Definition of combinations, Condition combinations Questions in English

(A) The problem asks for the number of ways to select $6$ coins from $3$ different types of coins,where each type has enough coins (at least $6$) to satisfy the selection.
This is a problem of finding the number of non-negative integer solutions to the equation $x_1 + x_2 + x_3 = 6$,where $x_1, x_2, x_3$ represent the number of one rupee,fifty paise,and twenty-five paise coins selected,respectively.
Using the stars and bars formula for combinations with repetition,the number of ways is given by $^{n + r - 1}C_r$,where $n = 3$ (types of coins) and $r = 6$ (number of coins to select).
Therefore,the number of ways is $^{3 + 6 - 1}C_6 = ^8C_6$.
Using the property $^nC_r = ^nC_{n-r}$,we have $^8C_6 = ^8C_2 = \frac{8 \times 7}{2 \times 1} = 28$.

(B) The problem is equivalent to finding the number of non-negative integer solutions to the equation $x_1 + x_2 + x_3 = 35$,where $x_i \ge 0$.
Using the stars and bars formula,the number of ways is given by $^{n+r-1}C_{r-1}$,where $n = 35$ and $r = 3$.
Number of ways $= ^{35+3-1}C_{3-1} = ^{37}C_2$.
$^{37}C_2 = \frac{37 \times 36}{2} = 37 \times 18 = 666$.

(B) To draw $5$ red balls from $10$ red balls,the number of ways is $^{10}C_{5}$.
To draw $4$ white balls from $8$ white balls,the number of ways is $^{8}C_{4}$.
Since these are independent events,the total number of ways is the product of the two combinations:
Total ways $= ^{10}C_{5} \times ^{8}C_{4}$.

(B) The given expression is $^{14}C_4 + \sum_{j=1}^{4} {^{18-j}C_3}$.
Expanding the summation,we get: $^{14}C_4 + (^{17}C_3 + ^{16}C_3 + ^{15}C_3 + ^{14}C_3)$.
Using the Pascal's identity,$^{n}C_r + ^{n}C_{r-1} = ^{n+1}C_r$,we have $^{14}C_4 + ^{14}C_3 = ^{15}C_4$.
Now,$^{15}C_4 + ^{15}C_3 = ^{16}C_4$.
Continuing this process,$^{16}C_4 + ^{16}C_3 = ^{17}C_4$.
Finally,$^{17}C_4 + ^{17}C_3 = ^{18}C_4$.

(A) The committee consists of $6$ members chosen from $8$ gentlemen and $4$ ladies,with at least $3$ ladies.
Case $1$: $3$ ladies and $3$ gentlemen.
Number of ways = $^4C_3 \times ^8C_3 = 4 \times 56 = 224$.
Case $2$: $4$ ladies and $2$ gentlemen.
Number of ways = $^4C_4 \times ^8C_2 = 1 \times 28 = 28$.
Total number of ways = $224 + 28 = 252$.

(A) The total number of ways to select $r$ coins from $(2n + 1)$ distinct coins is given by $\binom{2n+1}{r}$.
Given that the person selects at least $1$ and at most $n$ coins,the total number of ways $T$ is:
$T = \binom{2n+1}{1} + \binom{2n+1}{2} + \dots + \binom{2n+1}{n} = 255$.
We know that the sum of binomial coefficients is $\sum_{r=0}^{2n+1} \binom{2n+1}{r} = 2^{2n+1}$.
Since $\binom{2n+1}{r} = \binom{2n+1}{2n+1-r}$,we have $\binom{2n+1}{0} = \binom{2n+1}{2n+1} = 1$.
Thus,$\binom{2n+1}{0} + \binom{2n+1}{1} + \dots + \binom{2n+1}{n} + \binom{2n+1}{n+1} + \dots + \binom{2n+1}{2n+1} = 2^{2n+1}$.
This simplifies to $1 + T + T + 1 = 2^{2n+1}$,which is $2 + 2T = 2^{2n+1}$.
Dividing by $2$,we get $1 + T = 2^{2n}$.
Substituting $T = 255$,we get $1 + 255 = 2^{2n}$,so $256 = 2^{2n}$.
Since $256 = 2^8$,we have $2n = 8$,which implies $n = 4$.

