Definition of combinations, Condition combinations Questions in English

(A) Total number of questions $= 13$.
Number of questions to be selected $= 10$.
Restriction: At least $4$ from the first $5$ questions must be selected.
Case $I$: Exactly $4$ out of the first $5$ questions are selected.
Number of ways to select $4$ questions from $5 = {}^{5}C_{4} = 5$.
Remaining $10 - 4 = 6$ questions are selected from the last $13 - 5 = 8$ questions in ${}^{8}C_{6}$ ways.
Number of ways $= 5 \times {}^{8}C_{6} = 5 \times 28 = 140$.
Case $II$: Exactly $5$ out of the first $5$ questions are selected.
Number of ways to select $5$ out of $5$ questions $= {}^{5}C_{5} = 1$.
Remaining $10 - 5 = 5$ questions are selected from the last $13 - 5 = 8$ questions in ${}^{8}C_{5}$ ways.
Number of ways $= 1 \times {}^{8}C_{5} = 1 \times 56 = 56$.
Total number of ways $= 140 + 56 = 196$.

(A) Statement $I$: The number of ways of distributing $n$ identical items into $r$ distinct boxes such that no box is empty is given by the formula ${}^{n-1}C_{r-1}$.
Here,$n = 10$ and $r = 4$.
So,the number of ways is ${}^{10-1}C_{4-1} = {}^9C_3$.
Thus,Statement $I$ is true.
Statement $II$: The number of ways of choosing $r$ items from $n$ distinct items is ${}^nC_r$.
For choosing $3$ places from $9$ different places,the number of ways is ${}^9C_3$.
Thus,Statement $II$ is true.
Since Statement $II$ is a standard combinatorial formula and not the logical derivation for the specific distribution problem in Statement $I$,Statement $II$ is not the correct explanation for Statement $I$.

(A) Given,a group of $10$ men and $8$ women. We need to form a committee of $8$ members such that there are not more than $5$ men and not less than $5$ women.
This implies the possible combinations of (Women,Men) are:
$(5W, 3M), (6W, 2M), (7W, 1M), (8W, 0M)$.
The number of ways is calculated as:
$= \binom{8}{5} \times \binom{10}{3} + \binom{8}{6} \times \binom{10}{2} + \binom{8}{7} \times \binom{10}{1} + \binom{8}{8} \times \binom{10}{0}$
$= (56 \times 120) + (28 \times 45) + (8 \times 10) + (1 \times 1)$
$= 6720 + 1260 + 80 + 1 = 8061$ ways.

(B) To get a sum of $7$ with $5$ dice,each die must show at least $1$. Let the outcomes be $x_1, x_2, x_3, x_4, x_5$ where $x_i \ge 1$. The sum is $x_1 + x_2 + x_3 + x_4 + x_5 = 7$.
Since each $x_i \ge 1$,we can write $x_i = 1 + y_i$ where $y_i \ge 0$.
Substituting this,we get $(1+y_1) + (1+y_2) + (1+y_3) + (1+y_4) + (1+y_5) = 7$,which simplifies to $y_1 + y_2 + y_3 + y_4 + y_5 = 2$.
The number of non-negative integer solutions is given by the formula $\binom{n+r-1}{r-1}$ where $n=2$ and $r=5$.
This is $\binom{2+5-1}{5-1} = \binom{6}{4} = \binom{6}{2} = \frac{6 \times 5}{2} = 15$.
Alternatively,the cases are:
$(i)$ Four $1$s and one $3$: $\frac{5!}{4!1!} = 5$ ways.
(ii) Three $1$s and two $2$s: $\frac{5!}{3!2!} = 10$ ways.
Total ways = $5 + 10 = 15$.

(A) The number of ways to distribute $N = m \times n$ distinct items equally among $n$ persons is given by the multinomial coefficient formula: $\frac{(mn)!}{(m!)^n}$.
Here,$N = 500$,$n = 50$,and $m = 10$ (since $500 = 50 \times 10$).
Therefore,the number of ways is $\frac{500!}{(10!)^{50}}$.

