Definition of combinations, Condition combinations Questions in English

(A) To connect every city to every other city,we need to choose $2$ cities out of $10$ to form a road connection.
This is a combination problem where we need to calculate $^{10}C_2$.
$^{10}C_2 = \frac{10 \times 9}{2 \times 1} = 45$.
Therefore,the government needs to construct $45$ roads.

(B) To select $r$ items from $n$ items in a row such that no two are consecutive,the formula is given by $^{n-r+1}C_r$.
Here,$n = 10$ and $r = 4$.
Substituting the values,we get $^{10-4+1}C_4 = ^7C_4$.
Since $^7C_4 = ^7C_3$,we calculate:
$^7C_3 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.

(C) Let the number of apples received by the $4$ boys be $x_1, x_2, x_3, x_4$ respectively.
We are given the equation $x_1 + x_2 + x_3 + x_4 = 25$,where $x_i \ge 4$ for $i = 1, 2, 3, 4$.
Let $y_i = x_i - 4$,where $y_i \ge 0$.
Substituting $x_i = y_i + 4$ into the equation:
$(y_1 + 4) + (y_2 + 4) + (y_3 + 4) + (y_4 + 4) = 25$
$y_1 + y_2 + y_3 + y_4 + 16 = 25$
$y_1 + y_2 + y_3 + y_4 = 9$
Using the stars and bars formula,the number of non-negative integer solutions is given by $^{n+r-1}C_{r-1}$,where $n = 9$ and $r = 4$.
Number of ways = $^{9+4-1}C_{4-1} = ^{12}C_3$.

(C) To select at least one fruit of each kind,we must choose at least one from each group.
For $5$ apples,the number of ways to select at least one is $5$.
For $4$ mangoes,the number of ways to select at least one is $4$.
For $3$ oranges,the number of ways to select at least one is $3$.
For the $2$ other varieties,each having $1$ fruit,the number of ways to select at least one is $1$ for each.
Therefore,the total number of ways is $5 \times 4 \times 3 \times 1 \times 1 = 60$.

(D) To select one or more balls from $10$ white,$9$ green,and $7$ blue balls,we consider the choices for each color.
For white balls,we can choose $0, 1, 2, \dots, 10$ balls,which gives $(10 + 1) = 11$ options.
For green balls,we can choose $0, 1, 2, \dots, 9$ balls,which gives $(9 + 1) = 10$ options.
For blue balls,we can choose $0, 1, 2, \dots, 7$ balls,which gives $(7 + 1) = 8$ options.
The total number of ways to select any number of balls (including selecting zero balls) is $11 \times 10 \times 8 = 880$.
Since we must select at least one ball,we subtract the case where zero balls are selected from all three colors.
Required number of ways $= 880 - 1 = 879$.

(B) five-digit number with strictly increasing digits is uniquely determined by choosing $5$ distinct digits from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$. The total number of such integers is $\binom{9}{5} = 126$.
$1$. Numbers starting with $1$: We choose $4$ more digits from $\{2, 3, 4, 5, 6, 7, 8, 9\}$. Total = $\binom{8}{4} = 70$.
$2$. Numbers starting with $2$:
- Starting with $23$: We choose $3$ more digits from $\{4, 5, 6, 7, 8, 9\}$. Total = $\binom{6}{3} = 20$.
- Starting with $24$:
- Starting with $245$: We choose $2$ more digits from $\{6, 7, 8, 9\}$. Total = $\binom{4}{2} = 6$.
- Starting with $246$:
- $24678$ ($96^{th}$ number)
- $24679$ ($97^{th}$ number)
Thus,the $97^{th}$ number is $24679$. This number does not contain the digit $5$.

(B) The number of ways to select at least one candidate is given by $\sum_{k=1}^{n} {^{2n+1}C_k} = 63$.
We know that $\sum_{k=0}^{2n+1} {^{2n+1}C_k} = 2^{2n+1}$.
Since ${^{2n+1}C_k} = {^{2n+1}C_{2n+1-k}}$,we have $\sum_{k=0}^{n} {^{2n+1}C_k} = \sum_{k=n+1}^{2n+1} {^{2n+1}C_k} = \frac{2^{2n+1}}{2} = 2^{2n}$.
Thus,$\sum_{k=1}^{n} {^{2n+1}C_k} = (\sum_{k=0}^{n} {^{2n+1}C_k}) - {^{2n+1}C_0} = 2^{2n} - 1$.
Given $2^{2n} - 1 = 63$,we get $2^{2n} = 64 = 2^6$.
Therefore,$2n = 6$,which implies $n = 3$.
The maximum number of candidates that can be selected is $n = 3$.