(D) For each of the $10$ friends,the man has $2$ choices: either to invite them or not to invite them.
Since there are $10$ friends,the total number of ways to invite any number of friends (including zero) is $2 \times 2 \times \dots \times 2$ ($10$ times),which is $2^{10}$.
However,the question specifies that he must invite 'one or more' friends,so we must exclude the case where no friends are invited (i.e.,the case where he chooses not to invite all $10$ friends).
Therefore,the required number of ways is $2^{10} - 1$.

(B) The student needs to select $10$ questions out of $13$ with the condition that at least $4$ must be chosen from the first $5$ questions.
Case $I$: Selecting $4$ questions from the first $5$ and $6$ questions from the remaining $8$.
Number of ways $= {^5C_4} \times {^8C_6} = 5 \times 28 = 140$.
Case $II$: Selecting $5$ questions from the first $5$ and $5$ questions from the remaining $8$.
Number of ways $= {^5C_5} \times {^8C_5} = 1 \times 56 = 56$.
Total number of choices $= 140 + 56 = 196$.

(B) The student can select at least one book in $T$ ways,where $T = {}^{2n+1}C_1 + {}^{2n+1}C_2 + ... + {}^{2n+1}C_n = 63$.
We know that the sum of all binomial coefficients for $(2n+1)$ is ${}^{2n+1}C_0 + {}^{2n+1}C_1 + ... + {}^{2n+1}C_{2n+1} = 2^{2n+1}$.
Since ${}^{2n+1}C_r = {}^{2n+1}C_{2n+1-r}$,we have ${}^{2n+1}C_0 = {}^{2n+1}C_{2n+1} = 1$.
Thus,$1 + ({}^{2n+1}C_1 + ... + {}^{2n+1}C_n) + ({}^{2n+1}C_{n+1} + ... + {}^{2n+1}C_{2n}) + 1 = 2^{2n+1}$.
Since the sum of the first $n$ terms is $T$,and the sum of the next $n$ terms is also $T$,we have $1 + T + T + 1 = 2^{2n+1}$.
$2 + 2T = 2^{2n+1} \Rightarrow 1 + T = 2^{2n}$.
Given $T = 63$,we get $1 + 63 = 2^{2n} \Rightarrow 64 = 2^{2n}$.
$2^6 = 2^{2n}$ $\Rightarrow 2n = 6$ $\Rightarrow n = 3$.

(D) Given the equation: $^{n-1}C_r = (k^2 - 3) \cdot ^nC_{r+1}$
Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$,we have:
$\frac{(n-1)!}{r!(n-r-1)!} = (k^2 - 3) \cdot \frac{n!}{(r+1)!(n-r-1)!}$
Simplifying the factorials:
$\frac{1}{r!} = (k^2 - 3) \cdot \frac{n}{(r+1)r!}$
$1 = (k^2 - 3) \cdot \frac{n}{r+1}$
$k^2 - 3 = \frac{r+1}{n}$
$k^2 = \frac{r+1}{n} + 3$
Since $0 \le r \le n-1$,we have $1 \le r+1 \le n$,which implies $\frac{1}{n} \le \frac{r+1}{n} \le 1$.
Thus,$k^2 \in [\frac{1}{n} + 3, 4]$.
For $n \ge 2$,the value of $\frac{1}{n} + 3$ lies in the interval $(3, 3.5]$.
Therefore,$k \in [-2, -\sqrt{\frac{1}{n} + 3}] \cup [\sqrt{\frac{1}{n} + 3}, 2]$.
Comparing with the given options,the interval $(\sqrt{3}, 2)$ is the most appropriate subset.

