Mix Examples-Permutation and Combination Questions in English

(D) To find the number of positive integral solutions of $xyz = 60$,we first find the prime factorization of $60$.
$60 = 2^2 \times 3^1 \times 5^1$.
We need to distribute the factors $2^2, 3^1, 5^1$ among $x, y, z$.
For the factor $2^2$,the number of ways to distribute it among $3$ variables is given by the stars and bars formula $\binom{n+k-1}{k-1}$,where $n=2$ and $k=3$.
Number of ways for $2^2 = \binom{2+3-1}{3-1} = \binom{4}{2} = 6$.
For the factor $3^1$,the number of ways is $\binom{1+3-1}{3-1} = \binom{3}{2} = 3$.
For the factor $5^1$,the number of ways is $\binom{1+3-1}{3-1} = \binom{3}{2} = 3$.
Total number of solutions = $6 \times 3 \times 3 = 54$.

(C) $(i).$ The number of ways of placing $n$ distinct objects into $k$ distinct bins such that no bin is empty is given by the number of positive integral solutions to $x_1 + x_2 + \ldots + x_k = n$,which is ${}^{n-1}C_{k-1}$. This statement is true.
$(ii).$ The number of ways of writing a positive integer $n$ as a sum of $k$ positive integers is equivalent to the number of positive integral solutions to $x_1 + x_2 + \ldots + x_k = n$,which is ${}^{n-1}C_{k-1}$. This statement is true.
$(iii).$ The number of ways of placing $n$ distinct objects in $k$ distinct bins such that at least one bin is non-empty is $k^n - 1$ (if bins are distinct) or involves inclusion-exclusion. The formula ${}^{n-1}C_{k-1}$ is incorrect for this case. This statement is false.
$(iv).$ Using Pascal's Identity,${}^nC_k = {}^{n-1}C_k + {}^{n-1}C_{k-1}$. Rearranging gives ${}^nC_k - {}^{n-1}C_k = {}^{n-1}C_{k-1}$. This statement is true.
Therefore,all statements except $(iii)$ are correct.

(A) number is divisible by $6$ if it is divisible by both $2$ and $3$.
For a number to be divisible by $2$,the units digit must be $2, 4,$ or $6$.
For a number to be divisible by $3$,the sum of its digits must be divisible by $3$.
Let the $4$ digits be $d_1, d_2, d_3, d_4$ where $d_4 \in \{2, 4, 6\}$.
Case $I$: Units digit is $2$. The sum of the remaining $3$ digits must be $3k - 2$. Possible sets of $3$ digits from $\{1, 3, 4, 5, 6\}$ are $\{1, 4, 6\}, \{1, 5, 6\}, \{3, 4, 5\}$. Each set can be arranged in $3! = 6$ ways. Total $= 3 \times 6 = 18$.
Case $II$: Units digit is $4$. The sum of the remaining $3$ digits must be $3k - 4$. Possible sets of $3$ digits from $\{1, 2, 3, 5, 6\}$ are $\{1, 2, 6\}, \{1, 5, 6\}, \{2, 3, 6\}$. Each set can be arranged in $3! = 6$ ways. Total $= 3 \times 6 = 18$.
Case $III$: Units digit is $6$. The sum of the remaining $3$ digits must be $3k - 6$. Possible sets of $3$ digits from $\{1, 2, 3, 4, 5\}$ are $\{1, 2, 3\}, \{1, 3, 5\}, \{2, 3, 4\}, \{3, 4, 5\}$. Each set can be arranged in $3! = 6$ ways. Total $= 4 \times 6 = 24$.
Total numbers $= 18 + 18 + 24 = 60$.

(A) Given $a, b, c \in \mathbb{N}$ (positive integers) such that $a+b+c=5$. The possible triplets $(a, b, c)$ are $(1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1), (1, 2, 2), (2, 1, 2)$.
Calculating the values of $2^a 3^b 5^c$ for each triplet:
$(1, 1, 3) \implies 2^1 \cdot 3^1 \cdot 5^3 = 6 \cdot 125 = 750$
$(1, 3, 1) \implies 2^1 \cdot 3^3 \cdot 5^1 = 2 \cdot 27 \cdot 5 = 270$
$(3, 1, 1) \implies 2^3 \cdot 3^1 \cdot 5^1 = 8 \cdot 3 \cdot 5 = 120$
$(2, 2, 1) \implies 2^2 \cdot 3^2 \cdot 5^1 = 4 \cdot 9 \cdot 5 = 180$
$(1, 2, 2) \implies 2^1 \cdot 3^2 \cdot 5^2 = 2 \cdot 9 \cdot 25 = 450$
$(2, 1, 2) \implies 2^2 \cdot 3^1 \cdot 5^2 = 4 \cdot 3 \cdot 25 = 300$
The greatest value $M = 750$ and the least value $L = 120$.
Then $M - L = 750 - 120 = 630$.
Prime factorization of $630 = 2 \cdot 3^2 \cdot 5 \cdot 7$.
Thus,option $A$ is correct.