(D) Let $x_i$ be the number of coins given to player $A_i$. The total number of coins is $\sum_{i=1}^{11} x_i = 100$.
Given conditions: $x_i \ge i+1$ for $i=3, 4, \dots, 11$,$x_1 \ge 1+5=6$,and $x_2 \ge 2+3=5$.
Let $y_i = x_i - (i+1)$ for $i=3, \dots, 11$,$y_1 = x_1 - 6$,and $y_2 = x_2 - 5$.
Substituting these into the sum: $(y_1+6) + (y_2+5) + \sum_{i=3}^{11} (y_i + i + 1) = 100$.
$y_1 + y_2 + \dots + y_{11} + 11 + \sum_{i=1}^{11} i = 100$.
$y_1 + y_2 + \dots + y_{11} + 11 + \frac{11 \times 12}{2} = 100$.
$y_1 + y_2 + \dots + y_{11} + 11 + 66 = 100$.
$y_1 + y_2 + \dots + y_{11} = 100 - 77 = 23$.
Wait,re-evaluating the base distribution: $x_1 \ge 6, x_2 \ge 5, x_3 \ge 4, x_4 \ge 5, \dots, x_{11} \ge 12$.
Sum of minimums $= 6+5+4+5+6+7+8+9+10+11+12 = 83$.
Remaining coins $= 100 - 83 = 17$.
Using stars and bars,the number of ways to distribute $17$ coins among $11$ players is $^{17+11-1}C_{11-1} = ^{27}C_{10}$.

(B) The number of possible pairs of $n$ objects is given by $^nC_2 = \frac{n(n-1)}{2}$.
Each pair has a total weight of $2w$.
Therefore,the sum of the weights of all pairs is $\frac{n(n-1)}{2} \times 2w = n(n-1)w = 120$ --- $(1)$.
The number of possible triplets of $n$ objects is given by $^nC_3 = \frac{n(n-1)(n-2)}{6}$.
Each triplet has a total weight of $3w$.
Therefore,the sum of the weights of all triplets is $\frac{n(n-1)(n-2)}{6} \times 3w = \frac{n(n-1)(n-2)w}{2} = 480$ --- $(2)$.
Dividing equation $(2)$ by equation $(1)$:
$\frac{\frac{n(n-1)(n-2)w}{2}}{n(n-1)w} = \frac{480}{120}$
$\frac{n-2}{2} = 4$
$n-2 = 8$
$n = 10$.

(B) Let $n = 10$ be the total number of employees.
We need to form a team of size $r$ such that the team includes at least $3$ employees and excludes at least $3$ employees.
This means the number of employees in the team $r$ must satisfy $3 \le r \le 10 - 3$,which simplifies to $3 \le r \le 7$.
The number of ways to form such a team is the sum of combinations $\sum_{r=3}^{7} {^{10}C_r}$.
We know that $\sum_{r=0}^{10} {^{10}C_r} = 2^{10} = 1024$.
The required sum is $2^{10} - ({^{10}C_0} + {^{10}C_1} + {^{10}C_2} + {^{10}C_8} + {^{10}C_9} + {^{10}C_{10}})$.
Calculating the values: ${^{10}C_0} = 1$,${^{10}C_1} = 10$,${^{10}C_2} = \frac{10 \times 9}{2} = 45$.
Since ${^{10}C_8} = {^{10}C_2} = 45$,${^{10}C_9} = {^{10}C_1} = 10$,and ${^{10}C_{10}} = {^{10}C_0} = 1$.
The sum of excluded terms is $(1 + 10 + 45 + 45 + 10 + 1) = 112$.
Therefore,the number of ways is $1024 - 112 = 912$.

(A) Given: $^nC_{r-2} = 36$,$^nC_{r-1} = 84$,and $^nC_r = 126$.
Using the property $\frac{^nC_k}{^nC_{k-1}} = \frac{n-k+1}{k}$:
$\frac{^nC_{r-1}}{^nC_{r-2}} = \frac{84}{36} = \frac{7}{3}$ $\Rightarrow \frac{n-(r-1)+1}{r-1} = \frac{7}{3}$ $\Rightarrow \frac{n-r+2}{r-1} = \frac{7}{3}$
$3n - 3r + 6 = 7r - 7 \Rightarrow 3n - 10r = -13$ ....$(1)$
$\frac{^nC_r}{^nC_{r-1}} = \frac{126}{84} = \frac{3}{2} \Rightarrow \frac{n-r+1}{r} = \frac{3}{2}$
$2n - 2r + 2 = 3r \Rightarrow 2n - 5r = -2$ ....$(2)$
Multiplying equation $(2)$ by $2$: $4n - 10r = -4$ ....$(3)$
Subtracting $(1)$ from $(3)$: $(4n - 10r) - (3n - 10r) = -4 - (-13) \Rightarrow n = 9$
Substituting $n=9$ in $(2)$: $2(9) - 5r = -2$ $\Rightarrow 18 - 5r = -2$ $\Rightarrow 5r = 20$ $\Rightarrow r = 4$
We need to find $^nC_{2r} = ^9C_{2(4)} = ^9C_8 = ^9C_1 = 9$.

