Definition of combinations, Condition combinations Questions in English

(A) The number of ways to buy $6$ ice creams from $5$ types (with repetition allowed) is given by the formula $^{n+r-1}C_{r}$,where $n=5$ and $r=6$.
This equals $^{5+6-1}C_{6} = ^{10}C_6 = ^{10}C_4 = 210$.
Statement-$1$ says $^{10}C_5$,which is $252$. Thus,Statement-$1$ is false.
The number of ways to arrange $6$ '$A$'s and $4$ '$B$'s in a row is given by the permutation of multisets formula: $\frac{(6+4)!}{6! \times 4!} = \frac{10!}{6! \times 4!} = ^{10}C_4 = 210$.
This matches the calculation for the number of ways to buy the ice creams.
Therefore,Statement-$2$ is true.

(D) We know that the formula for permutation is $_n{P_r} = \frac{n!}{(n - r)!}$.
We know that the formula for combination is $\binom{n}{r} = \frac{n!}{r!(n - r)!}$.
Now,dividing the two:
$\frac{_n{P_r}}{\binom{n}{r}} = \frac{\frac{n!}{(n - r)!}}{\frac{n!}{r!(n - r)!}}$
$= \frac{n!}{(n - r)!} \times \frac{r!(n - r)!}{n!}$
$= r!$

(A) The number of candidates to be elected is $4$ out of $10$.
Since a voter can vote for any number of candidates from $1$ to $4$,the total number of ways is given by the sum of combinations:
Total ways $= ^{10}C_1 + ^{10}C_2 + ^{10}C_3 + ^{10}C_4$.
Calculating each term:
$^{10}C_1 = 10$
$^{10}C_2 = \frac{10 \times 9}{2 \times 1} = 45$
$^{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$
$^{10}C_4 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$
Total ways $= 10 + 45 + 120 + 210 = 385$.

(C) 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,$n = 10$ and $r = 4$.
So,the number of ways is $^{10-1}C_{4-1} = ^9C_3$.
Statement-$1$ is true.
Statement-$2$ states that $3$ positions out of $9$ can be selected in $^9C_3$ ways,which is the definition of combinations $(^nC_r)$.
This logic is exactly what is used in the stars and bars method to arrive at the formula for Statement-$1$.
Therefore,Statement-$2$ is the correct explanation for Statement-$1$.

(C) The number of ways to distribute $n$ identical items among $r$ recipients such that each recipient gets at least one item is given by the formula $^{n-1}C_{r-1}$.
Here,$n = 11$ and $r = 6$.
Number of ways $= ^{11-1}C_{6-1} = ^{10}C_5$.
Calculating the value: $^{10}C_5 = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252$.

(A) First,we arrange the $6$ '$+$' signs in a row: $+ + + + + +$.
There are $7$ possible gaps (including the ends) created by these $6$ '$+$' signs: $\_ + \_ + \_ + \_ + \_ + \_ \_$.
To ensure no two '$*$' signs occur together,we must place the $4$ '$*$' signs into these $7$ available gaps.
The number of ways to choose $4$ gaps out of $7$ is given by the combination formula $^nC_r = \binom{n}{r}$.
Therefore,the total number of ways is $^7C_4 = \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} = 35$.