(B) Given,$n = 1! + 4! + 7! + \ldots + 400!$.
We calculate the values of the factorials:
$1! = 1$
$4! = 24$
$7! = 5040$
$10! = 3628800$
For any $k \ge 10$,$k!$ ends in at least two zeros,so the last two digits of $k!$ are $00$.
Thus,the last two digits of $n$ are the same as the last two digits of $1! + 4! + 7! + 10! + \ldots$.
$1! + 4! + 7! = 1 + 24 + 5040 = 5065$.
Since all subsequent terms $10!, 13!, \ldots$ end in $00$,the sum $n$ ends in $65$.
Therefore,the ten's digit of $n$ is $6$.

(B) Given expression: ${ }^{34} C_5+\sum_{r=0}^4{ }^{(38-r)} C_4$
Expanding the summation: ${ }^{34} C_5+{ }^{38} C_4+{ }^{37} C_4+{ }^{36} C_4+{ }^{35} C_4+{ }^{34} C_4$
Using the identity ${ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r$:
${ }^{34} C_5+{ }^{34} C_4 = { }^{35} C_5$
Now,${ }^{35} C_5+{ }^{35} C_4 = { }^{36} C_5$
Then,${ }^{36} C_5+{ }^{36} C_4 = { }^{37} C_5$
Then,${ }^{37} C_5+{ }^{37} C_4 = { }^{38} C_5$
Finally,${ }^{38} C_5+{ }^{38} C_4 = { }^{39} C_5$
Thus,the result is ${ }^{39} C_5$.

(D) Given $x={ }^{16} C_5+{ }^{12} C_4, y=\sum_{r=1}^3{ }^{(20-r)} C_4, z=\sum_{k=1}^4{ }^{(16-k)} C_3$.
Then $x+y+z={ }^{16} C_5+{ }^{12} C_4+\left({ }^{19} C_4+{ }^{18} C_4+{ }^{17} C_4\right)+\left({ }^{15} C_3+{ }^{14} C_3+{ }^{13} C_3+{ }^{12} C_3\right)$.
Rearranging the terms:
$x+y+z={ }^{12} C_3+{ }^{12} C_4+{ }^{13} C_3+{ }^{14} C_3+{ }^{15} C_3+{ }^{16} C_5+{ }^{17} C_4+{ }^{18} C_4+{ }^{19} C_4$.
Using the identity ${ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r$:
$x+y+z={ }^{13} C_4+{ }^{13} C_3+{ }^{14} C_3+{ }^{15} C_3+{ }^{16} C_5+{ }^{17} C_4+{ }^{18} C_4+{ }^{19} C_4$.
Continuing the application of the identity:
$x+y+z={ }^{14} C_4+{ }^{14} C_3+{ }^{15} C_3+{ }^{16} C_5+{ }^{17} C_4+{ }^{18} C_4+{ }^{19} C_4$.
$x+y+z={ }^{15} C_4+{ }^{15} C_3+{ }^{16} C_5+{ }^{17} C_4+{ }^{18} C_4+{ }^{19} C_4$.
$x+y+z={ }^{16} C_4+{ }^{16} C_5+{ }^{17} C_4+{ }^{18} C_4+{ }^{19} C_4$.
$x+y+z={ }^{17} C_5+{ }^{17} C_4+{ }^{18} C_4+{ }^{19} C_4$.
$x+y+z={ }^{18} C_5+{ }^{18} C_4+{ }^{19} C_4$.
$x+y+z={ }^{19} C_5+{ }^{19} C_4={ }^{20} C_5$.
Calculating the value:
${ }^{20} C_5 = \frac{20 \times 19 \times 18 \times 17 \times 16}{5 \times 4 \times 3 \times 2 \times 1} = 19 \times 3 \times 17 \times 16 = 19 \times 17 \times 48$.

(B) For statement $I$: The number of positive integral solutions of $x_1+x_2+x_3+x_4=n$ is given by the formula $\binom{n-1}{r-1}$.
Here,$n=10$ and $r=4$,so the number of solutions is $\binom{10-1}{4-1} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.
Since $84 \neq 286$,statement $I$ is false.
For statement $II$: The exponent of a prime $p$ in $m!$ is given by Legendre's formula $\sum_{k=1}^{\infty} [\frac{m}{p^k}]$.
The exponent of $5$ in $25!$ is $[\frac{25}{5}] + [\frac{25}{25}] = 5 + 1 = 6$.
The exponent of $2$ in $25!$ is $[\frac{25}{2}] + [\frac{25}{4}] + [\frac{25}{8}] + [\frac{25}{16}] = 12 + 6 + 3 + 1 = 22$.
Since the exponent of $5$ is $6$ and the exponent of $2$ is $22$,the highest power of $10$ that divides $25!$ is $10^6$.
Thus,$25! = 10^6 \times k$,where $k$ is an integer not divisible by $10$.
Therefore,statement $II$ is true.

