Definition of combinations, Condition combinations Questions in English

(B) The total number of $m$-element subsets of a set with $n$ elements is given by $\binom{n}{m}$.
The number of $m$-element subsets containing a specific element $a_{4}$ is equivalent to choosing the remaining $(m-1)$ elements from the remaining $(n-1)$ elements,which is $\binom{n-1}{m-1}$.
According to the problem,$\binom{n}{m} = k \times \binom{n-1}{m-1}$.
Using the identity $\binom{n}{m} = \frac{n}{m} \binom{n-1}{m-1}$,we have:
$\frac{n}{m} \binom{n-1}{m-1} = k \binom{n-1}{m-1}$.
Dividing both sides by $\binom{n-1}{m-1}$ (assuming $n \ge m \ge 1$),we get:
$\frac{n}{m} = k \Rightarrow n = mk$.

(A) The candidate must select a total of $6$ questions from two parts,$A$ and $B$,with the constraint that no more than $4$ questions can be selected from either part. The possible combinations are as follows:

Part $A$	Part $B$
$4$ questions	$2$ questions
$3$ questions	$3$ questions
$2$ questions	$4$ questions

The total number of ways is given by:
$Ways = ({ }^{6}C_{4} \times { }^{6}C_{2}) + ({ }^{6}C_{3} \times { }^{6}C_{3}) + ({ }^{6}C_{2} \times { }^{6}C_{4})$
$Ways = (15 \times 15) + (20 \times 20) + (15 \times 15)$
$Ways = 225 + 400 + 225$
$Ways = 850$

(C) The number of ways to select $3$ consonants from $7$ is given by ${}^{7}C_{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$ ways.
The number of ways to select $2$ vowels from $4$ is given by ${}^{4}C_{2} = \frac{4 \times 3}{2 \times 1} = 6$ ways.
The total number of ways to select the $5$ letters is $35 \times 6 = 210$ ways.
Since these $5$ selected letters can be arranged among themselves in $5!$ ways,the total number of words is $210 \times 5! = 210 \times 120 = 25200$.

(A) Given the equation $x+y+z=10$,where $x, y, z \in \mathbb{Z}^+$.
This is a problem of finding the number of positive integer solutions to the equation $x_1+x_2+\dots+x_r=n$.
The formula for the number of positive integer solutions is given by $^{n-1}C_{r-1}$.
Here,$n=10$ and $r=3$.
Therefore,the number of solutions $= ^{10-1}C_{3-1} = ^{9}C_{2}$.
$^{9}C_{2} = \frac{9 \times 8}{2 \times 1} = 36$.

(A) We need to select $n$ objects from $2n$ objects,where $n$ objects are identical and $n$ objects are distinct.
Let $k$ be the number of distinct objects selected,where $0 \le k \le n$.
Then the remaining $(n-k)$ objects must be selected from the $n$ identical objects.
Since the $n$ identical objects are indistinguishable,there is only $1$ way to select any number of them.
Thus,for each $k$ from $0$ to $n$,the number of ways to select $k$ distinct objects is given by $\binom{n}{k}$.
The total number of ways is the sum of these possibilities:
$\sum_{k=0}^{n} \binom{n}{k} = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n} = 2^{n}$.

(C) Let $x_{1}, x_{2}, x_{3}, x_{4}$ be the marks assigned to the $4$ questions respectively.
We are given the equation: $x_{1} + x_{2} + x_{3} + x_{4} = 10$,where $x_{i} \geq 2$ for all $i \in \{1, 2, 3, 4\}$.
Let $y_{i} = x_{i} - 2$. Since $x_{i} \geq 2$,we have $y_{i} \geq 0$.
Substituting $x_{i} = y_{i} + 2$ into the equation:
$(y_{1} + 2) + (y_{2} + 2) + (y_{3} + 2) + (y_{4} + 2) = 10$
$y_{1} + y_{2} + y_{3} + y_{4} + 8 = 10$
$y_{1} + y_{2} + y_{3} + y_{4} = 2$.
The number of non-negative integer solutions is given by the formula $\binom{n+r-1}{r-1}$,where $n = 2$ and $r = 4$.
Number of ways $= \binom{2+4-1}{4-1} = \binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$.

(C) We are given the expression: ${}^{n}C_{r} + 2 \cdot {}^{n}C_{r+1} + {}^{n}C_{r+2}$
Split the middle term:
$= {}^{n}C_{r} + {}^{n}C_{r+1} + {}^{n}C_{r+1} + {}^{n}C_{r+2}$
Using the Pascal's identity ${}^{n}C_{r} + {}^{n}C_{r+1} = {}^{n+1}C_{r+1}$:
$= ({}^{n}C_{r} + {}^{n}C_{r+1}) + ({}^{n}C_{r+1} + {}^{n}C_{r+2})$
$= {}^{n+1}C_{r+1} + {}^{n+1}C_{r+2}$
Applying the identity again:
$= {}^{n+2}C_{r+2}$

(B) Given,$\frac{1}{{ }^{5}C_{r}} + \frac{1}{{ }^{6}C_{r}} = \frac{1}{{ }^{4}C_{r}}$
Using the formula ${ }^{n}C_{r} = \frac{n!}{r!(n-r)!}$,we have:
$\frac{r!(5-r)!}{5!} + \frac{r!(6-r)!}{6!} = \frac{r!(4-r)!}{4!}$
Dividing by $r!$ and multiplying by $4!$:
$\frac{(5-r)!}{5} + \frac{(6-r)(5-r)!}{6 \times 5} = (4-r)!$
Dividing by $(4-r)!$:
$\frac{(5-r)}{5} + \frac{(6-r)(5-r)}{30} = 1$
Multiply by $30$:
$6(5-r) + (6-r)(5-r) = 30$
$30 - 6r + 30 - 11r + r^{2} = 30$
$r^{2} - 17r + 30 = 0$
$(r-2)(r-15) = 0$
Since $r \leq 4$ (as ${ }^{4}C_{r}$ is defined),we have $r = 2$.