(B) Given $\binom{2n}{3} : \binom{n}{2} = 44 : 3$.
Using the formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$,we have:
$\frac{(2n)!}{3!(2n-3)!} \times \frac{2!(n-2)!}{n!} = \frac{44}{3}$
Expanding the factorials:
$\frac{2n(2n-1)(2n-2)(2n-3)!}{6(2n-3)!} \times \frac{2(n-2)!}{n(n-1)(n-2)!} = \frac{44}{3}$
$\frac{2n(2n-1) \cdot 2(n-1)}{6} \times \frac{2}{n(n-1)} = \frac{44}{3}$
$\frac{4n(2n-1)(n-1)}{6n(n-1)} = \frac{44}{3}$
$\frac{2(2n-1)}{3} = \frac{44}{3}$
$2(2n-1) = 44 \implies 2n-1 = 22 \implies 2n = 23$ (Wait,re-evaluating the simplification).
Correction: $\frac{2n(2n-1)(2n-2)}{6} \times \frac{2}{n(n-1)} = \frac{44}{3}$
$\frac{2n(2n-1) \cdot 2(n-1)}{6} \times \frac{2}{n(n-1)} = \frac{4(2n-1)}{6} = \frac{2(2n-1)}{3} = \frac{44}{3}$
$2(2n-1) = 44 \implies 2n-1 = 22 \implies 2n = 23$ (There is a slight error in the provided solution's arithmetic).
Let's re-calculate: $\frac{2n(2n-1)(2n-2)}{6} \cdot \frac{2}{n(n-1)} = \frac{2n(2n-1) \cdot 2(n-1)}{6} \cdot \frac{2}{n(n-1)} = \frac{4(2n-1)}{6} = \frac{2(2n-1)}{3}$.
If $\frac{2(2n-1)}{3} = \frac{44}{3}$,then $2n-1 = 22$,$n=11.5$.
Checking the ratio again: $\binom{2n}{3} / \binom{n}{2} = \frac{2n(2n-1)(2n-2)}{6} / \frac{n(n-1)}{2} = \frac{2n(2n-1)2(n-1)}{6} \cdot \frac{2}{n(n-1)} = \frac{8n(2n-1)(n-1)}{6n(n-1)} = \frac{4(2n-1)}{3} = \frac{44}{3}$.
$4(2n-1) = 44 \implies 2n-1 = 11 \implies 2n = 12 \implies n = 6$.
Now,$\binom{6}{r} = 15$.
Since $\binom{6}{2} = 15$ and $\binom{6}{4} = 15$,$r$ can be $2$ or $4$. Given the options,$r=4$ is correct.

(A) The word $INDEPENDENT$ contains $11$ letters: $3N, 3E, 2D, 1I, 1P, 1T$. The distinct letters are ${N, E, D, I, P, T}$.
We need to select $5$ letters. The possible cases are:
$(i)$ All $5$ letters are distinct: We choose $5$ from ${N, E, D, I, P, T}$. Number of ways = $^6C_5 = 6$.
$(ii)$ $2$ are alike (of one kind) and $3$ are distinct: We choose $1$ pair from ${N, E, D}$ and $3$ distinct letters from the remaining $5$ types. Number of ways = $^3C_1 \times ^5C_3 = 3 \times 10 = 30$.
$(iii)$ $3$ are alike (of one kind) and $2$ are distinct: We choose $1$ triplet from ${N, E}$ and $2$ distinct letters from the remaining $5$ types. Number of ways = $^2C_1 \times ^5C_2 = 2 \times 10 = 20$.
$(iv)$ $3$ are alike (of one kind) and $2$ are alike (of another kind): We choose $1$ triplet from ${N, E}$ and $1$ pair from the remaining $2$ types. Number of ways = $^2C_1 \times ^2C_1 = 2 \times 2 = 4$.
$(v)$ $2$ are alike (of one kind),$2$ are alike (of another kind),and $1$ is distinct: We choose $2$ pairs from ${N, E, D}$ and $1$ distinct letter from the remaining $4$ types. Number of ways = $^3C_2 \times ^4C_1 = 3 \times 4 = 12$.
Total number of ways = $6 + 30 + 20 + 4 + 12 = 72$.

(A) We know that if $^{n}C_{x} = ^{n}C_{y}$,then either $x = y$ or $x + y = n$.
Given $^{15}C_{3r} = ^{15}C_{r+3}$.
Case $1$: $3r = r + 3$ $\Rightarrow 2r = 3$ $\Rightarrow r = 1.5$.
Since $r$ must be an integer,this case is not possible.
Case $2$: $3r + (r + 3) = 15$ $\Rightarrow 4r + 3 = 15$ $\Rightarrow 4r = 12$ $\Rightarrow r = 3$.
Thus,the value of $r$ is $3$.

(C) There are two sections,$A$ and $B$,each with $5$ questions. The candidate must select $6$ questions in total,with at least $2$ questions from each section.
The possible cases are:
$(i)$ Selecting $2$ questions from section $A$ and $4$ questions from section $B$:
$^5C_2 \times ^5C_4 = 10 \times 5 = 50$ ways.
$(ii)$ Selecting $3$ questions from section $A$ and $3$ questions from section $B$:
$^5C_3 \times ^5C_3 = 10 \times 10 = 100$ ways.
$(iii)$ Selecting $4$ questions from section $A$ and $2$ questions from section $B$:
$^5C_4 \times ^5C_2 = 5 \times 10 = 50$ ways.
Total number of ways $= 50 + 100 + 50 = 200$.