(B) When a die is rolled three times,the total number of outcomes is $6^3 = 216$.
Let the outcomes be $x_1, x_2, x_3$ such that $1 \le x_1 < x_2 < x_3 \le 6$.
To satisfy the condition that each number is larger than the previous one,we need to select $3$ distinct numbers from the set $\{1, 2, 3, 4, 5, 6\}$.
The number of ways to choose $3$ distinct numbers is given by the combination formula $^6C_3$.
$^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
Once $3$ distinct numbers are chosen,there is only $1$ way to arrange them in strictly increasing order.
Therefore,the number of favorable outcomes is $20$.
The probability is $\frac{20}{216}$.

(A) To select $3$ objects from $n$ objects arranged in a row such that no two are consecutive,we use the gap method.
Let the $n-3$ objects that are not selected be represented by $X$.
$X \_ X \_ X \_ \dots \_ X$
There are $n-3$ objects,which create $(n-3) + 1 = n-2$ available gaps (including the ends).
We need to choose $3$ gaps out of these $n-2$ gaps to place the selected objects.
The number of ways to do this is given by the combination formula ${}^{n-2}C_3$.

(D) The given expression is $S = ^{20}C_1 + 3 ^{20}C_2 + 3 ^{20}C_3 + ^{20}C_4$.
We can rewrite the coefficients as:
$S = (^{20}C_1 + ^{20}C_2) + 2(^{20}C_2 + ^{20}C_3) + (^{20}C_3 + ^{20}C_4)$.
Using the Pascal's identity $^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r}$,we get:
$S = ^{21}C_2 + 2(^{21}C_3) + ^{21}C_4$.
Further,$S = (^{21}C_2 + ^{21}C_3) + (^{21}C_3 + ^{21}C_4)$.
Applying the identity again:
$S = ^{22}C_3 + ^{22}C_4 = ^{23}C_4$.

(B) There are $6$ married couples,so there are $12$ people in total.
We need to select $6$ people out of $12$ such that no two people form a couple.
First,we select $6$ couples out of $6$ available couples,which can be done in $\binom{6}{6} = 1$ way.
From each of these $6$ selected couples,we must choose exactly one person to be on the committee.
Since each couple has $2$ people,for each of the $6$ positions,we have $2$ choices.
Therefore,the total number of ways is $\binom{6}{6} \times 2^6 = 1 \times 64 = 64$.

(B) Given equation: $^{69}C_{3r-1} - ^{69}C_{r^2} = ^{69}C_{r^2-1} - ^{69}C_{3r}$
Rearranging the terms: $^{69}C_{3r-1} + ^{69}C_{3r} = ^{69}C_{r^2} + ^{69}C_{r^2-1}$
Using the identity $^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r}$:
$^{70}C_{3r} = ^{70}C_{r^2}$
This implies either $3r = r^2$ or $3r + r^2 = 70$.
Case $1$: $r^2 - 3r = 0$ $\Rightarrow r(r-3) = 0$ $\Rightarrow r = 0$ or $r = 3$.
Case $2$: $r^2 + 3r - 70 = 0$ $\Rightarrow (r+10)(r-7) = 0$ $\Rightarrow r = -10$ or $r = 7$.
Since $nCr$ is defined for $n \ge r \ge 0$,we check the values:
For $r=0$: $^{69}C_{-1}$ is undefined.
For $r=-10$: $^{69}C_{-31}$ is undefined.
For $r=3$: $^{69}C_{8} - ^{69}C_{9} = ^{69}C_{8} - ^{69}C_{9}$ (Valid).
For $r=7$: $^{69}C_{20} - ^{69}C_{49} = ^{69}C_{48} - ^{69}C_{21}$ (Valid,since $^{69}C_{49} = ^{69}C_{20}$ and $^{69}C_{48} = ^{69}C_{21}$).
Thus,the valid values for $r$ are $3$ and $7$.
The number of values is $2$.

