Definition of combinations, Condition combinations Questions in English

(B) Let the total number of persons in the room be $n$.
The number of handshakes is given by the combination formula ${}^{n}C_{2}$,as a handshake occurs between $2$ people.
Given,${}^{n}C_{2} = 45$.
Using the formula ${}^{n}C_{2} = \frac{n(n-1)}{2}$,we have:
$\frac{n(n-1)}{2} = 45$
$n(n-1) = 90$
$n^2 - n - 90 = 0$
$(n - 10)(n + 9) = 0$
Since the number of persons cannot be negative,$n = 10$.

(C) The set of digits is $\{1, 2, 3, 4, 5, 6, 7\}$.
Odd digits are $\{1, 3, 5, 7\}$ (total $4$) and even digits are $\{2, 4, 6\}$ (total $3$).
We need to select $2$ odd digits from $4$ and $2$ even digits from $3$.
Number of ways to select the digits $= {}^{4}C_{2} \times {}^{3}C_{2} = 6 \times 3 = 18$.
These $4$ selected digits can be arranged in $4!$ ways.
Total number of $4$-digit numbers $= 18 \times 4! = 18 \times 24 = 432$.

(A) We use the Pascal's identity: ${ }^{n} C_{r}+{ }^{n} C_{r+1}={ }^{n+1} C_{r+1}$.
Given expression: ${ }^{49} C_3+{ }^{48} C_3+{ }^{47} C_3+{ }^{46} C_3+{ }^{45} C_3+{ }^{45} C_4$.
Step $1$: Combine ${ }^{45} C_3+{ }^{45} C_4 = { }^{46} C_4$.
Step $2$: Combine ${ }^{46} C_3+{ }^{46} C_4 = { }^{47} C_4$.
Step $3$: Combine ${ }^{47} C_3+{ }^{47} C_4 = { }^{48} C_4$.
Step $4$: Combine ${ }^{48} C_3+{ }^{48} C_4 = { }^{49} C_4$.
Step $5$: Combine ${ }^{49} C_3+{ }^{49} C_4 = { }^{50} C_4$.
Thus,the final value is ${ }^{50} C_4$.

(D) Given that ${ }^{n} C_{12}={ }^{n} C_{8}$.
We know the property of combinations that ${ }^{n} C_{r}={ }^{n} C_{k}$ implies either $r=k$ or $n=r+k$.
Here,$r=12$ and $k=8$.
Since $12 \neq 8$,we must have $n=12+8$.
Therefore,$n=20$.

(A) We use the property of combinations: ${ }^{n} C_{r} = { }^{n} C_{n-r}$.
Applying this to the terms ${ }^{16} C_{6}$ and ${ }^{16} C_{7}$:
${ }^{16} C_{6} = { }^{16} C_{16-6} = { }^{16} C_{10}$
${ }^{16} C_{7} = { }^{16} C_{16-7} = { }^{16} C_{9}$
Substituting these into the original expression:
${ }^{16} C_{9} + { }^{16} C_{10} - { }^{16} C_{6} - { }^{16} C_{7} = { }^{16} C_{9} + { }^{16} C_{10} - { }^{16} C_{10} - { }^{16} C_{9}$
$= ({ }^{16} C_{9} - { }^{16} C_{9}) + ({ }^{16} C_{10} - { }^{16} C_{10}) = 0 + 0 = 0$.

(D) Step $1$: Choose $2$ distinct digits out of $9$ available digits ($1$ to $9$). This can be done in $^9C_2 = \frac{9 \times 8}{2} = 36$ ways.
Step $2$: For each pair of chosen digits,we need to form a $4$-digit number using both digits at least once.
Step $3$: The total number of ways to fill $4$ positions using $2$ chosen digits is $2^4 = 16$.
Step $4$: We must exclude the cases where only one digit is used (i.e.,all $4$ digits are the same). There are $2$ such cases (all $4$ digits are the first chosen digit or all $4$ digits are the second chosen digit).
Step $5$: The number of valid $4$-digit numbers for each pair is $(2^4 - 2) = 14$.
Step $6$: Total numbers = $36 \times 14 = 504$.