(A) Given the equation $\binom{a^2+a}{3} = \binom{a^2+a}{9}$.
Using the property $\binom{n}{r} = \binom{n}{n-r}$,if $\binom{n}{x} = \binom{n}{y}$,then either $x = y$ or $x + y = n$.
Here,$3 \neq 9$,so we must have $3 + 9 = a^2 + a$.
$a^2 + a = 12$.
$a^2 + a - 12 = 0$.
$(a + 4)(a - 3) = 0$.
Thus,$a = 3$ or $a = -4$.
Since $n$ in $\binom{n}{r}$ must be a non-negative integer and $n \geq r$,for $a = -4$,$n = (-4)^2 + (-4) = 16 - 4 = 12$,which is $\geq 9$. For $a = 3$,$n = 3^2 + 3 = 12$,which is $\geq 9$.
Both values are mathematically valid,but based on the provided options,$a = 3$ is the correct choice.

(C) The bag contains $3$ coins of one type,$4$ coins of another type,and $5$ coins of a third type.
The number of ways to select at least one coin is given by the formula $(n_1 + 1)(n_2 + 1)(n_3 + 1) - 1$.
Substituting the values: $(3 + 1)(4 + 1)(5 + 1) - 1 = 4 \times 5 \times 6 - 1 = 120 - 1 = 119$.

(D) Total persons available are $10$. We need to select $5$ persons.
Given that $A$ is always included and $G, H$ are excluded.
Remaining persons to choose from are $10 - 3 = 7$ (excluding $A, G, H$).
We need to select $4$ more persons from the remaining $7$ to complete the group of $5$. This can be done in $^7C_4$ ways.
Since $^7C_4 = ^7C_3$,the number of ways to select the group is $^7C_3$.
These $5$ persons can be arranged in a row in $5!$ ways.
Therefore,the total number of arrangements is $^7C_3 \times 5!$.

(B) To find the total number of handshakes,we need to choose $2$ people out of $15$ to perform a handshake.
This is a combination problem because the order of the two people shaking hands does not matter.
The number of ways to choose $2$ people out of $15$ is given by the combination formula $^{n}C_{r} = \frac{n!}{r!(n-r)!}$.
Here,$n = 15$ and $r = 2$.
Therefore,the total number of handshakes is $^{15}C_{2} = \frac{15 \times 14}{2 \times 1} = 105$.

(C) The total number of players is $14$,with $5$ bowlers and $9$ non-bowlers.
To select a team of $11$ players with at least $4$ bowlers,we have two cases:
Case $1$: Selecting $4$ bowlers and $7$ non-bowlers: $^5C_4 \times ^9C_7 = 5 \times 36 = 180$.
Case $2$: Selecting $5$ bowlers and $6$ non-bowlers: $^5C_5 \times ^9C_6 = 1 \times 84 = 84$.
Total ways $= 180 + 84 = 264$.

(B) The total number of ways to select any number of books (including zero) from $n$ distinct books is given by the sum of combinations: $\binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n} = 2^n$.
To select one or more books,we must exclude the case where zero books are selected (i.e.,$\binom{n}{0} = 1$).
Therefore,the number of ways to select one or more books from $n$ books is $2^n - 1$.
For $n = 6$,the number of ways is $2^6 - 1 = 64 - 1 = 63$.

(C) Total players = $13$,Bowlers = $4$,Others = $9$.
We need to select a team of $11$ players with at least $2$ bowlers.
The possible cases are:
Case $1$: $2$ bowlers and $9$ others: $\binom{4}{2} \times \binom{9}{9} = 6 \times 1 = 6$.
Case $2$: $3$ bowlers and $8$ others: $\binom{4}{3} \times \binom{9}{8} = 4 \times 9 = 36$.
Case $3$: $4$ bowlers and $7$ others: $\binom{4}{4} \times \binom{9}{7} = 1 \times 36 = 36$.
Total ways = $6 + 36 + 36 = 78$.

(C) We use the identity $^nC_r + ^nC_{r-1} = ^{n+1}C_r$.
Given expression: $^nC_{r+1} + ^nC_{r-1} + 2^nC_r$
$= (^nC_{r+1} + ^nC_r) + (^nC_r + ^nC_{r-1})$
Using the identity $^nC_r + ^nC_{r-1} = ^{n+1}C_r$:
$= ^{n+1}C_{r+1} + ^{n+1}C_r$
Again,applying the identity $^nC_r + ^nC_{r-1} = ^{n+1}C_r$ where $n$ is replaced by $n+1$:
$= ^{n+2}C_{r+1}$

(C) The word $DHOLPUR$ contains $7$ distinct letters: $D, H, O, L, P, U, R$.
We need to form a $4$-letter word that must include $L$ and $P$.
Since $L$ and $P$ are already selected,we need to choose $2$ more letters from the remaining $5$ letters $(D, H, O, U, R)$.
The number of ways to choose these $2$ letters is $^5C_2 = \frac{5 \times 4}{2 \times 1} = 10$.
Now,we have $4$ letters ($L, P$ and the $2$ chosen letters),which can be arranged in $4!$ ways.
$4! = 4 \times 3 \times 2 \times 1 = 24$.
Therefore,the total number of words is $10 \times 24 = 240$.