(B) Let $S$ be a set with $N = 2n + 1$ elements. We want to find the number of subsets with at most $n$ elements,which is given by the sum:
$S = \sum_{r=0}^{n} \binom{2n+1}{r} = \binom{2n+1}{0} + \binom{2n+1}{1} + \dots + \binom{2n+1}{n}$.
We know that the total number of subsets is $\sum_{r=0}^{2n+1} \binom{2n+1}{r} = 2^{2n+1}$.
By the property of combinations,$\binom{N}{r} = \binom{N}{N-r}$.
Thus,$\binom{2n+1}{0} = \binom{2n+1}{2n+1}$,$\binom{2n+1}{1} = \binom{2n+1}{2n}$,...,$\binom{2n+1}{n} = \binom{2n+1}{n+1}$.
Let $X = \sum_{r=0}^{n} \binom{2n+1}{r}$. Then $X = \binom{2n+1}{0} + \dots + \binom{2n+1}{n}$.
Also,$X = \binom{2n+1}{2n+1} + \dots + \binom{2n+1}{n+1}$.
Adding these two expressions: $2X = \sum_{r=0}^{2n+1} \binom{2n+1}{r} = 2^{2n+1}$.
Therefore,$X = \frac{2^{2n+1}}{2} = 2^{2n}$.

(B) For the function $f(x) = ^{9-x}C_{x-1}$ to be defined,the following conditions must be satisfied:
$1$. $9-x \geq 0 \Rightarrow x \leq 9$
$2$. $x-1 \geq 0 \Rightarrow x \geq 1$
$3$. $9-x \geq x-1 \Rightarrow 10 \geq 2x \Rightarrow x \leq 5$
Combining these,we get $1 \leq x \leq 5$. Since $x$ must be an integer for the combination to be defined,$x \in \{1, 2, 3, 4, 5\}$.
Thus,the domain has $m = 5$ elements.
Now,we calculate the values of $f(x)$ for each $x$:
$f(1) = ^{9-1}C_{1-1} = ^8C_0 = 1$
$f(2) = ^{9-2}C_{2-1} = ^7C_1 = 7$
$f(3) = ^{9-3}C_{3-1} = ^6C_2 = 15$
$f(4) = ^{9-4}C_{4-1} = ^5C_3 = 10$
$f(5) = ^{9-5}C_{5-1} = ^4C_4 = 1$
The range is the set of distinct values: $\{1, 7, 15, 10\}$.
Thus,the range has $n = 4$ elements.
Comparing $m=5$ and $n=4$,we see that $m = n + 1$.

(B) We need to select $4$ persons from $2$ ladies $(L)$,$2$ old men $(O)$,and $4$ young men $(Y)$ with the constraints: $L \ge 1$,$O \ge 1$,and $Y \le 2$.
The possible combinations $(L, O, Y)$ that sum to $4$ are:
$1. (1, 1, 2) \implies ^2C_1 \times ^2C_1 \times ^4C_2 = 2 \times 2 \times 6 = 24$
$2. (1, 2, 1) \implies ^2C_1 \times ^2C_2 \times ^4C_1 = 2 \times 1 \times 4 = 8$
$3. (2, 1, 1) \implies ^2C_2 \times ^2C_1 \times ^4C_1 = 1 \times 2 \times 4 = 8$
$4. (2, 2, 0) \implies ^2C_2 \times ^2C_2 \times ^4C_0 = 1 \times 1 \times 1 = 1$
Total number of ways $= 24 + 8 + 8 + 1 = 41$.

(C) Let the marks assigned to the $8$ questions be $x_1, x_2, \dots, x_8$.
We are given that $x_1 + x_2 + \dots + x_8 = 30$,where $x_i \ge 2$ for all $i = 1, 2, \dots, 8$.
Let $x_i = y_i + 2$,where $y_i \ge 0$.
Substituting this into the equation,we get:
$(y_1 + 2) + (y_2 + 2) + \dots + (y_8 + 2) = 30$
$y_1 + y_2 + \dots + y_8 + 16 = 30$
$y_1 + y_2 + \dots + y_8 = 14$.
The number of non-negative integer solutions to this equation 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) The set $A = \{a_1, a_2, \dots, a_{20}\}$ contains $20$ distinct elements.
The total number of $5$-element subsets is given by $^{20}C_5$.
The number of $5$-element subsets containing $a_4$ 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 $^nC_r = \frac{n}{r} \times ^{n-1}C_{r-1}$,we have $^{20}C_5 = \frac{20}{5} \times ^{19}C_4$.
Therefore,$k = \frac{20}{5} = 4$.