(B) We are given the expression $S = {}^{50}C_4 + \sum_{r=1}^{6} {}^{56-r}C_3$.
Expanding the summation,we get $S = {}^{50}C_4 + ({}^{55}C_3 + {}^{54}C_3 + {}^{53}C_3 + {}^{52}C_3 + {}^{51}C_3 + {}^{50}C_3)$.
Rearranging the terms,$S = ({}^{50}C_4 + {}^{50}C_3) + {}^{51}C_3 + {}^{52}C_3 + {}^{53}C_3 + {}^{54}C_3 + {}^{55}C_3$.
Using the Pascal's identity ${}^{n}C_r + {}^{n}C_{r-1} = {}^{n+1}C_r$,we have ${}^{50}C_4 + {}^{50}C_3 = {}^{51}C_4$.
Now,$S = ({}^{51}C_4 + {}^{51}C_3) + {}^{52}C_3 + {}^{53}C_3 + {}^{54}C_3 + {}^{55}C_3 = {}^{52}C_4 + {}^{52}C_3 + {}^{53}C_3 + {}^{54}C_3 + {}^{55}C_3$.
Continuing this process,$S = ({}^{52}C_4 + {}^{52}C_3) + {}^{53}C_3 + {}^{54}C_3 + {}^{55}C_3 = {}^{53}C_4 + {}^{53}C_3 + {}^{54}C_3 + {}^{55}C_3$.
$S = ({}^{53}C_4 + {}^{53}C_3) + {}^{54}C_3 + {}^{55}C_3 = {}^{54}C_4 + {}^{54}C_3 + {}^{55}C_3$.
$S = ({}^{54}C_4 + {}^{54}C_3) + {}^{55}C_3 = {}^{55}C_4 + {}^{55}C_3 = {}^{56}C_4$.
Thus,the correct option is $B$.

(B) Given $^nC_{12} = ^nC_6$.
Using the property $^nC_r = ^nC_k \Rightarrow r = k$ or $r + k = n$.
Since $12 \neq 6$,we have $n = 12 + 6 = 18$.
Now,calculate $^nC_2$ for $n = 18$:
$^nC_2 = ^{18}C_2 = \frac{18 \times 17}{2 \times 1} = 9 \times 17 = 153$.

(B) Let the amounts received by the $4$ persons be $x_1, x_2, x_3, x_4$ such that $x_1 + x_2 + x_3 + x_4 = 16$,where $x_i \ge 3$ for each $i \in \{1, 2, 3, 4\}$.
Let $y_i = x_i - 3$,where $y_i \ge 0$.
Substituting $x_i = y_i + 3$ into the equation:
$(y_1 + 3) + (y_2 + 3) + (y_3 + 3) + (y_4 + 3) = 16$
$y_1 + y_2 + y_3 + y_4 + 12 = 16$
$y_1 + y_2 + y_3 + y_4 = 4$
Using the stars and bars formula,the number of non-negative integer solutions is given by $\binom{n + k - 1}{k - 1}$,where $n = 4$ and $k = 4$.
Number of ways = $\binom{4 + 4 - 1}{4 - 1} = \binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.

(D) Let $S$ be the number of subsets containing at most $n$ elements. Then,$S = \binom{2n+1}{0} + \binom{2n+1}{1} + \dots + \binom{2n+1}{n}$.
We know that the total number of subsets of a set with $(2n+1)$ elements is $2^{2n+1}$.
Using the property $\binom{n}{r} = \binom{n}{n-r}$,we have:
$\binom{2n+1}{0} = \binom{2n+1}{2n+1}$
$\binom{2n+1}{1} = \binom{2n+1}{2n}$
$\dots$
$\binom{2n+1}{n} = \binom{2n+1}{n+1}$
Summing these,we get $2S = \binom{2n+1}{0} + \binom{2n+1}{1} + \dots + \binom{2n+1}{n} + \binom{2n+1}{n+1} + \dots + \binom{2n+1}{2n+1} = 2^{2n+1}$.
Therefore,$S = \frac{2^{2n+1}}{2} = 2^{2n}$.

(B) The number of positive integer solutions to the equation $x + y + z = n$ is given by the formula $\binom{n-1}{r-1}$,where $n = 100$ and $r = 3$.
Using the stars and bars method for positive integers:
Number of solutions $= \binom{100-1}{3-1} = \binom{99}{2}$.
Calculating the value:
$\binom{99}{2} = \frac{99 \times 98}{2 \times 1} = 99 \times 49 = 4851$.
Alternatively,using generating functions:
The number of solutions is the coefficient of $x^{100}$ in $(x + x^2 + x^3 + \dots)^3$.
$= \text{coefficient of } x^{100} \text{ in } x^3(1 - x)^{-3}$.
$= \text{coefficient of } x^{97} \text{ in } (1 - x)^{-3}$.
$= \binom{97 + 3 - 1}{3 - 1} = \binom{99}{2} = 4851$.