(B) Given that a person has $3$ coins of different denominations.
To form a sum,the person can choose $1, 2,$ or $3$ coins.
The number of ways to select $1$ coin out of $3$ is ${}^3C_1 = 3$.
The number of ways to select $2$ coins out of $3$ is ${}^3C_2 = 3$.
The number of ways to select $3$ coins out of $3$ is ${}^3C_3 = 1$.
Since each combination of coins results in a unique sum (as denominations are different),the total number of different sums is ${}^3C_1 + {}^3C_2 + {}^3C_3 = 3 + 3 + 1 = 7$.

(C) We are given the expression $\frac{{}^{n+1}C_{r+1}}{{}^{n+1}C_r} = \frac{n-r+1}{m}$.
Using the formula ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$,we have:
$\frac{{}^{n+1}C_{r+1}}{{}^{n+1}C_r} = \frac{\frac{(n+1)!}{(r+1)!(n-r)!}}{\frac{(n+1)!}{r!(n-r+1)!}} = \frac{r!(n-r+1)!}{(r+1)!(n-r)!}$.
Simplifying the factorials:
$= \frac{r! \times (n-r+1) \times (n-r)!}{(r+1) \times r! \times (n-r)!} = \frac{n-r+1}{r+1}$.
Comparing this with the given expression $\frac{n-r+1}{m}$,we get $m = r+1$.
Thus,option $C$ is correct.

(B) The expression is the product of four consecutive integers: $(n+16)(n+17)(n+18)(n+19)$.
Let $k = n+16$. Then the expression becomes $k(k+1)(k+2)(k+3)$.
This is equivalent to $4! \times \binom{k+3}{4}$.
Since $\binom{k+3}{4}$ is always an integer for any positive integer $k$,the product of any $r$ consecutive integers is always divisible by $r!$.
Here,$r = 4$,so the expression is divisible by $4! = 4 \times 3 \times 2 \times 1 = 24$.
Thus,the greatest positive integer that divides the product for all $n$ is $24$.

(C) We have $4$ married couples,which means $4$ men and $4$ women. We need to select $4$ persons such that no married couple is included.
First,we select $4$ couples out of $4$ available couples,which can be done in $\binom{4}{4} = 1$ way.
From these $4$ selected couples,we need to choose $1$ person from each couple such that we have $4$ persons in total. Since each couple has $2$ choices (either the husband or the wife),the number of ways to select $4$ persons is $2 \times 2 \times 2 \times 2 = 2^4 = 16$ ways.
Thus,the total number of ways is $16$.

(D) Step $1$: Select $4$ novels out of $6$ novels in ${}^6C_4$ ways.
${}^6C_4 = \frac{6 \times 5}{2 \times 1} = 15$ ways.
Step $2$: Select $1$ poetry book out of $3$ poetry books in ${}^3C_1$ ways.
${}^3C_1 = 3$ ways.
Step $3$: Arrange the $4$ selected novels and $1$ poetry book in a row such that the poetry book is in the middle. The arrangement looks like: $(N_1, N_2, P, N_3, N_4)$.
The $4$ novels can be arranged in the $4$ available positions in $4!$ ways.
$4! = 4 \times 3 \times 2 \times 1 = 24$ ways.
Step $4$: Total number of arrangements $= {}^6C_4 \times {}^3C_1 \times 4! = 15 \times 3 \times 24 = 1080$.

(A) To select $4$ pens from $8$ pens,the number of ways is ${ }^8 C_4 = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$.
To select $3$ pencils from $5$ pencils,the number of ways is ${ }^5 C_3 = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$.
Total number of ways $= { }^8 C_4 \times { }^5 C_3 = 70 \times 10 = 700$.

(C) To form an ordered pair $(p, q)$ such that $p > q$ from the set $S = \{1, 2, 3, \dots, 50\}$,we need to choose $2$ distinct numbers from the $50$ available numbers.
Let the chosen numbers be $x$ and $y$ where $x < y$.
For any such pair,there is exactly one way to assign them to $p$ and $q$ such that $p > q$,which is $p = y$ and $q = x$.
The number of ways to choose $2$ distinct numbers from $50$ is given by the combination formula $\binom{n}{r}$.
Here,$n = 50$ and $r = 2$.
Number of ways = $\binom{50}{2} = \frac{50 \times 49}{2 \times 1} = 25 \times 49 = 1225$.