(C) Total number of ways to select $2$ girls from $5$ and $3$ boys from $7$ without any restriction is given by $^5C_2 \times ^7C_3 = 10 \times 35 = 350$.
If both specific boys $A$ and $B$ are in the team,we need to select $1$ more boy from the remaining $5$ boys and $2$ girls from $5$ girls. The number of such teams is $^5C_1 \times ^5C_2 = 5 \times 10 = 50$.
Therefore,the number of teams where $A$ and $B$ are not together is $350 - 50 = 300$.

(B) The given expression is the sum of products of combinations:
$^{20}C_r ^{20}C_0 + ^{20}C_{r-1} ^{20}C_1 + ... + ^{20}C_0 ^{20}C_r = ^{40}C_r$
This is based on Vandermonde's Identity,which states that $\sum_{k=0}^{r} {^nC_k} {^mC_{r-k}} = ^{n+m}C_r$.
Here,$n=20$ and $m=20$,so the sum is $^{40}C_r$.
The value of $^{40}C_r$ is maximum when $r = \frac{n+m}{2} = \frac{40}{2} = 20$.

(A) We need to choose $3$ distinct balls from the set $\{1, 2, \dots, 10\}$ such that their labels satisfy $n_1 < n_2 < n_3$.
Since the order is strictly increasing,any selection of $3$ distinct balls from the $10$ available balls can be arranged in exactly one way to satisfy the condition $n_1 < n_2 < n_3$.
Therefore,the number of ways is given by the combination formula $^{10}C_3$.
$^{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.

(B) Given that $^nC_4, ^nC_5,$ and $^nC_6$ 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 both sides by $n!$ and multiplying by $6!(n-4)!$:
$\frac{2 \times 6 \times (n-4)}{5!} = \frac{6 \times 5 \times 4!}{4!} + \frac{(n-4)(n-5)}{6!/6!}$
$\frac{12}{5!(n-5)!} = \frac{1}{4!(n-4)!} + \frac{1}{6!(n-6)!}$
Dividing by $\frac{1}{4!(n-6)!}$:
$\frac{12}{5(n-5)} = \frac{1}{(n-4)(n-5)} + \frac{1}{30}$
Multiplying by $30(n-4)(n-5)$:
$72(n-4) = 30 + (n-4)(n-5)$
$72n - 288 = 30 + n^2 - 9n + 20$
$n^2 - 81n + 338 = 0$
$(n-14)(n-67) = 0$
Since $n \ge 6$ for $^nC_6$ to be defined,$n = 14$ or $n = 67$. Among the options,$n = 14$ is correct.

(C) Total members available are $8$ males and $5$ females,so total persons = $13$. We need to select $11$ members.
For $m$ (at least $6$ males):
Possible cases are ($6$ males,$5$ females),($7$ males,$4$ females),($8$ males,$3$ females).
$m = \binom{8}{6} \times \binom{5}{5} + \binom{8}{7} \times \binom{5}{4} + \binom{8}{8} \times \binom{5}{3} = (28 \times 1) + (8 \times 5) + (1 \times 10) = 28 + 40 + 10 = 78$.
For $n$ (at least $3$ females):
Possible cases are ($8$ males,$3$ females),($7$ males,$4$ females),($6$ males,$5$ females).
$n = \binom{5}{3} \times \binom{8}{8} + \binom{5}{4} \times \binom{8}{7} + \binom{5}{5} \times \binom{8}{6} = (10 \times 1) + (5 \times 8) + (1 \times 28) = 10 + 40 + 28 = 78$.
Thus,$m = n = 78$.

(A) Let the $10$ identical objects be $I$ and the $21$ distinct objects be $D_1, D_2, ..., D_{21}$.
To choose $10$ objects,we can select $k$ objects from the $21$ distinct objects and $(10-k)$ objects from the $10$ identical objects,where $0 \le k \le 10$.
Since the $10$ identical objects are indistinguishable,there is only $1$ way to choose any number of them.
Thus,the number of ways is the sum of the ways to choose $k$ distinct objects for all possible $k$ from $0$ to $10$:
Total ways = $\sum_{k=0}^{10} \binom{21}{k} = \binom{21}{0} + \binom{21}{1} + ... + \binom{21}{10}$.
We know that the sum of binomial coefficients is $\sum_{k=0}^{21} \binom{21}{k} = 2^{21}$.
Since $\binom{21}{k} = \binom{21}{21-k}$,we have $\sum_{k=0}^{10} \binom{21}{k} = \sum_{k=11}^{21} \binom{21}{k}$.
Let $S = \sum_{k=0}^{10} \binom{21}{k}$. Then $S + S = \sum_{k=0}^{21} \binom{21}{k} = 2^{21}$.
Therefore,$2S = 2^{21}$,which implies $S = 2^{20}$.