(A) The total number of balls in the box is $2 + 3 + 4 = 9$.
We need to select $3$ balls such that at least one is black.
This can be calculated by subtracting the number of ways to select $3$ balls with no black balls from the total number of ways to select $3$ balls.
Total ways to select $3$ balls from $9$ is $^{9}C_{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
Number of ways to select $3$ balls such that no black ball is chosen (i.e.,selecting from the $2$ white and $4$ red balls,total $6$ non-black balls) is $^{6}C_{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
Therefore,the number of ways to select at least one black ball is $84 - 20 = 64$.

(A) Let $n_i$ be the marks assigned to the $i^{th}$ question,where $i = 1, 2, \dots, 8$.
We are given $n_1 + n_2 + \dots + n_8 = 30$ with $n_i \ge 2$.
Let $x_i = n_i - 2$,so $x_i \ge 0$.
Substituting this into the sum: $(x_1 + 2) + (x_2 + 2) + \dots + (x_8 + 2) = 30$.
$x_1 + x_2 + \dots + x_8 + 16 = 30$,which simplifies to $x_1 + x_2 + \dots + x_8 = 14$.
The number of non-negative integer solutions is given by the formula $\binom{n+r-1}{r-1}$,where $n = 14$ and $r = 8$.
Number of ways = $\binom{14+8-1}{8-1} = \binom{21}{7} = ^{21}C_{7}$.

(B) The total number of ways to select $5$ members from $10$ people ($6$ men + $4$ women) is given by $^{10}C_5 = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$.
The number of ways to form a committee with no women (i.e.,all $5$ members are men) is given by $^6C_5 = 6$.
The number of ways to have at least one woman is the total number of ways minus the number of ways with no women:
$252 - 6 = 246$.

(D) Total persons available are $10$. We need to form a team of $5$ and arrange them in a line.
Since $A$ must be included,we have $1$ spot filled by $A$,leaving $4$ spots to be filled.
Since $G$ and $H$ must not be included,we exclude them from the remaining $9$ people,leaving $7$ people to choose from.
We need to select $4$ people from the remaining $7$,which can be done in $^7C_4$ ways.
Since $^7C_4 = ^7C_3$,the number of ways to select the team is $^7C_3$.
Once the $5$ members are selected (including $A$),they can be arranged in a line in $5!$ ways.
Therefore,the total number of arrangements is $^7C_3 \times 5!$.

(C) The word $MISSISSIPPI$ contains the following letters:
$M: 1, I: 4, S: 4, P: 2$.
To form a combination of one or more letters,we can choose any number of $M$s (from $0$ to $1$),$I$s (from $0$ to $4$),$S$s (from $0$ to $4$),and $P$s (from $0$ to $2$).
The total number of ways to select these letters is $(1+1)(4+1)(4+1)(2+1) = 2 \times 5 \times 5 \times 3 = 150$.
Since the question asks for combinations of one or more letters,we must exclude the case where no letters are selected (i.e.,the case where we choose $0$ of each letter).
Therefore,the total number of combinations is $150 - 1 = 149$.

(B) Let the total number of children be $n = 8$. The father takes $r = 3$ children at a time.
Total number of ways to select $3$ children out of $8$ is given by the combination formula ${^n}{C_r} = {^8}{C_3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.
Each trip involves $3$ children. Therefore,the total number of 'child-trips' is $56 \times 3 = 168$.
Since there are $8$ children and each child is treated equally,the number of times each child goes to the garden is $\frac{168}{8} = 21$.
Alternatively,for a specific child to be included in a group of $3$,the father must choose $2$ more children from the remaining $7$ children. This can be done in ${^7}{C_2} = \frac{7 \times 6}{2 \times 1} = 21$ ways.

