \times \frac{13!}{0! \cdot 13!}$
$= \frac{52!}{(13!)^4}$.

(B) Step $1$: Select $4$ consonants from $6$ available consonants. The number of ways is $^6C_4 = \frac{6 \times 5}{2 \times 1} = 15$.
Step $2$: Select $3$ vowels from $5$ available vowels. The number of ways is $^5C_3 = \frac{5 \times 4}{2 \times 1} = 10$.
Step $3$: The total number of ways to select the letters is $15 \times 10 = 150$.
Step $4$: These $7$ selected letters can be arranged among themselves in $7!$ ways.
$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$.
Step $5$: The total number of words that can be formed is $150 \times 5040 = 756000$.

(C) Total players = $13$,Bowlers = $4$,Non-bowlers = $9$. We need to select $11$ players including at least $2$ bowlers.
Case $1$: $2$ bowlers and $9$ non-bowlers
Number of ways = $^4C_2 \times ^9C_9 = 6 \times 1 = 6$
Case $2$: $3$ bowlers and $8$ non-bowlers
Number of ways = $^4C_3 \times ^9C_8 = 4 \times 9 = 36$
Case $3$: $4$ bowlers and $7$ non-bowlers
Number of ways = $^4C_4 \times ^9C_7 = 1 \times 36 = 36$
Total number of ways = $6 + 36 + 36 = 78$.

(C) To ensure that no two '$-$' signs come together,we first arrange the six '$+$' signs in a row.
This creates $7$ possible gaps (including the ends) where the '$-$' signs can be placed: $\_ + \_ + \_ + \_ + \_ + \_ \_$.
We have $4$ '$-$' signs to place in these $7$ available gaps.
The number of ways to choose $4$ gaps out of $7$ is given by the combination formula ${^n}C_r = \frac{n!}{r!(n-r)!}$.
Therefore,the total number of ways is ${^7}C_4 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.

(B) To form a group,we must select at least one green ball,at least one blue ball,and any number of red balls (including zero).
$1$. Number of ways to select at least one green ball from $5$ different green balls is $2^5 - 1 = 32 - 1 = 31$ ways.
$2$. Number of ways to select at least one blue ball from $4$ different blue balls is $2^4 - 1 = 16 - 1 = 15$ ways.
$3$. Number of ways to select any number of red balls from $3$ different red balls (including the case of selecting zero red balls) is $2^3 = 8$ ways.
Total number of ways = $31 \times 15 \times 8 = 3720$.

(C) The total number of ways to select $6$ persons from $12$ (i.e.,$4$ officers + $8$ constables) is given by $^{12}C_6$.
The number of ways to select $6$ persons such that no officer is included (i.e.,all $6$ are selected from $8$ constables) is $^8C_6$.
According to the condition,at least one officer must be included. Therefore,the required number of ways is:
Total ways - Ways with no officers = $^{12}C_6 - ^8C_6$.
Calculating the values:
$^{12}C_6 = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 924$.
$^8C_6 = ^8C_2 = \frac{8 \times 7}{2 \times 1} = 28$.
Required ways = $924 - 28 = 896$.

(A) Total candidates = $25$. Scheduled caste candidates = $5$. Other candidates = $25 - 5 = 20$.
Total vacancies = $12$. Reserved vacancies for scheduled caste = $3$. Open vacancies = $12 - 3 = 9$.
Step $1$: Selection for $3$ reserved vacancies from $5$ scheduled caste candidates can be done in $^5C_3$ ways.
Step $2$: After filling $3$ reserved seats,the remaining candidates are $25 - 3 = 22$ (since $2$ scheduled caste candidates were not selected for reserved seats,they are now eligible for the open seats along with the $20$ other candidates).
Step $3$: The remaining $9$ vacancies are open to all remaining $22$ candidates. This can be done in $^{22}C_9$ ways.
Total number of ways = $^5C_3 \times ^{22}C_9$.

(D) voter can vote for at least $1$ candidate and at most $3$ candidates.
Since there are $5$ candidates,the number of ways to choose the candidates is given by the sum of combinations:
Ways = $^5C_1 + ^5C_2 + ^5C_3$
Calculating each term:
$^5C_1 = 5$
$^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$
$^5C_3 = ^5C_2 = 10$
Total ways = $5 + 10 + 10 = 25$.
Therefore,the voter can cast their vote in $25$ ways.

(A) Total chairs = $9$. Total persons = $6$.
One person (the guest) has a specific chair reserved for them. This means the guest has only $1$ way to be seated.
After the guest is seated,there are $5$ persons remaining and $8$ chairs remaining.
The number of ways to seat the remaining $5$ persons on the remaining $8$ chairs is given by the permutation formula $P(n, r) = \frac{n!}{(n-r)!}$.
Here,$n = 8$ and $r = 5$.
Number of ways = $P(8, 5) = \frac{8!}{(8-5)!} = \frac{8!}{3!} = 8 \times 7 \times 6 \times 5 \times 4 = 6720$.
Therefore,the total number of ways is $1 \times 6720 = 6720$.

(C) To form a group of $7$ with a majority of boys,we consider the following cases:
Case $1$: $6$ boys and $1$ girl. The number of ways is $^6C_6 \times ^4C_1 = 1 \times 4 = 4$.
Case $2$: $5$ boys and $2$ girls. The number of ways is $^6C_5 \times ^4C_2 = 6 \times 6 = 36$.
Case $3$: $4$ boys and $3$ girls. The number of ways is $^6C_4 \times ^4C_3 = 15 \times 4 = 60$.
Total number of ways = $4 + 36 + 60 = 100$.

(B) Let the two particular persons be $P_1$ and $P_2$. Since $P_1$ and $P_2$ cannot be in the same boat,one must be in the first boat and the other in the second boat.
If $P_1$ is in the first boat,we need to select $4$ more persons from the remaining $8$ persons to join $P_1$. This can be done in $^8C_4$ ways.
The remaining $4$ persons will automatically go to the second boat with $P_2$.
Alternatively,if $P_2$ is in the first boat,we select $4$ persons from the remaining $8$ to join $P_2$,which can be done in $^8C_4$ ways.
Thus,the total number of ways is $^8C_4 + ^8C_4 = 2 \times ^8C_4$.