(C) We need to find the number of integers $n$ such that $10 < n < 10,000$ and the digits of $n$ are in strictly increasing order.
This means we are looking for integers with $2, 3,$ or $4$ digits.
For any set of $k$ distinct digits chosen from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$,there is exactly one way to arrange them in strictly increasing order.
Note that $0$ cannot be included because if $0$ is present,it would have to be the first digit to maintain the increasing order,which is not allowed for integers between $10$ and $10,000$.
$1$. For $2$-digit integers: We choose $2$ digits from $9$ available digits. The number of ways is $\binom{9}{2} = \frac{9 \times 8}{2} = 36$.
$2$. For $3$-digit integers: We choose $3$ digits from $9$ available digits. The number of ways is $\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
$3$. For $4$-digit integers: We choose $4$ digits from $9$ available digits. The number of ways is $\binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$.
Total number of such integers $= 36 + 84 + 126 = 246$.

(D) We need to select $11$ members from $6$ batsmen $(B)$,$6$ bowlers $(Bo)$,$4$ all-rounders $(A)$,and $4$ wicket keepers $(W)$.
Constraints: $B \ge 4$,$Bo \ge 3$,$A \ge 2$,$W = 1$.
Total members selected = $11$.
Let $b, bo, a, w$ be the number of players selected from each category.
$w = 1$,so $b + bo + a = 10$.
Possible cases $(b, bo, a)$ such that $b \ge 4, bo \ge 3, a \ge 2$:
Case $1$: $(5, 3, 2) \implies \binom{6}{5} \times \binom{6}{3} \times \binom{4}{2} \times \binom{4}{1} = 6 \times 20 \times 6 \times 4 = 2880$.
Case $2$: $(4, 4, 2) \implies \binom{6}{4} \times \binom{6}{4} \times \binom{4}{2} \times \binom{4}{1} = 15 \times 15 \times 6 \times 4 = 5400$.
Case $3$: $(4, 3, 3) \implies \binom{6}{4} \times \binom{6}{3} \times \binom{4}{3} \times \binom{4}{1} = 15 \times 20 \times 4 \times 4 = 4800$.
Total ways = $2880 + 5400 + 4800 = 13080$.

(D) Using the identity ${ }^n C_r + { }^n C_{r-1} = { }^{n+1} C_r$,we have ${ }^{(n-1)} C_2 + { }^{(n-1)} C_3 = { }^n C_3$.
Given the inequality ${ }^n C_3 > { }^n C_2$.
Expanding the combinations: $\frac{n!}{3!(n-3)!} > \frac{n!}{2!(n-2)!}$.
Simplifying the expression: $\frac{1}{3(n-3)!} > \frac{1}{2(n-2)(n-3)!}$.
$\frac{1}{3} > \frac{1}{2(n-2)}$.
$2(n-2) > 3$.
$2n - 4 > 3$.
$2n > 7$.
$n > 3.5$.
Since $n$ must be an integer and $n \geq 3$ for the combinations to be defined,the smallest integer $n$ satisfying the inequality is $n = 4$. However,checking the original expression for $n=4$: ${ }^3 C_2 + { }^3 C_3 = 3 + 1 = 4$ and ${ }^4 C_2 = 6$. $4 > 6$ is false.
Checking $n=5$: ${ }^4 C_2 + { }^4 C_3 = 6 + 4 = 10$ and ${ }^5 C_2 = 10$. $10 > 10$ is false.
Checking $n=6$: ${ }^5 C_2 + { }^5 C_3 = 10 + 10 = 20$ and ${ }^6 C_2 = 15$. $20 > 15$ is true.
Thus,the least value is $n = 6$.