(C) The given expression is $^x{C_r} ^y{C_0} + ^x{C_{r-1}} ^y{C_1} + ^x{C_{r-2}} ^y{C_2} + \dots + ^x{C_0} ^y{C_r}$.
This is a standard identity known as Vandermonde's Identity.
According to Vandermonde's Identity,the sum of the products of combinations $\sum_{k=0}^{r} {^x{C_{r-k}} ^y{C_k}}$ is equal to $^{x+y}{C_r}$.
Therefore,the correct option is $C$.

(A) Let $x_i$ be the marks assigned to the $i$-th question,where $i = 1, 2, \dots, 8$.
We are given that $x_1 + x_2 + \dots + x_8 = 30$ and $x_i \ge 2$ for all $i$.
Let $y_i = x_i - 2$,so $y_i \ge 0$.
Substituting $x_i = y_i + 2$ into the equation:
$(y_1 + 2) + (y_2 + 2) + \dots + (y_8 + 2) = 30$
$y_1 + y_2 + \dots + y_8 + 16 = 30$
$y_1 + y_2 + \dots + y_8 = 14$
The number of non-negative integer solutions is given by the formula $^{n+r-1}C_{r-1}$,where $n = 14$ and $r = 8$.
Number of ways = $^{14+8-1}C_{8-1} = ^{21}C_7$.

(C) Total players = $22$.
$4$ players are always excluded,so remaining players = $22 - 4 = 18$.
$2$ specific players must be included in every team,so we need to select $11 - 2 = 9$ more players.
These $9$ players must be selected from the remaining $18 - 2 = 16$ players.
The number of ways to select $9$ players from $16$ is $^{16}C_9$.

(A) Using the Pascal's identity $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$,we have:
$\binom{n}{3} + \binom{n}{4} = \binom{n+1}{4}$
Given the inequality:
$\binom{n+1}{4} > \binom{n+1}{3}$
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 $n-2 > 0$ (as $n \ge 4$ for the combinations to be defined),
$n-2 > 4$
$n > 6$

(D) Given the equation $\binom{n-1}{r} = (k^2 - 3) \binom{n}{r+1}$.
We know that $\binom{n}{r+1} = \frac{n}{r+1} \binom{n-1}{r}$.
Substituting this into the equation,we get $\binom{n-1}{r} = (k^2 - 3) \frac{n}{r+1} \binom{n-1}{r}$.
Assuming $\binom{n-1}{r} \neq 0$,we have $1 = (k^2 - 3) \frac{n}{r+1}$,which implies $k^2 - 3 = \frac{r+1}{n}$.
Since $0 \leq r \leq n-1$,we have $1 \leq r+1 \leq n$.
Dividing by $n$,we get $\frac{1}{n} \leq \frac{r+1}{n} \leq 1$.
As $n$ can be any positive integer,the range of $\frac{r+1}{n}$ is $(0, 1]$.
Thus,$0 < k^2 - 3 \leq 1$,which simplifies to $3 < k^2 \leq 4$.
Solving for $k$,we get $k \in [-2, -\sqrt{3}) \cup (\sqrt{3}, 2]$.

(A) Total number of ways to form a committee of $5$ members from $6$ men and $4$ women ($10$ people total) is given by $^{10}C_5 = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$.
The number of ways to form a committee with no women (i.e.,all $5$ members are men) is given by $^6C_5 = 6$.
Therefore,the number of ways to form a committee with at least one woman is $252 - 6 = 246$.

(B) We use the Pascal's identity: $\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}$.
Given expression: $\binom{n}{r+1} + 2\binom{n}{r} + \binom{n}{r-1}$.
This can be rewritten as: $\left[ \binom{n}{r+1} + \binom{n}{r} \right] + \left[ \binom{n}{r} + \binom{n}{r-1} \right]$.
Applying the identity:
$= \binom{n+1}{r+1} + \binom{n+1}{r}$.
Applying the identity again:
$= \binom{n+2}{r+1}$.

(B) The number of ways to select $3$ men from $5$ men is given by $^5C_3 = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$.
The number of ways to select $2$ women from $4$ women is given by $^4C_2 = \frac{4 \times 3}{2 \times 1} = 6$.
Therefore,the total number of ways to form the committee is $^5C_3 \times ^4C_2 = 10 \times 6 = 60$.