(B) Each of the $10$ bulbs has $2$ possibilities: it can be either switched on or switched off.
Since there are $10$ bulbs,the total number of ways to switch them is $2^{10} = 1024$.
However,the room is illuminated only if at least one bulb is switched on.
Therefore,we must exclude the case where all bulbs are switched off.
The number of ways the room can be illuminated is $2^{10} - 1 = 1024 - 1 = 1023$.

(C) The boys are in the majority if the group consists of $(4 \text{ boys}, 3 \text{ girls})$,$(5 \text{ boys}, 2 \text{ girls})$,or $(6 \text{ boys}, 1 \text{ girl})$.
Total number of combinations:
$= \binom{6}{4} \times \binom{4}{3} + \binom{6}{5} \times \binom{4}{2} + \binom{6}{6} \times \binom{4}{1}$
$= 15 \times 4 + 6 \times 6 + 1 \times 4$
$= 60 + 36 + 4 = 100$
Hence,option $(C)$ is correct.

(B) Let the two specific people be $P_1$ and $P_2$. Since they cannot be in the same boat,one must be in Boat $A$ and the other in Boat $B$.
Boat $A$ already has $P_1$,so we need to choose $4$ more people from the remaining $8$ people to fill Boat $A$. This can be done in $\binom{8}{4}$ ways.
Boat $B$ will then automatically contain $P_2$ and the remaining $4$ people from the $8$ people.
Since the boats are distinct (or if we consider the arrangement of the two groups),the total number of ways is $\binom{8}{4} + \binom{8}{4} = 2 \binom{8}{4}$.

(B) The student must select $10$ questions out of $13$ total questions,with the constraint of choosing at least $4$ from the first $5$ questions.
Case $I$: Choosing $4$ 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 $II$: Choosing $5$ 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 = $140 + 56 = 196$.

(A) The committee must have at least $3$ women. Since the total committee size is $6$,the possible compositions are:
Case $1$: $3$ women and $3$ men.
The number of ways is $^4C_3 \times ^8C_3 = 4 \times 56 = 224$.
Case $2$: $4$ women and $2$ men.
The number of ways is $^4C_4 \times ^8C_2 = 1 \times 28 = 28$.
Total number of ways = $224 + 28 = 252$.

(B) We need to select $7$ members such that the number of boys $(B)$ is greater than the number of girls $(G)$. The total members are $B + G = 7$.
Possible cases are:
Case $1$: $4$ boys and $3$ girls. Number of ways = $\binom{6}{4} \times \binom{4}{3} = 15 \times 4 = 60$.
Case $2$: $5$ boys and $2$ girls. Number of ways = $\binom{6}{5} \times \binom{4}{2} = 6 \times 6 = 36$.
Case $3$: $6$ boys and $1$ girl. Number of ways = $\binom{6}{6} \times \binom{4}{1} = 1 \times 4 = 4$.
Total number of ways = $60 + 36 + 4 = 100$.

(A) Using the Pascal's identity,$\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$.
Given expression: $\left( \binom{15}{8} + \binom{15}{9} \right) - \left( \binom{15}{6} + \binom{15}{7} \right)$.
Applying the identity:
$\binom{15}{8} + \binom{15}{9} = \binom{16}{9}$.
$\binom{15}{6} + \binom{15}{7} = \binom{16}{7}$.
So,the expression becomes $\binom{16}{9} - \binom{16}{7}$.
Using the property $\binom{n}{r} = \binom{n}{n-r}$,we have $\binom{16}{7} = \binom{16}{16-7} = \binom{16}{9}$.
Therefore,$\binom{16}{9} - \binom{16}{9} = 0$.

(C) We know that $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ and $\binom{n}{n-1} = \binom{n}{1} = \frac{n!}{1!(n-1)!} = n$.
Alternatively,using the formula:
$\binom{n}{r} \div \binom{n}{n-1} = \frac{n!}{r!(n-r)!} \div \frac{n!}{(n-1)!(n-(n-1))!} = \frac{n!}{r!(n-r)!} \times \frac{(n-1)!1!}{n!} = \frac{(n-1)!}{r!(n-r)!} \times 1 = \frac{n!}{r(r-1)!(n-r)!} \times \frac{(r-1)!(n-r+1)(n-r)!}{n!} = \frac{n-r+1}{r}$.