(C) Given the equation: $6 \cdot ^{35}C_{r} = (k^{2} - 3) \cdot ^{36}C_{r+1}$.
Using the identity $^{n}C_{r} = \frac{n}{r} \cdot ^{n-1}C_{r-1}$ or $^{n+1}C_{r+1} = \frac{n+1}{r+1} \cdot ^{n}C_{r}$,we have $^{36}C_{r+1} = \frac{36}{r+1} \cdot ^{35}C_{r}$.
Substituting this into the equation:
$6 \cdot ^{35}C_{r} = (k^{2} - 3) \cdot \frac{36}{r+1} \cdot ^{35}C_{r}$.
Since $^{35}C_{r} \neq 0$,we can divide both sides by $^{35}C_{r}$:
$6 = (k^{2} - 3) \cdot \frac{36}{r+1}$.
Rearranging for $(k^{2} - 3)$:
$k^{2} - 3 = \frac{6(r+1)}{36} = \frac{r+1}{6}$.
Thus,$k^{2} = \frac{r+1}{6} + 3 = \frac{r+19}{6}$.
Since $k$ is an integer,$k^{2}$ must be a perfect square. Also,$0 \le r \le 35$.
For $r=5$,$k^{2} = \frac{5+19}{6} = 4 \Rightarrow k = \pm 2$.
For $r=35$,$k^{2} = \frac{35+19}{6} = 9 \Rightarrow k = \pm 3$.
These give the ordered pairs $(5, 2), (5, -2), (35, 3), (35, -3)$.
There are $4$ such ordered pairs.

(D) Total number of marbles $= 5 + 4 + 3 = 12$.
We need to draw $4$ marbles from $12$ marbles.
The total number of ways to draw $4$ marbles is $^{12}C_{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495$.
We want to find the number of ways such that at most $3$ marbles are red.
This is equivalent to: (Total ways) - (Ways where all $4$ marbles are red).
The number of ways to choose $4$ red marbles from $5$ red marbles is $^{5}C_{4} = 5$.
Therefore,the number of ways to draw $4$ marbles such that at most $3$ are red is $495 - 5 = 490$.

(A) We know that $n! = n \times (n-1) \times (n-2) \times \dots \times 1$.
Thus,$\frac{12!}{10! \times 2!} = \frac{12 \times 11 \times 10!}{10! \times (2 \times 1)}$.
Canceling $10!$ from the numerator and denominator,we get $\frac{12 \times 11}{2}$.
This simplifies to $6 \times 11 = 66$.

(A) Given the expression $\frac{n!}{r!(n-r)!}$ with $n=5$ and $r=2$.
Substituting the values,we get $\frac{5!}{2!(5-2)!}$.
This simplifies to $\frac{5!}{2! \times 3!}$.
Expanding the factorials,we have $\frac{5 \times 4 \times 3!}{2 \times 1 \times 3!}$.
Canceling $3!$ from the numerator and denominator,we get $\frac{5 \times 4}{2 \times 1} = \frac{20}{2} = 10$.

(A) We know that $n! = n \times (n-1) \times (n-2) \times \dots \times 1$.
Thus,$\frac{8!}{6! \times 2!} = \frac{8 \times 7 \times 6!}{6! \times (2 \times 1)}$.
Canceling $6!$ from the numerator and denominator,we get $\frac{8 \times 7}{2}$.
This simplifies to $\frac{56}{2} = 28$.

(A) We know the property of combinations: if $^{n}C_{r} = ^{n}C_{k}$,then either $r = k$ or $n = r + k$.
Given $^{n}C_{9} = ^{n}C_{8}$,since $9 \neq 8$,we must have $n = 9 + 8 = 17$.
Now,we need to find $^{n}C_{17}$,which is $^{17}C_{17}$.
Using the formula $^{n}C_{n} = 1$,we get $^{17}C_{17} = 1$.

(A) The total number of persons is $2 + 3 = 5$. We need to select $3$ persons out of $5$. Since the order does not matter,we use combinations.
Total ways $= ^{5}C_{3} = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10$.
To form a committee with $1$ man and $2$ women:
We select $1$ man from $2$ men in $^{2}C_{1}$ ways.
We select $2$ women from $3$ women in $^{3}C_{2}$ ways.
Number of ways $= ^{2}C_{1} \times ^{3}C_{2} = 2 \times 3 = 6$.