(C) We use the Pascal's identity for combinations: $\binom{n}{r-1} + \binom{n}{r} = \binom{n+1}{r}$.
Comparing this with the given equation $\binom{189}{35} + \binom{189}{x} = \binom{190}{x}$,we identify $n = 189$ and $r = x$.
For the identity to hold,the lower term in the second binomial coefficient must be $r = x$ and the lower term in the first must be $r-1 = 35$.
Thus,$x - 1 = 35$,which gives $x = 36$.

(A) Since $\binom{n-1}{4}$,$\binom{n-1}{5}$,and $\binom{n-1}{6}$ are in arithmetic progression,we have:
$2 \binom{n-1}{5} = \binom{n-1}{4} + \binom{n-1}{6}$
Using the property $\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}$,we can rewrite the equation:
$\binom{n-1}{4} + \binom{n-1}{5} + \binom{n-1}{5} + \binom{n-1}{6} = 4 \binom{n-1}{5}$
$\binom{n}{5} + \binom{n}{6} = 4 \binom{n-1}{5}$
$\binom{n+1}{6} = 4 \binom{n-1}{5}$
Expanding the combinations:
$\frac{(n+1)n(n-1)(n-2)(n-3)(n-4)}{6!} = 4 \frac{(n-1)(n-2)(n-3)(n-4)(n-5)}{5!}$
$\frac{(n+1)n}{6} = 4(n-5)$
$n^2 + n = 24(n-5)$
$n^2 - 23n + 120 = 0$
$(n-15)(n-8) = 0$
Thus,$n = 15$ or $n = 8$.

(C) The word $CORGOO$ contains $6$ letters: $C, O, R, G, O, O$. The distinct letters are $C, R, G, O$. The frequency is $O: 3, C: 1, R: 1, G: 1$.
We need to select $4$ letters. The possible cases are:
$(i)$ All $4$ letters are distinct: This is not possible as there are only $4$ distinct letters available $(C, R, G, O)$. Number of ways = $^4C_4 = 1$.
$(ii)$ $2$ letters are same and $2$ are distinct: We select one pair of $O$'s $(^1C_1 = 1)$ and $2$ letters from the remaining $3$ distinct letters $(C, R, G)$. Number of ways = $1 \times ^3C_2 = 3$.
$(iii)$ $3$ letters are same and $1$ is distinct: We select the three $O$'s $(^1C_1 = 1)$ and $1$ letter from the remaining $3$ distinct letters $(C, R, G)$. Number of ways = $1 \times ^3C_1 = 3$.
Total number of ways = $1 + 3 + 3 = 7$.

(B) Let the amount received by the $4$ persons be $x_1, x_2, x_3, x_4$ respectively.
We are given $x_1 + x_2 + x_3 + x_4 = 16$,where $x_i \ge 3$ for all $i \in \{1, 2, 3, 4\}$.
Let $y_i = x_i - 3$. Then $y_i \ge 0$.
Substituting $x_i = y_i + 3$ into the equation:
$(y_1 + 3) + (y_2 + 3) + (y_3 + 3) + (y_4 + 3) = 16$
$y_1 + y_2 + y_3 + y_4 + 12 = 16$
$y_1 + y_2 + y_3 + y_4 = 4$
The number of non-negative integer solutions is given by the formula $\binom{n+r-1}{r-1}$,where $n=4$ and $r=4$.
Number of ways $= \binom{4+4-1}{4-1} = \binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.

(B) For each of the $n$ distinct books,there are $p$ copies available.
For a specific book,we can choose $0, 1, 2, \dots, p$ copies.
This gives $(p + 1)$ choices for each book.
Since there are $n$ such books,the total number of ways to select any number of books (including selecting zero books) is $(p + 1)^n$.
To select one or more books,we must exclude the case where zero books are selected from all $n$ types.
Therefore,the required number of ways is $(p + 1)^n - 1$.