(C) To form a combination of $5$ cards with exactly one ace from a deck of $52$ cards:
$1$. Select $1$ ace from the $4$ available aces: $^4C_1 = 4$.
$2$. Select the remaining $4$ cards from the $48$ non-ace cards: $^{48}C_4$.
$3$. The total number of combinations is $^4C_1 \times ^{48}C_4$.
$\begin{aligned} & = 4 \times \frac{48 \times 47 \times 46 \times 45}{4 \times 3 \times 2 \times 1} \\ & = 4 \times 194580 \\ & = 778320 \end{aligned}$

(B) The total number of subsets with $8$ elements from a set of $n$ elements is given by $\binom{n}{8}$.
The number of subsets with $8$ elements that contain $a_4$ is equivalent to choosing $7$ more elements from the remaining $(n-1)$ elements,which is $\binom{n-1}{7}$.
According to the problem,$\binom{n}{8} = 5 \times \binom{n-1}{7}$.
Using the identity $\binom{n}{r} = \frac{n}{r} \binom{n-1}{r-1}$,we have $\binom{n}{8} = \frac{n}{8} \binom{n-1}{7}$.
Substituting this into the equation: $\frac{n}{8} \binom{n-1}{7} = 5 \times \binom{n-1}{7}$.
Since $\binom{n-1}{7} \neq 0$,we can divide both sides by $\binom{n-1}{7}$ to get $\frac{n}{8} = 5$.
Therefore,$n = 5 \times 8 = 40$.

(C) For the binomial coefficient $^{n}C_{r}$ to be a natural number,we must have $n \geq r \geq 0$ and $n, r \in \mathbb{N}_0$.
Given $^{20-2x}C_{x-3} \in N$,the following conditions must be satisfied:
$1) \; x-3 \geq 0 \Rightarrow x \geq 3$
$2) \; 20-2x \geq x-3$ $\Rightarrow 23 \geq 3x$ $\Rightarrow x \leq \frac{23}{3} \approx 7.66$
$3) \; 20-2x \geq 0 \Rightarrow x \leq 10$
Combining these inequalities,we get $3 \leq x \leq 7.66$.
Since $x \in N$,the possible values for $x$ are $3, 4, 5, 6, 7$.
Thus,there are $5$ elements in the set.

(C) Using the identity ${}^{n}C_{r} + {}^{n}C_{r-1} = {}^{n+1}C_{r}$,we have:
${}^{(n-1)}C_6 + {}^{(n-1)}C_7 = {}^{n}C_7$
Given inequality: ${}^{n}C_7 < {}^{n}C_8$
Using the property $\frac{{}^{n}C_{r}}{{}^{n}C_{r-1}} = \frac{n-r+1}{r}$,we get:
$\frac{{}^{n}C_8}{{}^{n}C_7} > 1$
$\frac{n-8+1}{8} > 1$
$\frac{n-7}{8} > 1$
$n-7 > 8$
$n > 15$
Therefore,the least integer value of $n$ is $16$.

(C) Given the inequality: ${ }^{(n-1)} C_3 + { }^{(n-1)} C_4 > { }^n C_3$
Using the Pascal's identity ${ }^n C_{r-1} + { }^n C_r = { }^{n+1} C_r$,we have:
${ }^{(n-1)} C_3 + { }^{(n-1)} C_4 = { }^n C_4$
Substituting this into the inequality:
${ }^n C_4 > { }^n C_3$
Expanding the combinations:
$\frac{n!}{4!(n-4)!} > \frac{n!}{3!(n-3)!}$
Since $n! > 0$,we can divide both sides by $n!$:
$\frac{1}{4!(n-4)!} > \frac{1}{3!(n-3)!}$
$\frac{1}{4 \times 3! \times (n-4)!} > \frac{1}{3! \times (n-3) \times (n-4)!}$
$\frac{1}{4} > \frac{1}{n-3}$
Since $n-3 > 0$ (as $n \ge 4$),we can cross-multiply:
$n - 3 > 4$
$n > 7$
Therefore,the least integer value of $n$ is $8$.

(D) Given the equation: $10 \cdot ^nC_2 = 3 \cdot ^{n+1}C_3$
Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$:
$10 \cdot \frac{n!}{2!(n-2)!} = 3 \cdot \frac{(n+1)!}{3!(n+1-3)!}$
$10 \cdot \frac{n(n-1)}{2} = 3 \cdot \frac{(n+1)n(n-1)}{6}$
$5n(n-1) = \frac{(n+1)n(n-1)}{2}$
Since $n > 2$,we can divide both sides by $n(n-1)$:
$5 = \frac{n+1}{2}$
$10 = n + 1$
$n = 9$

(A) Given $\frac{{ }^{2n}C_3}{{ }^{n}C_3} = \frac{12}{1}$.
Using the formula ${ }^{n}C_r = \frac{n!}{r!(n-r)!}$,we have:
$\frac{\frac{(2n)!}{3!(2n-3)!}}{\frac{n!}{3!(n-3)!}} = 12$
$\Rightarrow \frac{2n(2n-1)(2n-2)(2n-3)!}{3!(2n-3)!} \times \frac{3!(n-3)!}{n(n-1)(n-2)(n-3)!} = 12$
$\Rightarrow \frac{2n(2n-1) \cdot 2(n-1)}{n(n-1)(n-2)} = 12$
$\Rightarrow \frac{4n(2n-1)}{n(n-2)} = 12$
$\Rightarrow \frac{4(2n-1)}{n-2} = 12$
$\Rightarrow 2n-1 = 3(n-2)$
$\Rightarrow 2n-1 = 3n-6$
$\Rightarrow n = 5$.