(B) Let the two friends be $F_1$ and $F_2$.
Case $1$: $F_1$ gets $8$ balls and $F_2$ gets $4$ balls. The number of ways is $\binom{12}{8} \times \binom{4}{4} = \frac{12!}{8!4!}$.
Case $2$: $F_1$ gets $4$ balls and $F_2$ gets $8$ balls. The number of ways is $\binom{12}{4} \times \binom{8}{8} = \frac{12!}{4!8!}$.
Since these are two distinct cases,the total number of ways is $\frac{12!}{8!4!} + \frac{12!}{4!8!} = 2 \times \frac{12!}{8!4!} = \frac{2! \times 12!}{8!4!}$.

(B) We use the property $\frac{^nC_k}{^nC_{k-1}} = \frac{n-k+1}{k}$.
For the first two equations:
$\frac{^nC_r}{^nC_{r-1}} = \frac{84}{36} = \frac{7}{3} \implies \frac{n-r+1}{r} = \frac{7}{3} \implies 3n - 3r + 3 = 7r \implies 3n - 10r = -3$ $(1)$
For the next two equations:
$\frac{^nC_{r+1}}{^nC_r} = \frac{126}{84} = \frac{3}{2} \implies \frac{n-(r+1)+1}{r+1} = \frac{3}{2} \implies \frac{n-r}{r+1} = \frac{3}{2} \implies 2n - 2r = 3r + 3 \implies 2n - 5r = 3$ $(2)$
Multiply equation $(2)$ by $2$ to get $4n - 10r = 6$ $(3)$.
Subtract equation $(1)$ from $(3)$:
$(4n - 10r) - (3n - 10r) = 6 - (-3) \implies n = 9$.

(C) The word $MISSISSIPPI$ contains the following letters: $M: 1, I: 4, S: 4, P: 2$.
To form a combination,we can choose any number of $M$s ($0$ or $1$),any number of $I$s ($0$ to $4$),any number of $S$s ($0$ to $4$),and any number of $P$s ($0$ to $2$).
The total number of ways to select letters is $(1+1)(4+1)(4+1)(2+1) = 2 \times 5 \times 5 \times 3 = 150$.
Since we must select at least one letter,we exclude the case where no letters are selected (the empty set).
Thus,the total number of distinct combinations is $150 - 1 = 149$.

(C) Let the number of candidates be $n$.
Since the number of candidates is $1$ more than the number of persons to be elected,the number of persons to be elected is $n - 1$.
The number of ways a voter can cast their vote is the sum of combinations of choosing $1, 2, \dots, n - 1$ candidates out of $n$.
This is given by: $^nC_1 + ^nC_2 + \dots + ^nC_{n - 1} = 254$.
We know that the sum of binomial coefficients is $\sum_{k=0}^{n} {^nC_k} = 2^n$.
Therefore,$^nC_0 + ^nC_1 + \dots + ^nC_{n - 1} + ^nC_n = 2^n$.
Substituting $^nC_0 = 1$ and $^nC_n = 1$,we get $1 + \left(\sum_{k=1}^{n-1} {^nC_k}\right) + 1 = 2^n$.
$2 + 254 = 2^n$.
$256 = 2^n$.
$2^8 = 2^n$.
Thus,$n = 8$.

(B) Total ways to choose $6$ guests out of $10$ such that two specific friends (let them be $A$ and $B$) do not attend together is calculated by considering the following cases:
Case $1$: Neither $A$ nor $B$ is invited.
We choose $6$ guests from the remaining $8$ friends: $\binom{8}{6} = \binom{8}{2} = \frac{8 \times 7}{2 \times 1} = 28$.
Case $2$: Exactly one of $A$ or $B$ is invited.
We choose $1$ friend from ${A, B}$ and $5$ friends from the remaining $8$: $\binom{2}{1} \times \binom{8}{5} = 2 \times \binom{8}{3} = 2 \times \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 2 \times 56 = 112$.
Total ways = $28 + 112 = 140$.