(B) Given the ratio: $\frac{^{2n}C_4}{^nC_3} = \frac{99}{4}$
Using the formula $^nC_r = \frac{n!}{r!(n-r)!}$:
$\frac{\frac{(2n)!}{4!(2n-4)!}}{\frac{n!}{3!(n-3)!}} = \frac{99}{4}$
$\frac{(2n)(2n-1)(2n-2)(2n-3)}{4 \times 3 \times 2 \times 1} \times \frac{3 \times 2 \times 1}{n(n-1)(n-2)} = \frac{99}{4}$
$\frac{(2n)(2n-1) \times 2(n-1)(2n-3)}{4 \times n(n-1)(n-2)} = \frac{99}{4}$
$\frac{2(2n-1)(2n-3)}{n-2} = 99$
$4(4n^2 - 6n - 2n + 3) = 99(n-2)$
$16n^2 - 32n + 12 = 99n - 198$
$16n^2 - 131n + 210 = 0$
$(n-6)(16n-35) = 0$
Since $n$ must be an integer,$n = 6$.

(D) We are given the combinations: ${}^nC_{r-1}=36$,${}^nC_r=84$,and ${}^nC_{r+1}=126$.
Taking the ratio $\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{84}{36} = \frac{7}{3}$.
Using the formula $\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n-r+1}{r}$,we get $\frac{n-r+1}{r} = \frac{7}{3}$ $\Rightarrow 3n-3r+3 = 7r$ $\Rightarrow 3n+3 = 10r$ (Equation $i$).
Taking the ratio $\frac{{}^nC_{r+1}}{{}^nC_r} = \frac{126}{84} = \frac{3}{2}$.
Using the formula $\frac{{}^nC_{r+1}}{{}^nC_r} = \frac{n-r}{r+1}$,we get $\frac{n-r}{r+1} = \frac{3}{2}$ $\Rightarrow 2n-2r = 3r+3$ $\Rightarrow 2n-3 = 5r$ (Equation $ii$).
From Equation $ii$,$r = \frac{2n-3}{5}$. Substituting this into Equation $i$:
$3n+3 = 10 \left( \frac{2n-3}{5} \right)$ $\Rightarrow 3n+3 = 2(2n-3)$ $\Rightarrow 3n+3 = 4n-6$ $\Rightarrow n = 9$.
Substituting $n=9$ into Equation $ii$: $2(9)-3 = 5r$ $\Rightarrow 15 = 5r$ $\Rightarrow r = 3$.
Therefore,$nr^2 = 9 \times (3)^2 = 9 \times 9 = 81$.

(D) Selection of two vowels $\Rightarrow {}^{5}C_{2} = \frac{5 \times 4}{2 \times 1} = 10$.
Selection of two consonants $\Rightarrow {}^{21}C_{2} = \frac{21 \times 20}{2 \times 1} = 210$.
Total selection of four letters $= 10 \times 210 = 2100$.
Arrangements of these four distinct letters $= 4!$.
$\therefore$ Total words $= 2100 \times 4!$.

(B) We know the relationship between permutations and combinations is given by:
${}^{n}P_r = {}^{n}C_r \times r!$
Substituting the given values:
$604800 = 120 \times r!$
$r! = \frac{604800}{120}$
$r! = 5040$
Since $7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$,we have:
$r! = 7!$
Therefore,$r = 7$.

(A) Section $1$ has $3$ questions. The number of ways to select one or more questions from this section is $2^3 - 1 = 7$.
Section $2$ has $4$ questions. The number of ways to select one or more questions from this section is $2^4 - 1 = 15$.
Since the candidate must select at least one question from each section,the total number of ways is the product of the ways to select from each section.
Total ways $= 7 \times 15 = 105$.

(B) The total number of elements in the set is $n = 11$.
To find the number of subsets containing at most $5$ elements,we sum the number of ways to choose $k$ elements for $k = 0, 1, 2, 3, 4, 5$.
This is given by the sum: ${ }^{11}C_0 + { }^{11}C_1 + { }^{11}C_2 + { }^{11}C_3 + { }^{11}C_4 + { }^{11}C_5$.