(A) The number of ways to choose $4$ cards from $52$ cards is given by the combination formula $^{52}C_{4}$.
$^{52}C_{4} = \frac{52!}{4!48!} = \frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1} = 270725$.
To choose $4$ cards such that each belongs to a different suit,we must select $1$ card from each of the $4$ suits (diamonds,hearts,clubs,spades).
There are $13$ cards in each suit. The number of ways to choose $1$ card from each suit is $^{13}C_{1} \times ^{13}C_{1} \times ^{13}C_{1} \times ^{13}C_{1} = 13 \times 13 \times 13 \times 13 = 13^{4} = 28561$.
Thus,the total ways are $270725$ and the ways to choose $4$ cards from different suits are $13^{4}$.

(B) The number of ways to choose $4$ cards from $52$ cards is given by the combination formula $^{n}C_{r} = \frac{n!}{r!(n-r)!}$.
For choosing $4$ cards from $52$,we have $^{52}C_{4} = \frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1} = 270725$.
There are $12$ face cards in a deck (Jack,Queen,King of each of the $4$ suits).
The number of ways to select $4$ face cards out of $12$ is $^{12}C_{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495$.

(A) The total number of ways to choose $4$ cards from $52$ is given by the combination formula $^{n}C_{r} = \frac{n!}{r!(n-r)!}$.
Total ways $= ^{52}C_{4} = \frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1} = 270725$.
There are $26$ red cards and $26$ black cards in a deck.
To choose $2$ red cards from $26$ and $2$ black cards from $26$,the number of ways is $^{26}C_{2} \times ^{26}C_{2}$.
$^{26}C_{2} = \frac{26 \times 25}{2 \times 1} = 325$.
So,the required number of ways $= 325 \times 325 = 105625$.

(A) The total number of ways to choose $4$ cards from $52$ is given by the combination formula $^{n}C_{r} = \frac{n!}{r!(n-r)!}$.
Total ways $= ^{52}C_{4} = \frac{52 \times 51 \times 50 \times 49}{4 \times 3 \times 2 \times 1} = 270725$.
For the cards to be of the same colour,they must be either all red or all black.
There are $26$ red cards and $26$ black cards.
Ways to choose $4$ red cards $= ^{26}C_{4} = \frac{26 \times 25 \times 24 \times 23}{4 \times 3 \times 2 \times 1} = 14950$.
Ways to choose $4$ black cards $= ^{26}C_{4} = 14950$.
Total ways for same colour $= 14950 + 14950 = 29900$.

(A) We know that if $^{n}C_{a} = ^{n}C_{b}$,then either $a = b$ or $n = a + b$.
Given $^{n}C_{8} = ^{n}C_{2}$,since $8 \neq 2$,we must have $n = 8 + 2 = 10$.
Now,we need to find $^{n}C_{2} = ^{10}C_{2}$.
$^{10}C_{2} = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45$.

(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:
$\frac{(2n)!}{3!(2n-3)!} \times \frac{3!(n-3)!}{n!} = 12$.
Simplifying the factorials:
$\frac{(2n)(2n-1)(2n-2)(2n-3)!}{(2n-3)!} \times \frac{(n-3)!}{n(n-1)(n-2)(n-3)!} = 12$.
$\frac{(2n)(2n-1)(2(n-1))}{n(n-1)(n-2)} = 12$.
$\frac{2(2n-1)(2)}{n-2} = 12$.
$\frac{4(2n-1)}{n-2} = 12$.
$\frac{2n-1}{n-2} = 3$.
$2n - 1 = 3(n - 2)$.
$2n - 1 = 3n - 6$.
$n = 5$.

(B) Given $\frac{^{2n}C_{3}}{^{n}C_{3}} = \frac{11}{1}$.
Using the formula $^{n}C_{r} = \frac{n!}{r!(n-r)!}$,we have:
$\frac{(2n)!}{3!(2n-3)!} \times \frac{3!(n-3)!}{n!} = 11$
$\Rightarrow \frac{(2n)(2n-1)(2n-2)(2n-3)!}{(2n-3)!} \times \frac{(n-3)!}{n(n-1)(n-2)(n-3)!} = 11$
$\Rightarrow \frac{(2n)(2n-1) \cdot 2(n-1)}{n(n-1)(n-2)} = 11$
$\Rightarrow \frac{2(2n-1) \cdot 2}{n-2} = 11$
$\Rightarrow \frac{4(2n-1)}{n-2} = 11$
$\Rightarrow 8n - 4 = 11n - 22$
$\Rightarrow 3n = 18$
$\Rightarrow n = 6$.