(D) Step $1$: Select $2$ balls from Urn $A$ containing $3$ distinct red balls. The number of ways is $^3C_2 = 3$.
Step $2$: Select $2$ balls from Urn $B$ containing $9$ blue balls. The number of ways is $^9C_2 = \frac{9 \times 8}{2} = 36$.
Step $3$: Since the operations are independent,the total number of ways to select the balls is $3 \times 36 = 108$.
Step $4$: After selection,the $2$ balls from $A$ are placed into $B$,and the $2$ balls from $B$ are placed into $A$. This is a single way of performing the swap for each specific selection.
Therefore,the total number of ways is $108$.

(C) Mahesh can invite one,two,three,four,five,or six friends for dinner.
This is equivalent to finding the total number of non-empty subsets of a set of $6$ friends.
The number of ways is given by:
$^6C_1 + ^6C_2 + ^6C_3 + ^6C_4 + ^6C_5 + ^6C_6 = 2^6 - 1$.
$2^6 - 1 = 64 - 1 = 63$.

(A) The product of $r$ consecutive natural numbers is given by $n(n+1)(n+2)...(n+r-1)$.
This product is equal to $r! \times \binom{n+r-1}{r}$.
Since $\binom{n+r-1}{r}$ is an integer,the product is always divisible by $r!$.

(A) The number of ways to distribute $n$ identical items among $r$ persons such that each person receives at least one item is given by the formula $^{n-1}C_{r-1}$.
Here,$n = 20$ and $r = 4$.
Therefore,the number of ways $= ^{20-1}C_{4-1} = ^{19}C_3$.
Calculating the value: $^{19}C_3 = \frac{19 \times 18 \times 17}{3 \times 2 \times 1} = 19 \times 3 \times 17 = 969$.

(B) Let $n = 9$ be the total number of papers. Let $p$ be the number of papers passed and $f$ be the number of papers failed,where $p + f = 9$.
The candidate is successful if $p > f$. Since $p = 9 - f$,the condition $p > f$ becomes $9 - f > f$,which implies $2f < 9$,or $f < 4.5$.
Thus,the candidate is successful if they fail in $0, 1, 2, 3,$ or $4$ papers.
The candidate is unsuccessful if they fail in $5, 6, 7, 8,$ or $9$ papers.
The number of ways to be unsuccessful is the sum of combinations: $\binom{9}{5} + \binom{9}{6} + \binom{9}{7} + \binom{9}{8} + \binom{9}{9}$.
Using the property $\binom{n}{r} = \binom{n}{n-r}$,this is equal to $\binom{9}{4} + \binom{9}{3} + \binom{9}{2} + \binom{9}{1} + \binom{9}{0}$.
We know that the sum of all binomial coefficients $\sum_{k=0}^{9} \binom{9}{k} = 2^9 = 512$.
Since $\sum_{k=0}^{9} \binom{9}{k} = (\binom{9}{0} + \binom{9}{1} + \binom{9}{2} + \binom{9}{3} + \binom{9}{4}) + (\binom{9}{5} + \binom{9}{6} + \binom{9}{7} + \binom{9}{8} + \binom{9}{9}) = 512$,and both sums are equal,each sum is $\frac{512}{2} = 256$.
Therefore,the number of ways to be unsuccessful is $256$.

(C) The total number of $5$-element subsets of a set with $20$ elements is given by $^{20}C_5$.
The number of $5$-element subsets containing a specific element $a_4$ is equivalent to choosing $4$ more elements from the remaining $19$ elements,which is $^{19}C_4$.
According to the problem,$^{20}C_5 = k \times ^{19}C_4$.
Using the formula $^{n}C_r = \frac{n}{r} \times ^{n-1}C_{r-1}$,we have:
$^{20}C_5 = \frac{20}{5} \times ^{19}C_4 = 4 \times ^{19}C_4$.
Comparing this with $k \times ^{19}C_4$,we get $k = 4$.

(C) Total number of candidates $= 10$.
Number of candidates to be elected $= 4$.
$A$ voter can vote for at most $4$ candidates and at least $1$ candidate.
The number of ways to vote for $r$ candidates is given by $^{10}C_{r}$.
Number of ways to vote for $1$ candidate $= ^{10}C_{1} = 10$.
Number of ways to vote for $2$ candidates $= ^{10}C_{2} = \frac{10 \times 9}{2 \times 1} = 45$.
Number of ways to vote for $3$ candidates $= ^{10}C_{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.
Number of ways to vote for $4$ candidates $= ^{10}C_{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$.
Total number of ways $= 10 + 45 + 120 + 210 = 385$.