(B) The number of ways to distribute $n$ identical items into $r$ distinct boxes such that no box is empty is given by the formula $^{n-1}C_{r-1}$.
Here,$n = 8$ and $r = 3$.
Substituting the values,we get:
$^{8-1}C_{3-1} = ^7C_2$.
Calculating the value:
$^7C_2 = \frac{7 \times 6}{2 \times 1} = 21$.

(A) We know that if $\binom{n}{x} = \binom{n}{y}$,then either $x = y$ or $x + y = n$.
Case $1$: $3r = r + 3$
$2r = 3$
$r = 1.5$ (Not an integer,so this case is rejected).
Case $2$: $3r + (r + 3) = 15$
$4r + 3 = 15$
$4r = 12$
$r = 3$.

(C) Given the equation: $\binom{n+1}{3} = 2 \times \binom{n}{2}$
Using the formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$,we have:
$\frac{(n+1)!}{3!(n-2)!} = 2 \times \frac{n!}{2!(n-2)!}$
Since $(n+1)! = (n+1) \times n!$ and $3! = 6$,$2! = 2$:
$\frac{(n+1) \times n!}{6 \times (n-2)!} = 2 \times \frac{n!}{2 \times (n-2)!}$
Canceling $n!$ and $(n-2)!$ from both sides:
$\frac{n+1}{6} = \frac{2}{2}$
$\frac{n+1}{6} = 1$
$n+1 = 6$
$n = 5$

(A) The voter can cast their vote for $1$,$2$,or $3$ candidates. \\ The number of ways to choose $1$ candidate from $6$ is $^6C_1 = 6$. \\ The number of ways to choose $2$ candidates from $6$ is $^6C_2 = \frac{6 \times 5}{2 \times 1} = 15$. \\ The number of ways to choose $3$ candidates from $6$ is $^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$. \\ Total number of ways $= ^6C_1 + ^6C_2 + ^6C_3 = 6 + 15 + 20 = 41$.

(C) Given $\binom{n}{r-1} = 36$,$\binom{n}{r} = 84$,and $\binom{n}{r+1} = 126$.
Taking the ratio of the first two terms:
$\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{84}{36} = \frac{7}{3}$
$\frac{n!}{r!(n-r)!} \times \frac{(r-1)!(n-r+1)!}{n!} = \frac{7}{3}$
$\frac{n-r+1}{r} = \frac{7}{3} \implies 3n - 3r + 3 = 7r \implies 3n - 10r = -3$ (Equation $1$)
Taking the ratio of the next two terms:
$\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{126}{84} = \frac{3}{2}$
$\frac{n!}{(r+1)!(n-r-1)!} \times \frac{r!(n-r)!}{n!} = \frac{3}{2}$
$\frac{n-r}{r+1} = \frac{3}{2} \implies 2n - 2r = 3r + 3 \implies 2n - 5r = 3$ (Equation $2$)
Multiply Equation $2$ by $2$:
$4n - 10r = 6$ (Equation $3$)
Subtract Equation $1$ from Equation $3$:
$(4n - 10r) - (3n - 10r) = 6 - (-3)$
$n = 9$
Substitute $n=9$ into Equation $2$:
$2(9) - 5r = 3 \implies 18 - 5r = 3 \implies 5r = 15 \implies r = 3$.

(B) The number of ways to distribute $n$ identical items among $r$ recipients such that each recipient can receive any number of items (including zero) is given by the formula $\binom{n+r-1}{r-1}$.
Here,$n = 35$ and $r = 3$.
So,the number of ways $= \binom{35+3-1}{3-1} = \binom{37}{2}$.
Calculating the value: $\binom{37}{2} = \frac{37 \times 36}{2 \times 1} = 37 \times 18 = 666$.

(A) To form a committee of $5$ members where the number of gentlemen is greater than the number of ladies,we consider the following cases:
Case $1$: $4$ gentlemen and $1$ lady. The number of ways is $^4C_4 \times ^6C_1 = 1 \times 6 = 6$.
Case $2$: $3$ gentlemen and $2$ ladies. The number of ways is $^4C_3 \times ^6C_2 = 4 \times 15 = 60$.
Total number of ways = $6 + 60 = 66$.