(B) We use the property ${ }^n C_r={ }^n C_{n-r}$.
${ }^9 C_5={ }^9 C_{9-5}={ }^9 C_4$.
Now,the expression becomes ${ }^9 C_3+{ }^9 C_4$.
Using the Pascal's identity ${ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r$,we get:
${ }^9 C_4+{ }^9 C_3={ }^{10} C_4$.
Comparing this with ${ }^{10} C_r$,we find $r=4$.

(B) We use the Pascal's identity: ${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$.
The given expression is ${}^{34}C_5 + \sum_{i=0}^4 {}^{38-i}C_4 = {}^{34}C_5 + {}^{38}C_4 + {}^{37}C_4 + {}^{36}C_4 + {}^{35}C_4 + {}^{34}C_4$.
Rearranging the terms,we get $({}^{34}C_5 + {}^{34}C_4) + {}^{35}C_4 + {}^{36}C_4 + {}^{37}C_4 + {}^{38}C_4$.
Applying the identity ${}^{34}C_5 + {}^{34}C_4 = {}^{35}C_5$,the expression becomes ${}^{35}C_5 + {}^{35}C_4 + {}^{36}C_4 + {}^{37}C_4 + {}^{38}C_4$.
Continuing this process: ${}^{35}C_5 + {}^{35}C_4 = {}^{36}C_5$,then ${}^{36}C_5 + {}^{36}C_4 = {}^{37}C_5$,then ${}^{37}C_5 + {}^{37}C_4 = {}^{38}C_5$,and finally ${}^{38}C_5 + {}^{38}C_4 = {}^{39}C_5$.

(C) We use the identity ${ }^n C_r + { }^n C_{r-1} = { }^{n+1} C_r$.
Given: ${ }^{n-3} C_r + B \cdot { }^{n-3} C_{r-1} + B^{\prime} \cdot { }^{n-3} C_{r-2} + { }^{n-3} C_{r-3} = { }^n C_r$.
We know that ${ }^n C_r = { }^{n-1} C_r + { }^{n-1} C_{r-1} = ({ }^{n-2} C_r + { }^{n-2} C_{r-1}) + ({ }^{n-2} C_{r-1} + { }^{n-2} C_{r-2}) = { }^{n-2} C_r + 2 \cdot { }^{n-2} C_{r-1} + { }^{n-2} C_{r-2}$.
Expanding further: ${ }^{n-2} C_r + 2({ }^{n-3} C_{r-1} + { }^{n-3} C_{r-2}) + ({ }^{n-3} C_{r-2} + { }^{n-3} C_{r-3}) = { }^{n-3} C_r + { }^{n-3} C_{r-1} + 2 \cdot { }^{n-3} C_{r-1} + 2 \cdot { }^{n-3} C_{r-2} + { }^{n-3} C_{r-2} + { }^{n-3} C_{r-3} = { }^{n-3} C_r + 3 \cdot { }^{n-3} C_{r-1} + 3 \cdot { }^{n-3} C_{r-2} + { }^{n-3} C_{r-3}$.
Comparing this with the given equation ${ }^{n-3} C_r + B \cdot { }^{n-3} C_{r-1} + B^{\prime} \cdot { }^{n-3} C_{r-2} + { }^{n-3} C_{r-3} = { }^n C_r$,we get $B = 3$ and $B^{\prime} = 3$.
Thus,$(B, B^{\prime}) = (3, 3)$.

(C) We use the identity ${}^{n} C_r + {}^{n} C_{r-1} = {}^{n+1} C_r$.
Given expression: $\sum_{r=0}^4 {}^{19-r} C_3 + {}^{15} C_4$
$= {}^{19} C_3 + {}^{18} C_3 + {}^{17} C_3 + {}^{16} C_3 + {}^{15} C_3 + {}^{15} C_4$
$= {}^{19} C_3 + {}^{18} C_3 + {}^{17} C_3 + {}^{16} C_3 + ({}^{15} C_3 + {}^{15} C_4)$
$= {}^{19} C_3 + {}^{18} C_3 + {}^{17} C_3 + ({}^{16} C_3 + {}^{16} C_4)$
$= {}^{19} C_3 + {}^{18} C_3 + ({}^{17} C_3 + {}^{17} C_4)$
$= {}^{19} C_3 + ({}^{18} C_3 + {}^{18} C_4)$
$= {}^{19} C_3 + {}^{19} C_4$
$= {}^{20} C_4$

(D) Given,${}^n C_7 = {}^n C_6$.
Using the property: If ${}^n C_x = {}^n C_y$,then either $x = y$ or $x + y = n$.
Since $7 \neq 6$,we must have $n = 7 + 6 = 13$.
Therefore,${}^n C_2 = {}^{13} C_2 = \frac{13 \times 12}{2 \times 1} = 13 \times 6 = 78$.
Hence,option $D$ is correct.