(A) Given that ${}^nC_4, {}^nC_5, {}^nC_6$ are in $A.P.$
Therefore,$2({}^nC_5) = {}^nC_4 + {}^nC_6$.
Using the formula ${}^nC_r = \frac{n!}{r!(n-r)!}$,we have:
$2 \times \frac{n!}{5!(n-5)!} = \frac{n!}{4!(n-4)!} + \frac{n!}{6!(n-6)!}$
Dividing by $n!$ and multiplying by $6!(n-4)!$:
$2 \times 6(n-4) = 30 + (n-4)(n-5)$
$12n - 48 = 30 + n^2 - 9n + 20$
$n^2 - 21n + 98 = 0$
$(n-7)(n-14) = 0$
Thus,$n = 7$ or $n = 14$.

(D) Using the Pascal's identity,${}^{n-1}C_r + {}^{n-1}C_{r-1} = {}^{n}C_r$,we have:
${}^{n-1}C_3 + {}^{n-1}C_4 = {}^{n}C_4$.
Given the inequality:
${}^{n}C_4 > {}^{n}C_3$.
Expanding the combinations:
$\frac{n!}{4!(n-4)!} > \frac{n!}{3!(n-3)!}$.
Simplifying the factorials:
$\frac{1}{4(n-4)!} > \frac{1}{(n-3)(n-4)!}$.
$\frac{1}{4} > \frac{1}{n-3}$.
Since $n-3 > 0$,we get:
$n-3 > 4$,which implies $n > 7$.
Thus,$n$ is just greater than $7$.

(D) Let the number of oranges given to the four children be $x_1, x_2, x_3, x_4$ respectively.
Since the oranges are identical,we need to find the number of positive integer solutions to the equation:
$x_1 + x_2 + x_3 + x_4 = 16$,where $x_i \geq 1$ for $i = 1, 2, 3, 4$.
Using the substitution $x_i = x_i^{\prime} + 1$,where $x_i^{\prime} \geq 0$,the equation becomes:
$(x_1^{\prime} + 1) + (x_2^{\prime} + 1) + (x_3^{\prime} + 1) + (x_4^{\prime} + 1) = 16$
$x_1^{\prime} + x_2^{\prime} + x_3^{\prime} + x_4^{\prime} = 12$.
The number of non-negative integer solutions is given by the formula $\binom{n+k-1}{k-1}$,where $n = 12$ and $k = 4$.
Number of ways $= \binom{12+4-1}{4-1} = \binom{15}{3}$.
$\binom{15}{3} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = 5 \times 7 \times 13 = 455$.

(B) Given equation: $^{36}C_{r+1} = \frac{6(^{35}C_r)}{k^2-3}$.
Using the property $^{n}C_r = \frac{n}{r} \cdot ^{n-1}C_{r-1}$,we have $^{36}C_{r+1} = \frac{36}{r+1} \cdot ^{35}C_r$.
Substituting this into the equation: $\frac{36}{r+1} \cdot ^{35}C_r = \frac{6(^{35}C_r)}{k^2-3}$.
Assuming $^{35}C_r \neq 0$,we get $\frac{36}{r+1} = \frac{6}{k^2-3}$,which simplifies to $\frac{6}{r+1} = \frac{1}{k^2-3}$.
Thus,$k^2-3 = \frac{r+1}{6}$,or $k^2 = \frac{r+1}{6} + 3$.
Since $k \in Z$,$k^2$ must be a perfect square integer. This implies $(r+1)$ must be a multiple of $6$.
Given $0 \leq r \leq 35$,we have $1 \leq r+1 \leq 36$.
Possible values for $r+1$ are $6, 12, 18, 24, 30, 36$.
For $r+1 = 6$,$k^2 = 1+3 = 4 \implies k = \pm 2$. Pairs: $(5, 2), (5, -2)$.
For $r+1 = 12$,$k^2 = 2+3 = 5$ (not a perfect square).
For $r+1 = 18$,$k^2 = 3+3 = 6$ (not a perfect square).
For $r+1 = 24$,$k^2 = 4+3 = 7$ (not a perfect square).
For $r+1 = 30$,$k^2 = 5+3 = 8$ (not a perfect square).
For $r+1 = 36$,$k^2 = 6+3 = 9 \implies k = \pm 3$. Pairs: $(35, 3), (35, -3)$.
Total pairs $(r, k)$ are $(5, 2), (5, -2), (35, 3), (35, -3)$.
There are $4$ such elements.

(A) For player $A$ to win the series,they must win the $5$th game. This implies that in the previous $n-1$ games,player $A$ must have won exactly $4$ games,and player $B$ must have won $n-5$ games.
The series can end in $5, 6, 7, 8,$ or $9$ games.
If the series ends in $5$ games: $A$ wins $5$ games,$B$ wins $0$. Ways = $\binom{4}{4} = 1$.
If the series ends in $6$ games: $A$ wins $4$ of the first $5$ games and the $6$th game. Ways = $\binom{5}{4} = 5$.
If the series ends in $7$ games: $A$ wins $4$ of the first $6$ games and the $7$th game. Ways = $\binom{6}{4} = 15$.
If the series ends in $8$ games: $A$ wins $4$ of the first $7$ games and the $8$th game. Ways = $\binom{7}{4} = 35$.
If the series ends in $9$ games: $A$ wins $4$ of the first $8$ games and the $9$th game. Ways = $\binom{8}{4} = 70$.
Total ways = $1 + 5 + 15 + 35 + 70 = 126$.