(A) team of $3$ boys and $3$ girls is to be selected from $5$ boys and $4$ girls.
$3$ boys can be selected from $5$ boys in $^{5}C_{3}$ ways.
$3$ girls can be selected from $4$ girls in $^{4}C_{3}$ ways.
Therefore,by the multiplication principle,the number of ways in which a team of $3$ boys and $3$ girls can be selected is given by:
$^{5}C_{3} \times ^{4}C_{3} = \frac{5!}{3!2!} \times \frac{4!}{3!1!}$
$= \frac{5 \times 4}{2 \times 1} \times \frac{4}{1}$
$= 10 \times 4 = 40$ ways.

(A) We need to select $9$ balls such that $3$ balls are of each colour (red,white,and blue).
Number of ways to select $3$ red balls from $6$ is $^{6}C_{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
Number of ways to select $3$ white balls from $5$ is $^{5}C_{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$.
Number of ways to select $3$ blue balls from $5$ is $^{5}C_{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10$.
By the multiplication principle,the total number of ways is $20 \times 10 \times 10 = 2000$.

(A) In a deck of $52$ cards,there are $4$ aces and $48$ non-ace cards.
We need to select $5$ cards such that there is exactly $1$ ace.
First,select $1$ ace from $4$ available aces in $^{4}C_{1}$ ways.
Next,select the remaining $4$ cards from the $48$ non-ace cards in $^{48}C_{4}$ ways.
Using the multiplication principle,the total number of combinations is $^{4}C_{1} \times ^{48}C_{4}$.
$^{4}C_{1} = 4$.
$^{48}C_{4} = \frac{48 \times 47 \times 46 \times 45}{4 \times 3 \times 2 \times 1} = 194580$.
Total combinations $= 4 \times 194580 = 778320$.

(A) Total players = $17$. Number of bowlers = $5$. Number of non-bowlers = $17 - 5 = 12$.
We need to select a team of $11$ players including exactly $4$ bowlers.
Number of ways to select $4$ bowlers from $5$ is $^{5}C_{4} = \frac{5!}{4!1!} = 5$.
Number of ways to select the remaining $11 - 4 = 7$ players from $12$ non-bowlers is $^{12}C_{7} = \frac{12!}{7!5!} = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = 792$.
Total number of ways = $^{5}C_{4} \times ^{12}C_{7} = 5 \times 792 = 3960$.

(A) There are $5$ black and $6$ red balls in the bag.
$2$ black balls can be selected out of $5$ black balls in $^{5}C_{2}$ ways.
$3$ red balls can be selected out of $6$ red balls in $^{6}C_{3}$ ways.
Thus,by the multiplication principle,the required number of ways of selecting $2$ black and $3$ red balls is:
$= ^{5}C_{2} \times ^{6}C_{3} = \frac{5!}{2!3!} \times \frac{6!}{3!3!} = \frac{5 \times 4}{2 \times 1} \times \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 10 \times 20 = 200$.

(A) There are $9$ courses available,out of which $2$ specific courses are compulsory for every student.
Since $2$ courses are already selected,the student needs to choose $5 - 2 = 3$ more courses.
The number of remaining courses to choose from is $9 - 2 = 7$.
Therefore,the number of ways to choose the remaining courses is given by the combination formula $^{n}C_{r} = \frac{n!}{r!(n-r)!}$.
Number of ways $= ^{7}C_{3} = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.

(A) Since the team will not include any girl,only boys are to be selected.
We need to select $5$ boys out of $7$ boys.
The number of ways to select $5$ boys out of $7$ is given by the combination formula $^{n}C_{r} = \frac{n!}{r!(n-r)!}$.
Therefore,the number of ways $= ^{7}C_{5} = ^{7}C_{7-5} = ^{7}C_{2}$.
$^{7}C_{2} = \frac{7 \times 6}{2 \times 1} = 21$.
Thus,there are $21$ ways to select the team.

(A) The total number of members to be selected is $5$. The team must contain at least one boy and one girl.
The possible compositions of the team are:
$(a)$ $1$ boy and $4$ girls: $^{7}C_{1} \times ^{4}C_{4} = 7 \times 1 = 7$ ways.
$(b)$ $2$ boys and $3$ girls: $^{7}C_{2} \times ^{4}C_{3} = 21 \times 4 = 84$ ways.
$(c)$ $3$ boys and $2$ girls: $^{7}C_{3} \times ^{4}C_{2} = 35 \times 6 = 210$ ways.
$(d)$ $4$ boys and $1$ girl: $^{7}C_{4} \times ^{4}C_{1} = 35 \times 4 = 140$ ways.
Total number of ways = $7 + 84 + 210 + 140 = 441$.