(D) Given that,${ }^{12} C_{2 k-1}={ }^{12} C_{k+1}$.
We know that if ${ }^n C_x={ }^n C_y$,then either $x=y$ or $x+y=n$.
Case $1$: $2k-1 = k+1$
$k = 2$.
Case $2$: $(2k-1) + (k+1) = 12$
$3k = 12$
$k = 4$.
Since $k=4$ is one of the given options,the correct value is $4$.

(C) We are given the equation: $^{n-1}C_r = (k^2 - 3) \cdot ^nC_{r+1}$.
Using the identity $^nC_r = \frac{n}{r} \cdot ^{n-1}C_{r-1}$,we can write $^nC_{r+1} = \frac{n}{r+1} \cdot ^{n-1}C_r$.
Substituting this into the given equation:
$^{n-1}C_r = (k^2 - 3) \cdot \frac{n}{r+1} \cdot ^{n-1}C_r$.
Assuming $^{n-1}C_r \neq 0$,we divide both sides by $^{n-1}C_r$:
$1 = (k^2 - 3) \cdot \frac{n}{r+1}$.
Rearranging for $k^2 - 3$:
$k^2 - 3 = \frac{r+1}{n}$.
Since $0 \le r \le n-1$,the ratio $\frac{r+1}{n}$ lies in the interval $(0, 1]$.
Thus,$0 < k^2 - 3 \le 1$.
Adding $3$ to all parts:
$3 < k^2 \le 4$.
Taking the square root,we get $k \in [-2, -\sqrt{3}) \cup (\sqrt{3}, 2]$.
Comparing this with the given options,the interval $[\sqrt{3}, 2]$ is a subset of the possible values for $k$.

(A) Given $N(n) = n \prod_{r=1}^{2023} (n^2 - r^2) = n \cdot \left[ \prod_{r=1}^{2023} (n-r) \right] \left[ \prod_{r=1}^{2023} (n+r) \right]$.
For $n = 2024$,we have:
$N(2024) = 2024 \cdot [(2023)(2022) \dots (1)] \cdot [(2025)(2026) \dots (4047)]$.
Rearranging the terms,we get:
$N(2024) = (4047)(4046) \dots (2025)(2024)(2023) \dots (1) = (4047)!$.
We need to find ${}^{N}C_{N-1}$ where $N = N(2024) = (4047)!$.
Using the property ${}^{n}C_{r} = {}^{n}C_{n-r}$,we have ${}^{N}C_{N-1} = {}^{N}C_{1} = N$.
Therefore,${}^{N}C_{N-1} = (4047)!$.

(D) The given equation is $x_1+x_2+x_3+x_4=10$.
For non-negative integral solutions of the equation $x_1+x_2+...+x_r=n$,the formula is given by $^{n+r-1}C_{r-1}$.
Here,$n=10$ and $r=4$.
Substituting the values,we get:
$^{10+4-1}C_{4-1} = ^{13}C_3$.
Calculating the value:
$^{13}C_3 = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 13 \times 2 \times 11 = 286$.

(C) Given $n(A) = 8$.
The number of subsets of $A$ containing at least $6$ elements is given by the sum of combinations:
$^8C_6 + ^8C_7 + ^8C_8$
Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$:
$^8C_6 = \frac{8 \times 7}{2 \times 1} = 28$
$^8C_7 = \frac{8}{1} = 8$
$^8C_8 = 1$
Total subsets $= 28 + 8 + 1 = 37$.

(C) Since $2$ particular students are always to be included in the team,we have already selected $2$ members.
Remaining members to be selected $= 5 - 2 = 3$.
Remaining students available to choose from $= 12 - 2 = 10$.
Therefore,the number of ways to select the remaining $3$ students from $10$ available students is given by the combination formula ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$.
Number of ways $= {}^{10}C_{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 10 \times 3 \times 4 = 120$.