(C) We use the identity $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$.
Given expression: $\binom{18}{15} + \binom{18}{16} + \binom{18}{16} + \binom{17}{16} + 1 = \binom{n}{3}$.
Note that $1 = \binom{17}{17}$.
So,$\binom{18}{15} + \binom{18}{16} + \binom{18}{16} + \binom{17}{16} + \binom{17}{17} = \binom{n}{3}$.
Using the identity $\binom{17}{16} + \binom{17}{17} = \binom{18}{17}$,we get:
$\binom{18}{15} + \binom{18}{16} + \binom{18}{16} + \binom{18}{17} = \binom{n}{3}$.
Using the identity $\binom{18}{16} + \binom{18}{17} = \binom{19}{17}$,we get:
$\binom{18}{15} + \binom{18}{16} + \binom{19}{17} = \binom{n}{3}$.
Using the identity $\binom{18}{15} + \binom{18}{16} = \binom{19}{16}$,we get:
$\binom{19}{16} + \binom{19}{17} = \binom{n}{3}$.
Using the identity $\binom{19}{16} + \binom{19}{17} = \binom{20}{17}$,we get:
$\binom{20}{17} = \binom{n}{3}$.
Since $\binom{20}{17} = \binom{20}{20-17} = \binom{20}{3}$,we have $\binom{20}{3} = \binom{n}{3}$.
Therefore,$n = 20$.

(D) The $3$ reserved positions must be filled by the $5$ reserved category candidates,which can be done in $\binom{5}{3}$ ways.
After filling the $3$ reserved positions,there are $12 - 3 = 9$ positions remaining.
The total number of candidates remaining is $25 - 3 = 22$ (since $3$ reserved candidates are already selected,but the remaining $2$ reserved candidates and $20$ open candidates are available for the $9$ open positions).
Thus,the $9$ open positions can be filled from the remaining $22$ candidates in $\binom{22}{9}$ ways.
The total number of ways for the selection is $\binom{5}{3} \cdot \binom{22}{9}$.

(B) Given that $^nC_{15} = ^nC_8$.
Using the property $^nC_x = ^nC_y \implies x + y = n$,we get $n = 15 + 8 = 23$.
Now,we need to find the value of $^nC_{21} = ^{23}C_{21}$.
Using the property $^nC_r = ^nC_{n-r}$,we have $^{23}C_{21} = ^{23}C_{23-21} = ^{23}C_2$.
Calculating $^{23}C_2 = \frac{23 \times 22}{2 \times 1} = 23 \times 11 = 253$.

(A) The number of ways to visit the zoo is equal to the number of ways to select $3$ children out of $8$ children.
Therefore,the number of ways is given by the combination formula:
$^8C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.

(D) We need to choose $4$ people out of $8$ ($4$ husbands and $4$ wives) such that no couple is selected.
Let the couples be $(H_1, W_1), (H_2, W_2), (H_3, W_3), (H_4, W_4)$.
First,we select $4$ couples out of $4$ from which we will pick one person each. This can be done in $^4C_4 = 1$ way.
From each of the $4$ selected couples,we can choose either the husband or the wife. This gives $2^4$ ways.
Total number of ways $= ^4C_4 \times 2^4 = 1 \times 16 = 16$.

(C) Total number of persons to be selected is $6$ from $4$ officers and $8$ constables.
To include at least one officer,we can use the complement method: Total ways - Ways with no officers.
Total ways to select $6$ persons from $12$ $(4+8)$ is $\binom{12}{6} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 924$.
Ways to select $6$ persons with no officers (i.e.,all $6$ are constables) is $\binom{8}{6} = \binom{8}{2} = \frac{8 \times 7}{2} = 28$.
Number of ways with at least one officer = $924 - 28 = 896$.

(C) To select two balls of the same color,we have two mutually exclusive cases:
Case $1$: Selecting $2$ black balls from $4$ distinct black balls.
Number of ways = $^4C_2 = \frac{4 \times 3}{2 \times 1} = 6$.
Case $2$: Selecting $2$ white balls from $3$ distinct white balls.
Number of ways = $^3C_2 = ^3C_1 = 3$.
Total number of ways = $6 + 3 = 9$.