(D) This is a problem of combinations with repetition allowed (stars and bars method).
Let $n = 5$ be the number of types of ice-creams and $r = 6$ be the number of ice-creams to be bought.
The number of ways to choose $r$ items from $n$ types with repetition is given by the formula $^{n+r-1}C_r$.
Substituting the values,we get $^{5+6-1}C_6 = ^{10}C_6$.
Since $^{10}C_6 = ^{10}C_{10-6} = ^{10}C_4$,and $^{10}C_4 = 210$,while $^{10}C_5 = 252$.
Thus,$Statement-1$ is false.
For $Statement-2$,the number of ways to arrange $6$ $A$'s and $4$ $B$'s is the number of ways to choose $6$ positions out of $10$ for the $A$'s,which is $^{10}C_6$.
This is equal to the number of ways to buy the ice-creams.
Therefore,$Statement-2$ is true.

(C) Urn $A$ contains $3$ distinct red balls. The number of ways to select $2$ balls from urn $A$ is $^3C_2 = \frac{3!}{2!1!} = 3$.
Urn $B$ contains $9$ distinct blue balls. The number of ways to select $2$ balls from urn $B$ is $^9C_2 = \frac{9 \times 8}{2 \times 1} = 36$.
Since the selection of balls from urn $A$ and urn $B$ are independent events,the total number of ways to perform these transfers is the product of the number of ways to select balls from each urn.
Total ways $= ^3C_2 \times ^9C_2 = 3 \times 36 = 108$.

(D) For Statement-$1$: 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,$n = 10$ and $r = 4$.
So,the number of ways is $^{10-1}C_{4-1} = ^9C_3$.
Thus,Statement-$1$ is true.
For Statement-$2$: The number of ways to choose $r$ items from $n$ distinct items is $^nC_r$.
Here,$n = 9$ and $r = 3$,so the number of ways is $^9C_3$.
Thus,Statement-$2$ is true.
Since the formula for distributing identical items into distinct boxes (stars and bars method) is derived using the concept of choosing positions,Statement-$2$ is the correct explanation for Statement-$1$.

(C) Since the $5$ balls are identical and the $10$ boxes are identical,distributing the balls such that no box contains more than one ball is equivalent to selecting $5$ boxes out of $10$ to place the balls in.
Because the boxes are identical,the arrangement does not depend on the specific labels of the boxes,but rather on the selection of the positions.
However,in standard combinatorial problems of this type where boxes are considered distinct positions,the number of ways is given by the combination formula $C(n, r) = \binom{n}{r}$.
If the boxes are truly identical (indistinguishable),there is only $1$ way to place the balls. Given the options provided,the question implies the boxes are distinct positions,leading to the calculation $\binom{10}{5} = \frac{10!}{5!5!}$.

(D) Total committee size is $12$. Let $W$ be the number of women and $M$ be the number of men. We require $W + M = 12$ and $W \ge 5$.
$(i)$ Women are in the majority if $W > M$. Since $W + M = 12$,this implies $W > 6$. Possible values for $W$ are $7, 8, 9$.
Number of ways = $^9C_7 \times ^8C_5 + ^9C_8 \times ^8C_4 + ^9C_9 \times ^8C_3$
$= (36 \times 56) + (9 \times 70) + (1 \times 56) = 2016 + 630 + 56 = 2702$.
(ii) Men are in the majority if $M > W$. Since $W + M = 12$,this implies $M > 6$. Possible values for $W$ are $5, 6$ (since $W \ge 5$).
Number of ways = $^9C_5 \times ^8C_7 + ^9C_6 \times ^8C_6$
$= (126 \times 8) + (84 \times 28) = 1008 + 2352 = 3360$.
Wait,re-evaluating: The question asks for committees where $W \ge 5$. If $M > W$,then $W$ can be $5$ or $6$.
For $W=5, M=7$: $^9C_5 \times ^8C_7 = 126 \times 8 = 1008$.
For $W=6, M=6$: This is not a majority. Thus,only $W=5$ gives a male majority.
Correct answer is $2702, 1008$.

(C) Using the Pascal's identity,$^{n-1}C_r + ^{n-1}C_{r-1} = ^nC_r$,we have:
$^{n-1}C_3 + ^{n-1}C_4 = ^nC_4$
Given the inequality: $^{n-1}C_3 + ^{n-1}C_4 > ^nC_3$
Substituting the identity: $^nC_4 > ^nC_3$
Using the formula $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$,we get:
$\frac{^nC_4}{^nC_3} > 1$
$\frac{n-4+1}{4} > 1$
$\frac{n-3}{4} > 1$
$n-3 > 4$
$n > 7$

(B) Total number of ways to select $6$ guests from $10$ friends without any restriction is $^{10}C_6 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$.
Let the two friends who will not attend together be $A$ and $B$.
The number of ways in which both $A$ and $B$ attend the party is the number of ways to select the remaining $4$ guests from the other $8$ friends,which is $^8C_4 = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$.
The number of ways in which $A$ and $B$ do not attend together is the total ways minus the ways where both attend: $210 - 70 = 140$.