(B) We are looking for the number of triplets $(x, y, z)$ such that $x, y, z \in N$,$x < y < z$,and $x+y+z=12$.
Since $x < y < z$,we have $x+y+z > x+x+x = 3x$,so $3x < 12$,which implies $x < 4$. Thus,$x$ can be $1, 2,$ or $3$.
Case $1$: If $x=1$,then $y+z=11$ with $1 < y < z$. Possible pairs $(y, z)$ are $(2, 9), (3, 8), (4, 7), (5, 6)$. (Total $4$)
Case $2$: If $x=2$,then $y+z=10$ with $2 < y < z$. Possible pairs $(y, z)$ are $(3, 7), (4, 6)$. (Total $2$)
Case $3$: If $x=3$,then $y+z=9$ with $3 < y < z$. The only possible pair $(y, z)$ is $(4, 5)$. (Total $1$)
Total number of triplets $= 4 + 2 + 1 = 7$.

(C) The student must select $10$ questions out of $13$ such that at least $5$ are from the first $6$ questions. This implies two possible cases:
Case $I$: Selecting $5$ questions from the first $6$ and $5$ questions from the remaining $7$.
Number of ways $= {}^{6}C_{5} \times {}^{7}C_{5} = 6 \times 21 = 126$.
Case $II$: Selecting $6$ questions from the first $6$ and $4$ questions from the remaining $7$.
Number of ways $= {}^{6}C_{6} \times {}^{7}C_{4} = 1 \times 35 = 35$.
Total number of ways $= 126 + 35 = 161$.

(B) committee of $12$ members is to be formed such that women are in majority. Since there are $9$ women and $8$ men,the possible cases for women to be in majority are:
Case $I$: $9$ women and $3$ men
Number of ways $= {^9C_9} \times {^8C_3} = 1 \times \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$
Case $II$: $8$ women and $4$ men
Number of ways $= {^9C_8} \times {^8C_4} = 9 \times \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 9 \times 70 = 630$
Case $III$: $7$ women and $5$ men
Number of ways $= {^9C_7} \times {^8C_5} = \frac{9 \times 8}{2 \times 1} \times \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 36 \times 56 = 2016$
Total number of ways $= 56 + 630 + 2016 = 2702$

(D) The total number of ways to select any number of alternatives from $5$ available alternatives is given by the sum of combinations: $\binom{5}{0} + \binom{5}{1} + \binom{5}{2} + \binom{5}{3} + \binom{5}{4} + \binom{5}{5} = 2^5 = 32$.
Since the student must choose two or more than two alternatives,we must exclude the cases where $0$ or $1$ alternative is chosen.
The number of ways to choose $0$ alternatives is $\binom{5}{0} = 1$.
The number of ways to choose $1$ alternative is $\binom{5}{1} = 5$.
Therefore,the number of ways to choose two or more alternatives is $32 - (1 + 5) = 32 - 6 = 26$.

(A) Let the $3$-digit number be represented as $abc$,where $a$ is the hundreds digit,$b$ is the tens digit,and $c$ is the units digit.
Given the condition $a + b + c = 10$,where $1 \leq a \leq 9$ and $0 \leq b, c \leq 9$.
Let $a' = a - 1$,so $a = a' + 1$. Substituting this into the equation:
$(a' + 1) + b + c = 10 \Rightarrow a' + b + c = 9$,where $a', b, c \geq 0$.
The number of non-negative integral solutions is given by the formula $^{n+r-1}C_{r-1}$,where $n=9$ and $r=3$:
$^{9+3-1}C_{3-1} = ^{11}C_2 = \frac{11 \times 10}{2} = 55$.
However,we must exclude cases where any digit exceeds $9$.
Since $a = a' + 1$,if $a' = 9$,then $a = 10$,which is not possible for a digit.
This corresponds to the solution $(a', b, c) = (9, 0, 0)$,which gives $a = 10, b = 0, c = 0$.
Thus,we subtract this $1$ invalid case: $55 - 1 = 54$.
Therefore,the total number of such $3$-digit numbers is $54$.

(D) The student needs to select $10$ questions out of $13$. The first $5$ questions are in one group and the remaining $8$ questions are in another group.
He must answer at least $4$ questions from the first $5$ questions.
Case $1$: He selects $4$ 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 $2$: He selects $5$ 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 ways = $140 + 56 = 196$.

(C) To select a committee of $30$ persons such that the number of boys,girls,and teachers is equal,we must select $10$ boys,$10$ girls,and $10$ teachers.
The number of ways to select $10$ boys from $20$ is $^{20}C_{10} = \frac{20!}{10!10!}$.
The number of ways to select $10$ girls from $20$ is $^{20}C_{10} = \frac{20!}{10!10!}$.
The number of ways to select $10$ teachers from $20$ is $^{20}C_{10} = \frac{20!}{10!10!}$.
Since these selections are independent,the total number of ways is the product of these individual combinations:
Total ways $= \frac{20!}{10!10!} \times \frac{20!}{10!10!} \times \frac{20!}{10!10!} = \frac{(20!)^3}{(10!)^6}$.