(C) Given inequality: ${ }^5 C_{x-1} > 2 \cdot { }^5 C_x$
For the combinations to be defined,we must have $0 \le x-1 \le 5$ and $0 \le x \le 5$. Thus,$x \in \{1, 2, 3, 4, 5\}$.
Using the formula ${ }^n C_r = \frac{n!}{r!(n-r)!}$,we have:
$\frac{{ }^5 C_{x-1}}{{ }^5 C_x} > 2$
$\frac{5!}{(x-1)!(5-(x-1))!} \cdot \frac{x!(5-x)!}{5!} > 2$
$\frac{x!(5-x)!}{(x-1)!(6-x)!} > 2$
$\frac{x(x-1)!(5-x)!}{(x-1)!(6-x)(5-x)!} > 2$
$\frac{x}{6-x} > 2$
$\frac{x}{6-x} - 2 > 0$
$\frac{x - 2(6-x)}{6-x} > 0$
$\frac{x - 12 + 2x}{6-x} > 0$
$\frac{3x - 12}{6-x} > 0$
$\frac{3(x-4)}{-(x-6)} > 0$
$\frac{x-4}{x-6} < 0$
This inequality holds for $4 < x < 6$.
Since $x$ must be an integer,the only possible value is $x = 5$.
Therefore,the solution set is $\{5\}$.

(B) The student must select $10$ questions out of $13$. The first $5$ questions are in one group and the remaining $8$ questions are in another group. The student must select at least $3$ from the first $5$ questions.
Case $1$: Select $3$ from the first $5$ and $7$ from the remaining $8$.
Number of ways $= {}^{5}C_{3} \times {}^{8}C_{7} = 10 \times 8 = 80$.
Case $2$: Select $4$ from the first $5$ and $6$ from the remaining $8$.
Number of ways $= {}^{5}C_{4} \times {}^{8}C_{6} = 5 \times 28 = 140$.
Case $3$: Select $5$ from the first $5$ and $5$ from the remaining $8$.
Number of ways $= {}^{5}C_{5} \times {}^{8}C_{5} = 1 \times 56 = 56$.
Total number of ways $= 80 + 140 + 56 = 276$.

(B) We know that $k \times k! = (k+1-1) \times k! = (k+1)! - k!$.
Summing this from $k=1$ to $n$:
$\sum_{k=1}^{n} k \times k! = \sum_{k=1}^{n} ((k+1)! - k!) = (2! - 1!) + (3! - 2!) + \ldots + ((n+1)! - n!) = (n+1)! - 1!$.
Given that the sum is $11! - 1$,we have $(n+1)! - 1 = 11! - 1$,which implies $n+1 = 11$,so $n = 10$.
The maximum value of ${}^n C_r$ occurs at $r = n/2$ when $n$ is even.
For $n = 10$,the maximum value is ${}^{10} C_5 = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$.

(A) match between two kids is equivalent to selecting a pair of $2$ kids out of $30$ kids.
Since the order of selection does not matter in a match,we use the combination formula.
The number of ways to choose $2$ kids out of $30$ is given by $^{30}C_2 = \frac{30 \times 29}{2 \times 1} = 435$ matches.

(B) We use the identity ${}^n C_r + {}^n C_{r-1} = {}^{n+1} C_r$.
The given expression is ${}^n C_{r+1} + {}^n C_{r-1} + 2{}^n C_r$.
This can be rewritten as $({}^n C_{r+1} + {}^n C_r) + ({}^n C_r + {}^n C_{r-1})$.
Applying the identity,we get ${}^{n+1} C_{r+1} + {}^{n+1} C_r$.
Applying the identity again,we get ${}^{n+2} C_{r+1}$.
Hence,option $B$ is correct.

(C) Since the committee must always include a specified member,we have already filled $1$ spot out of the $6$ required spots.
Therefore,we need to select the remaining $6 - 1 = 5$ members from the remaining $10 - 1 = 9$ members.
The number of ways to choose $5$ members from $9$ is given by the combination formula ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$.
Thus,the number of ways is ${}^{9}C_{5}$.
Hence,option $C$ is correct.