(C) Total players available = $22$.
We need to select a team of $10$ players.
$6$ particular players must always be included,so we have already selected $6$ players.
Remaining players to be selected = $10 - 6 = 4$.
$4$ particular players must always be excluded,so we remove them from the total pool.
Remaining players available to choose from = $22 - 6 - 4 = 12$.
Therefore,the number of ways to select the remaining $4$ players from $12$ is $^{12}C_4$.

(A) The number of solutions to $x_1 + x_2 + ... + x_k = n$ with $x_i \ge i$ is the coefficient of $t^n$ in the product $(t^1 + t^2 + ...) (t^2 + t^3 + ...) ... (t^k + t^{k+1} + ...)$.
This is the coefficient of $t^n$ in $t^{1+2+...+k} (1 + t + t^2 + ...)^k$.
Let $r = 1 + 2 + ... + k = \frac{k(k+1)}{2}$.
The expression becomes the coefficient of $t^{n-r}$ in $(1-t)^{-k}$.
Using the binomial expansion $(1-t)^{-k} = \sum_{j=0}^{\infty} \binom{k+j-1}{k-1} t^j$,the coefficient of $t^{n-r}$ is $\binom{k+(n-r)-1}{k-1}$.
Let $m = k + n - r - 1 = k + n - \frac{k(k+1)}{2} - 1 = \frac{2k + 2n - k^2 - k - 2}{2} = \frac{2n - k^2 + k - 2}{2}$.
Thus,the number of solutions is $\binom{m}{k-1}$.

(C) For the function $f(x)$ to be defined,we must satisfy three conditions:
$1$. The binomial coefficients and permutations must have non-negative integer parameters:
For $^{16-x}C_{2x-1}$,we need $16-x \ge 2x-1 \ge 0$,which implies $17 \ge 3x$ $(x \le 17/3)$ and $2x \ge 1$ $(x \ge 1/2)$.
For $^{20-3x}P_{2x-5}$,we need $20-3x \ge 2x-5 \ge 0$,which implies $25 \ge 5x$ $(x \le 5)$ and $2x \ge 5$ $(x \ge 2.5)$.
Combining these,we get $2.5 \le x \le 5$.
$2$. The base of the logarithm $[x + \frac{1}{x}]$ must be greater than $0$ and not equal to $1$.
For $x \in [2.5, 5]$,$x + \frac{1}{x}$ ranges from $2.5 + 0.4 = 2.9$ to $5 + 0.2 = 5.2$.
Thus,$[x + \frac{1}{x}]$ can take values $2, 3, 4, 5$.
Since $[x + \frac{1}{x}] \neq 1$ is satisfied (as the minimum value is $2$),the condition is satisfied for all $x \in [2.5, 5]$.
$3$. The argument of the logarithm $|x^2 - x - 6|$ must be greater than $0$.
$|x^2 - x - 6| > 0 \implies x^2 - x - 6 \neq 0 \implies (x-3)(x+2) \neq 0$.
So $x \neq 3$ and $x \neq -2$.
Given $x \in [2.5, 5]$,we must exclude $x=3$.
Checking integer values in the range $[2.5, 5]$: $x$ can be $3, 4, 5$.
However,the binomial coefficients require $x$ to be such that $2x-1$ and $2x-5$ are integers,implying $2x$ must be an integer.
Testing $x=3$: $f(3) = \log_{[3+1/3]} |9-3-6| + ^{13}C_5 + ^{11}P_1 = \log_3(0) + \dots$,which is undefined.
Testing $x=4$: $f(4) = \log_{[4+0.25]} |16-4-6| + ^{12}C_7 + ^{8}P_3 = \log_4(6) + 792 + 336$,which is defined.
Testing $x=5$: $f(5) = \log_{[5+0.2]} |25-5-6| + ^{11}C_9 + ^{5}P_5 = \log_5(14) + 55 + 120$,which is defined.
Since the domain must be discrete due to the nature of $nCr$ and $nPr$ with integer constraints,the values are $\{4, 5\}$.
Re-evaluating the options,there appears to be a discrepancy. Given the standard constraints,the set is $\{4, 5\}$.