(A) The voter can vote for $1, 2, 3,$ or $4$ candidates out of $12$ available candidates.
The number of ways to choose $k$ candidates out of $12$ is given by the combination formula ${}^{12}C_k$.
Since the voter must vote for at least one candidate,the total number of ways is the sum of selecting $1, 2, 3,$ or $4$ candidates:
$\text{Total ways} = {}^{12}C_1 + {}^{12}C_2 + {}^{12}C_3 + {}^{12}C_4$
$= 12 + \frac{12 \times 11}{2 \times 1} + \frac{12 \times 11 \times 10}{3 \times 2 \times 1} + \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1}$
$= 12 + 66 + 220 + 495 = 793$

(B) Given the equation $x+y+z+w=25$ with constraints $x, y, z \geq -1$ and $w \geq 1$.
Let $a = x+1 \geq 0$,$b = y+1 \geq 0$,$c = z+1 \geq 0$,and $d = w-1 \geq 0$.
Substituting these into the equation: $(a-1) + (b-1) + (c-1) + (d+1) = 25$.
This simplifies to $a+b+c+d-2 = 25$,or $a+b+c+d = 27$.
The number of non-negative integral solutions for $a+b+c+d = n$ is given by the formula ${}^{n+k-1}C_{k-1}$,where $k$ is the number of variables.
Here $n=27$ and $k=4$,so the number of solutions is ${}^{27+4-1}C_{4-1} = {}^{30}C_3$.

(A) Given the ratio: $\frac{{ }^{n+2} C_2}{{ }^{n+3} C_1} = \frac{4}{2} = 2$.
Using the formula ${ }^n C_r = \frac{n!}{r!(n-r)!}$,we have:
$\frac{(n+2)!}{2! n!} \div (n+3) = 2$.
$\frac{(n+2)(n+1)}{2} \times \frac{1}{n+3} = 2$.
$(n+2)(n+1) = 4(n+3)$.
$n^2 + 3n + 2 = 4n + 12$.
$n^2 - n - 10 = 0$.
Solving for $n$ using the quadratic formula $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$n = \frac{1 \pm \sqrt{1 - 4(1)(-10)}}{2} = \frac{1 \pm \sqrt{41}}{2}$.
Since $\sqrt{41}$ is not an integer,$n$ cannot be a natural number $(n \notin N)$.
Therefore,the number of values of $n$ is $0$.

(D) We need to form a committee of $5$ members from $7$ Indians $(I)$,$6$ Americans $(A)$,$5$ Russians $(R)$,and $4$ Australians $(AU)$ such that each country is represented at least once.
Since the total number of members is $5$ and there are $4$ countries,one country must have $2$ representatives,and the other three countries must have $1$ representative each.
The possible distributions are:
$1$. $2I, 1A, 1R, 1AU: {^7C_2} \times {^6C_1} \times {^5C_1} \times {^4C_1} = 21 \times 6 \times 5 \times 4 = 2520$
$2$. $1I, 2A, 1R, 1AU: {^7C_1} \times {^6C_2} \times {^5C_1} \times {^4C_1} = 7 \times 15 \times 5 \times 4 = 2100$
$3$. $1I, 1A, 2R, 1AU: {^7C_1} \times {^6C_1} \times {^5C_2} \times {^4C_1} = 7 \times 6 \times 10 \times 4 = 1680$
$4$. $1I, 1A, 1R, 2AU: {^7C_1} \times {^6C_1} \times {^5C_1} \times {^4C_2} = 7 \times 6 \times 5 \times 6 = 1260$
Total ways $= 2520 + 2100 + 1680 + 1260 = 7560$.

(D) The word $TSEAMCET$ contains $8$ letters: $T, T, E, E, S, A, M, C$.
There are $6$ distinct letters: $\{T, E, S, A, M, C\}$.
We need to select $4$ letters.
Case $1$: All $4$ letters are different.
Number of ways $= {}^{6}C_{4} = \frac{6 \times 5 \times 4 \times 3}{4 \times 3 \times 2 \times 1} = 15$.
Case $2$: One pair of identical letters and $2$ different letters.
Number of ways $= {}^{2}C_{1} \times {}^{5}C_{2} = 2 \times \frac{5 \times 4}{2 \times 1} = 20$.
Case $3$: Two pairs of identical letters.
Number of ways $= {}^{2}C_{2} = 1$.
Total number of ways $= 15 + 20 + 1 = 36$.