(A) Let the total number of ways to select at most $n$ books be $x$. Since the student must select at least one book,we have:
$x = {}^{2n+1}C_1 + {}^{2n+1}C_2 + \dots + {}^{2n+1}C_n = 255$
We know that the sum of all combinations for $2n+1$ items is:
${}^{2n+1}C_0 + {}^{2n+1}C_1 + \dots + {}^{2n+1}C_n + {}^{2n+1}C_{n+1} + \dots + {}^{2n+1}C_{2n+1} = 2^{2n+1}$
Using the property ${}^{m}C_r = {}^{m}C_{m-r}$,we have ${}^{2n+1}C_0 = {}^{2n+1}C_{2n+1} = 1$ and ${}^{2n+1}C_1 = {}^{2n+1}C_{2n}$,etc.
Thus,the sum can be written as:
$2({}^{2n+1}C_1 + {}^{2n+1}C_2 + \dots + {}^{2n+1}C_n) + {}^{2n+1}C_0 + {}^{2n+1}C_{2n+1} = 2^{2n+1}$
$2x + 1 + 1 = 2^{2n+1}$
$2x + 2 = 2^{2n+1}$
$x + 1 = 2^{2n}$
Given $x = 255$,we get:
$255 + 1 = 2^{2n}$
$256 = 2^{2n}$
$2^8 = 2^{2n}$
$2n = 8 \implies n = 4$
Therefore,the value of $n$ is $4$.

(B) To form a committee of $6$ members with at least one member from each of the $3$ countries (Indians,Americans,Australians),we consider the possible distributions of members $(n_I, n_A, n_{Au})$ such that $n_I + n_A + n_{Au} = 6$ and $n_I, n_A, n_{Au} \ge 1$.
The possible partitions of $6$ into $3$ parts are:
$1. (4, 1, 1)$ and its permutations: $(4, 1, 1), (1, 4, 1), (1, 1, 4)$. There are $3$ such cases.
Number of ways $= 3 \times \binom{5}{4} \times \binom{5}{1} \times \binom{5}{1} = 3 \times 5 \times 5 \times 5 = 375$.
$2. (3, 2, 1)$ and its permutations: $(3, 2, 1), (3, 1, 2), (2, 3, 1), (2, 1, 3), (1, 3, 2), (1, 2, 3)$. There are $3! = 6$ such cases.
Number of ways $= 6 \times \binom{5}{3} \times \binom{5}{2} \times \binom{5}{1} = 6 \times 10 \times 10 \times 5 = 3000$.
$3. (2, 2, 2)$. There is $1$ such case.
Number of ways $= 1 \times \binom{5}{2} \times \binom{5}{2} \times \binom{5}{2} = 10 \times 10 \times 10 = 1000$.
Total ways $= 375 + 3000 + 1000 = 4375$.

(A) Let the $10$ intermediate stations be $S_1, S_2, S_3, \dots, S_{10}$.
We need to select $3$ stations such that no two are consecutive.
This is a classic problem of selecting $r$ objects from $n$ objects arranged in a row such that no two are consecutive,which is given by the formula $^{n-r+1}C_r$.
Here,$n = 10$ and $r = 3$.
Number of ways $= ^{10-3+1}C_3 = ^8C_3$.
Calculating the value: $^8C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.

(C) Given,${}^nC_{r-1}=330$,${}^nC_r=462$,and ${}^nC_{r+1}=462$.
We know that $\frac{{}^nC_{r+1}}{{}^nC_r} = \frac{n-r}{r+1}$.
Since ${}^nC_{r+1} = {}^nC_r = 462$,we have $\frac{462}{462} = 1$.
Therefore,$\frac{n-r}{r+1} = 1 \implies n-r = r+1 \implies n = 2r+1$.
Now,consider the ratio $\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{462}{330}$.
Using the formula $\frac{{}^nC_r}{{}^nC_{r-1}} = \frac{n-r+1}{r}$,we get $\frac{n-r+1}{r} = \frac{462}{330} = \frac{7}{5}$.
Substituting $n = 2r+1$ into the equation:
$\frac{(2r+1)-r+1}{r} = \frac{7}{5} \implies \frac{r+2}{r} = \frac{7}{5}$.
$5(r+2) = 7r \implies 5r + 10 = 7r \implies 2r = 10 \implies r = 5$.

(D) Given,${ }^{n-1} C_3+{ }^{n-1} C_4>{ }^n C_3$
Using the identity ${ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r$,we have:
${ }^n C_4>{ }^n C_3$
Expanding the combinations:
$\frac{n!}{4!(n-4)!} > \frac{n!}{3!(n-3)!}$
$\frac{1}{4(n-4)!} > \frac{1}{(n-3)(n-4)!}$
$\frac{1}{4} > \frac{1}{n-3}$
$n-3 > 4$
$n > 7$
Since $n$ must be an integer greater than $7$,the minimum value of $n$ is $8$.