(B) The total number of ways to select at least $(n+1)$ books from $(2n+1)$ books is given by the sum of combinations:
$^{2n+1}C_{n+1} + ^{2n+1}C_{n+2} + \dots + ^{2n+1}C_{2n} = 255$.
Note that the student cannot select all books,so the term $^{2n+1}C_{2n+1}$ is excluded.
We know that the sum of all combinations from $^{2n+1}C_0$ to $^{2n+1}C_{2n+1}$ is $2^{2n+1}$.
Since $^{2n+1}C_k = ^{2n+1}C_{2n+1-k}$,the sum of the first half of the terms is equal to the sum of the second half.
Specifically,$\sum_{k=n+1}^{2n} {^{2n+1}C_k} = \frac{2^{2n+1}}{2} - 1 = 2^{2n} - 1$.
Given $2^{2n} - 1 = 255$,we have $2^{2n} = 256$.
Since $256 = 2^8$,we get $2n = 8$,which implies $n = 4$.
The total number of books is $2n + 1 = 2(4) + 1 = 9$.

(C) The student needs to select $10$ questions out of $13$ total questions,with the constraint of selecting at least $4$ from the first $5$ questions.
Case $I$: Selecting $4$ questions from the first $5$ and $6$ questions from the remaining $8$.
Number of ways $= {}^{5}C_{4} \times {}^{8}C_{6} = 5 \times 28 = 140$.
Case $II$: Selecting $5$ questions from the first $5$ and $5$ questions from the remaining $8$.
Number of ways $= {}^{5}C_{5} \times {}^{8}C_{5} = 1 \times 56 = 56$.
Total number of choices $= 140 + 56 = 196$.

(A) The total number of ways to choose the team is calculated by first selecting $5$ players from $10$ and then selecting $1$ captain from the remaining $5$ players.
This is given by the expression:
$({ }^{10} C_5) \times ({ }^5 C_1)$
$= \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \times 5$
$= 252 \times 5 = 1260$.

(C) Each of the $6$ distinct objects can be placed in either of the $2$ boxes in $2$ ways.
Since there are $6$ objects,the total number of ways to distribute them is $2^6 = 64$.
However,this includes the $2$ cases where one of the boxes remains empty (i.e.,all $6$ objects are in the first box,or all $6$ objects are in the second box).
Since the condition is that no box is empty,we subtract these $2$ cases.
Required number of ways $= 2^6 - 2 = 64 - 2 = 62$.

(B) The problem is equivalent to finding the number of non-negative integer solutions to the equation $x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 = 3$,where $x_i \ge 0$.
Using the stars and bars formula,the number of ways is given by $\binom{n+r-1}{r-1}$,where $n = 3$ (identical balls) and $r = 7$ (distinct bins).
Number of ways = $\binom{7+3-1}{7-1} = \binom{9}{6}$.
Since $\binom{9}{6} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 3 \times 4 \times 7 = 84$.

(B) Given the equation $x+y+z+t=10$ with constraints $x \geq 2$ and $z \geq 5$.
Let $x = x' + 2$ where $x' \geq 0$.
Let $z = z' + 5$ where $z' \geq 0$.
Substituting these into the equation: $(x' + 2) + y + (z' + 5) + t = 10$.
$x' + y + z' + t + 7 = 10$.
$x' + y + z' + t = 3$.
The number of non-negative integral solutions is given by the formula $\binom{n+r-1}{r-1}$,where $n=3$ and $r=4$.
Number of solutions = $\binom{3+4-1}{4-1} = \binom{6}{3}$.
$\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.

(B) Let $x_1, x_2, x_3$ be the number of coins received by the $3$ persons.
We have the condition $x_1 + x_2 + x_3 = 15$ where $x_i \geq 3$ for $i = 1, 2, 3$.
Let $y_i = x_i - 3$,then $y_i \geq 0$.
Substituting $x_i = y_i + 3$ into the equation:
$(y_1 + 3) + (y_2 + 3) + (y_3 + 3) = 15$
$y_1 + y_2 + y_3 + 9 = 15$
$y_1 + y_2 + y_3 = 6$
The number of non-negative integer solutions is given by the formula $\binom{n+r-1}{r-1}$,where $n = 6$ and $r = 3$.
Number of ways $= \binom{6+3-1}{3-1} = \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28$.