(B) Using the Pascal's identity,${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$,we have ${}^nC_5 + {}^nC_6 = {}^{n+1}C_6$.
Given 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: $\frac{1}{6!(n-5)!} > \frac{1}{5!(n-4)!}$.
Since $6! = 6 \times 5!$ and $(n-4)! = (n-4) \times (n-5)!$,we get: $\frac{1}{6 \times 5!(n-5)!} > \frac{1}{5!(n-4)(n-5)!}$.
Canceling common terms: $\frac{1}{6} > \frac{1}{n-4}$.
This implies $n-4 > 6$,so $n > 10$.
The least natural number $n$ satisfying this is $n = 11$.

(A) We need to form a committee of $8$ members from $10$ men and $8$ women such that there are at most $5$ men and at least $5$ women.
Since the total number of members is $8$,the possible combinations of (women,men) are:
$(5, 3), (6, 2), (7, 1), (8, 0)$.
The number of ways is given by:
$\sum_{k=5}^{8} {}^{8}C_{k} \times {}^{10}C_{8-k}$
$= {}^{8}C_{5} \times {}^{10}C_{3} + {}^{8}C_{6} \times {}^{10}C_{2} + {}^{8}C_{7} \times {}^{10}C_{1} + {}^{8}C_{8} \times {}^{10}C_{0}$
$= (56 \times 120) + (28 \times 45) + (8 \times 10) + (1 \times 1)$
$= 6720 + 1260 + 80 + 1 = 8061$.

(D) To form a committee of $10$ members where men are in majority,the number of men must be greater than the number of women. Since the total members are $10$,the possible cases for (men,women) are $(6, 4), (7, 3), (8, 2)$.
Number of ways = $^8C_6 \times ^6C_4 + ^8C_7 \times ^6C_3 + ^8C_8 \times ^6C_2$.
Calculating each term:
$^8C_6 \times ^6C_4 = 28 \times 15 = 420$.
$^8C_7 \times ^6C_3 = 8 \times 20 = 160$.
$^8C_8 \times ^6C_2 = 1 \times 15 = 15$.
Total ways = $420 + 160 + 15 = 595$.

(C) To solve this,we use the gap method.
Since the $7$ white balls are identical,they can be arranged in only $1$ way.
To ensure no two black balls are adjacent,we place the $3$ black balls in the gaps created by the $7$ white balls.
Representing white balls as $W$,the arrangement is: $\_ W \_ W \_ W \_ W \_ W \_ W \_ W \_$.
There are $8$ available slots for the $3$ identical black balls.
The number of ways to choose $3$ slots out of $8$ is given by ${}^8C_3$.
${}^8C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.
Thus,the total number of distinguishable arrangements is $56$.

(A) To distribute $n$ identical items among $r$ distinct recipients such that each recipient gets at least $1$ item,we use the stars and bars method (or the formula for positive integer solutions to $x_1 + x_2 + \dots + x_r = n$).
Here,$n = 8$ and $r = 3$.
The number of ways is given by the formula $\binom{n-1}{r-1}$.
Substituting the values,we get $\binom{8-1}{3-1} = \binom{7}{2}$.
$\binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21$.

(C) The total number of ways to select $4$ balls from $10$ balls ($6$ black + $4$ green) is given by $^{10}C_4 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$.
The number of ways to select $4$ balls such that no black ball is selected (i.e.,all $4$ balls are green) is $^{4}C_4 = 1$.
Therefore,the number of ways to select at least one black ball is (Total ways) - (Ways with no black ball) = $210 - 1 = 209$.

(C) Let $m$ be the number of men and $w$ be the number of women in the committee.
We are given that $w = 2m$ and $m \ge 2$.
Since there are $4$ men and $6$ women available,we have the constraints $m \le 4$ and $w \le 6$.
Substituting $w = 2m$ into $w \le 6$,we get $2m \le 6$,which implies $m \le 3$.
Thus,the possible values for $m$ are $2$ and $3$.
Case $1$: If $m = 2$,then $w = 2(2) = 4$. The number of ways is $^4C_2 \times ^6C_4 = 6 \times 15 = 90$.
Case $2$: If $m = 3$,then $w = 2(3) = 6$. The number of ways is $^4C_3 \times ^6C_6 = 4 \times 1 = 4$.
Total number of ways = $90 + 4 = 94$.