(B) Given,$S_r = \{(x, y, z) : x + y + z = 11, x \geq r, y \geq r, z \geq r\}$.
For $S_2$: $x+y+z=11, x \geq 2, y \geq 2, z \geq 2$.
Let $x-2=a, y-2=b, z-2=c$,where $a, b, c \geq 0$.
Then $(a+2)+(b+2)+(c+2)=11 \Rightarrow a+b+c=5$.
The number of non-negative integer solutions is given by $\binom{n+k-1}{k-1} = \binom{5+3-1}{3-1} = \binom{7}{2} = 21$.
So,$n(S_2) = 21$.
For $S_3$: $x+y+z=11, x \geq 3, y \geq 3, z \geq 3$.
Let $x-3=a, y-3=b, z-3=c$,where $a, b, c \geq 0$.
Then $(a+3)+(b+3)+(c+3)=11 \Rightarrow a+b+c=2$.
The number of solutions is $\binom{2+3-1}{3-1} = \binom{4}{2} = 6$.
So,$n(S_3) = 6$.
For $S_4$: $x+y+z=11, x \geq 4, y \geq 4, z \geq 4$.
Let $x-4=a, y-4=b, z-4=4$,where $a, b, c \geq 0$.
Then $(a+4)+(b+4)+(c+4)=11 \Rightarrow a+b+c=-1$.
Since $a, b, c \geq 0$,this is not possible,so $n(S_4) = 0$.
Therefore,$n(S_2) + n(S_3) + n(S_4) = 21 + 6 + 0 = 27$.

(A) To distribute $n$ identical items among $r$ persons,we use the stars and bars formula: $\binom{n+r-1}{r-1}$.
Here,$n = 8$ (identical apples) and $r = 3$ (persons).
The number of ways is $\binom{8+3-1}{3-1} = \binom{10}{2}$.
Calculating the combination: $\binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45$.

(A) The number of ways to select at most $n$ books from $(2n+1)$ books is given by the sum of combinations:
${}^{2n+1}C_1 + {}^{2n+1}C_2 + \dots + {}^{2n+1}C_n = 255$ $(i)$
Using the property ${}^mC_r = {}^mC_{m-r}$,we have:
${}^{2n+1}C_{2n} + {}^{2n+1}C_{2n-1} + \dots + {}^{2n+1}C_{n+1} = 255$ $(ii)$
Adding $(i)$ and $(ii)$:
$({}^{2n+1}C_1 + {}^{2n+1}C_2 + \dots + {}^{2n+1}C_{2n}) = 510$
Adding ${}^{2n+1}C_0$ and ${}^{2n+1}C_{2n+1}$ (both equal to $1$) to both sides:
${}^{2n+1}C_0 + {}^{2n+1}C_1 + \dots + {}^{2n+1}C_{2n+1} = 510 + 1 + 1 = 512$
Since the sum of binomial coefficients $\sum_{k=0}^{m} {}^mC_k = 2^m$,we have:
$2^{2n+1} = 512 = 2^9$
Equating the exponents:
$2n + 1 = 9$
$2n = 8$
$n = 4$

(B) The expression $(p+1)(p+2)(p+3) \ldots (p+q)$ represents the product of $q$ consecutive natural numbers starting from $(p+1)$.
We know that the product of $q$ consecutive natural numbers is always divisible by $q!$.
This is because the expression can be written as $\frac{(p+q)!}{p!} = q! \times \binom{p+q}{q}$.
Since $\binom{p+q}{q}$ is an integer for all $p, q \in N$,the expression is always divisible by $q!$.
Thus,the greatest integer that divides this product for all $p \in N$ is $q!$.

(A) Let the $r$ consecutive natural numbers be $(n+1), (n+2), \dots, (n+r)$.
Their product is $P = (n+1)(n+2) \dots (n+r)$.
We can write this as:
$P = \frac{(n+r)!}{n!}$.
Multiplying and dividing by $r!$,we get:
$P = \frac{(n+r)!}{n! r!} \times r! = \binom{n+r}{r} \times r!$.
Since $\binom{n+r}{r}$ is the number of ways to choose $r$ objects from $n+r$ objects,it is always an integer.
Therefore,the product $P$ is always divisible by $r!$.

(C) The word '$EQUATION$' has $8$ distinct letters: $5$ vowels $(E, U, A, I, O)$ and $3$ consonants $(Q, T, N)$.
We need to select $3$ vowels out of $5$ and $2$ consonants out of $3$.
Number of ways to select the letters = $^5C_3 \times ^3C_2 = 10 \times 3 = 30$.
Since all $3$ vowels must be together,we treat the $3$ vowels as a single block. Now we have $1$ block of vowels and $2$ consonants,totaling $3$ units to arrange.
Number of ways to arrange these $3$ units = $3! = 6$.
Within the vowel block,the $3$ vowels can be arranged in $3! = 6$ ways.
Total number of words = $30 \times 6 \times 6 = 1080$.