(A) We need to form words using $3$ consonants and $2$ vowels from $5$ consonants and $5$ vowels.
First,we select the consonants and vowels:
Number of ways to select $3$ consonants from $5$ is ${}^5C_3 = \frac{5 \times 4}{2 \times 1} = 10$.
Number of ways to select $2$ vowels from $5$ is ${}^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$.
Total ways to select the letters = $10 \times 10 = 100$.
Now,these $5$ selected letters can be arranged among themselves in $5!$ ways.
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.
Total number of words formed = $100 \times 120 = 12000$.

(D) Let $x_1, x_2, x_3, x_4$ be the number of apples received by the $4$ guests respectively. We need to find the number of integer solutions to $x_1 + x_2 + x_3 + x_4 = 17$ where $x_i \ge 3$ for each $i \in \{1, 2, 3, 4\}$.
Let $y_i = x_i - 3$,then $y_i \ge 0$.
Substituting $x_i = y_i + 3$ into the equation: $(y_1 + 3) + (y_2 + 3) + (y_3 + 3) + (y_4 + 3) = 17$.
$y_1 + y_2 + y_3 + y_4 + 12 = 17 \implies y_1 + y_2 + y_3 + y_4 = 5$.
The number of non-negative integer solutions is given by the formula $\binom{n+r-1}{r-1}$,where $n=5$ and $r=4$.
Number of ways $= \binom{5+4-1}{4-1} = \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.

(B) Since the items in each category are alike,the number of ways to select items from each category is equal to the number of items available plus one (for the case of selecting zero items).
For $6$ alike fruits,the number of ways to select is $(6+1) = 7$.
For $7$ alike vegetables,the number of ways to select is $(7+1) = 8$.
For $8$ alike biscuits,the number of ways to select is $(8+1) = 9$.
To ensure at least one item from each category is selected,we must select at least $1$ fruit,$1$ vegetable,and $1$ biscuit.
The number of ways to select at least one fruit is $6$.
The number of ways to select at least one vegetable is $7$.
The number of ways to select at least one biscuit is $8$.
Therefore,the total number of ways is $6 \times 7 \times 8 = 336$.

(D) The student needs to select $10$ questions in total,with at least $4$ from section-$A$ ($8$ available) and at least $3$ from section-$B$ ($6$ available).
Possible combinations $(A, B)$ are:
$(i)$ $4$ from $A$ and $6$ from $B$: $\binom{8}{4} \times \binom{6}{6} = 70 \times 1 = 70$
(ii) $5$ from $A$ and $5$ from $B$: $\binom{8}{5} \times \binom{6}{5} = 56 \times 6 = 336$
(iii) $6$ from $A$ and $4$ from $B$: $\binom{8}{6} \times \binom{6}{4} = 28 \times 15 = 420$
(iv) $7$ from $A$ and $3$ from $B$: $\binom{8}{7} \times \binom{6}{3} = 8 \times 20 = 160$
Total ways $= 70 + 336 + 420 + 160 = 986$.

(D) The number of ways to select at least one green toy from $5$ is $2^5 - 1 = 31$ ways.
The number of ways to select at least one blue toy from $4$ is $2^4 - 1 = 15$ ways.
The number of ways to select any number of red toys from $3$ (including zero) is $2^3 = 8$ ways.
Since these selections are independent,the total number of combinations is $31 \times 15 \times 8$.

(D) Considering fruits of the same kind to be identical.
If there are $p$ identical items of the $1^{\text{st}}$ kind,$q$ identical items of the $2^{\text{nd}}$ kind,and $r$ identical items of the $3^{\text{rd}}$ kind,then the total number of ways of selecting any number of objects is given by $(p+1)(q+1)(r+1)$.
In this case,$p=4$,$q=5$,and $r=7$.
Total number of ways including the case of selecting zero fruits $= (4+1)(5+1)(7+1) = 5 \times 6 \times 8 = 240$.
Since we must select at least one fruit,we exclude the case where $0$ oranges,$0$ apples,and $0$ mangoes are selected.
Total number of ways of selecting at least one fruit $= 240 - 1 = 239$.