(A) We need to find the unit digit of the sum $S = 1! + 2! + 3! + 4! + 5! + \ldots + 99!$.
First,calculate the factorials:
$1! = 1$
$2! = 2 \times 1 = 2$
$3! = 3 \times 2 \times 1 = 6$
$4! = 4 \times 3 \times 2 \times 1 = 24$
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$
For any $n \ge 5$,$n!$ contains the factors $5$ and $2$,so $n!$ ends in $0$.
Thus,for all $n \ge 5$,the unit digit of $n!$ is $0$.
The sum is $S = 1! + 2! + 3! + 4! + (5! + 6! + \ldots + 99!)$.
The unit digit of $S$ is the unit digit of $(1! + 2! + 3! + 4!) + (0 + 0 + \ldots + 0)$.
$1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33$.
The unit digit of $33$ is $3$.

(C) The total number of possible combinations for the three rings is $15^3 = 3375$.
There is only $1$ correct combination to open the lock.
Therefore,the number of unsuccessful attempts $N$ is given by $N = 15^3 - 1$.
Using the algebraic identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$,we have $N = (15 - 1)(15^2 + 15 \times 1 + 1^2) = 14 \times (225 + 15 + 1) = 14 \times 241$.
Since $14 = 2 \times 7$,we get $N = 2 \times 7 \times 241$.
Here,$2$,$7$,and $241$ are all prime numbers.
Thus,$N$ is the product of three distinct prime numbers.

(A) The value of $17!$ is $355687428096000$.
Comparing this with the given expression $3556xy428096000$,we identify the digits $x = 8$ and $y = 7$.
Therefore,the sum $x + y = 8 + 7 = 15$.

(A) Let $|P| = k$. Then $Q$ must have $k+1$ elements.
For a fixed $k$,the number of ways to choose $P$ is $^{n}C_k$.
The number of ways to choose $Q$ is $^{n}C_{k+1}$.
Since $P$ and $Q$ are chosen independently,the total number of ways is the sum over all possible values of $k$ from $0$ to $n-1$:
$\sum_{k=0}^{n-1} (^{n}C_k \cdot ^{n}C_{k+1})$
Using the identity $^{n}C_k = ^{n}C_{n-k}$,we have:
$\sum_{k=0}^{n-1} (^{n}C_{n-k} \cdot ^{n}C_{k+1})$
This is the coefficient of $x^{n+1}$ in the expansion of $(1+x)^n \cdot (1+x)^n = (1+x)^{2n}$,which is $^{2n}C_{n+1}$.
Note that $^{2n}C_{n+1} = ^{2n}C_{2n-(n+1)} = ^{2n}C_{n-1}$.

(B) The number of ways to select $3$ consonants out of $7$ is ${}^{7}C_{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.
The number of ways to select $2$ vowels out of $4$ is ${}^{4}C_{2} = \frac{4 \times 3}{2 \times 1} = 6$.
The total number of letters selected is $3 + 2 = 5$.
These $5$ letters can be arranged among themselves in $5!$ ways,where $5! = 120$.
Therefore,the total number of words formed is ${}^{7}C_{3} \times {}^{4}C_{2} \times 5! = 35 \times 6 \times 120 = 25200$.

(D) The total number of ways to arrange $n$ distinct numbers is $n!$.
To ensure the digits $1, 2, 3, \ldots, k$ appear as a single block in that specific order,we treat this block as a single entity.
Now,we have $(n - k)$ remaining digits plus the one block,totaling $(n - k + 1)$ entities to arrange.
The number of ways to arrange these $(n - k + 1)$ entities is $(n - k + 1)!$.
Therefore,the probability of the required event is $\frac{(n - k + 1)!}{n!}$.

(C) To arrange $n$ white and $n$ black balls such that no two balls of the same colour are adjacent,the balls must alternate in colour.
There are two possible patterns for the arrangement:
$1$. $B, W, B, W, \ldots, B, W$ (starting with Black)
$2$. $W, B, W, B, \ldots, W, B$ (starting with White)
For each pattern,the $n$ black balls can be arranged in $n!$ ways and the $n$ white balls can be arranged in $n!$ ways.
Thus,the number of ways for the first pattern is $n! \times n! = (n!)^2$.
Similarly,the number of ways for the second pattern is $n! \times n! = (n!)^2$.
Total number of ways = $(n!)^2 + (n!)^2 = 2(n!)^2$.

(A) The total number of ways to select $4$ numbers from $20$ is given by $^{20}C_{4}$.
The number of ways to select $4$ consecutive numbers from $20$ is $17$ (these are $(1,2,3,4), (2,3,4,5), \ldots, (17,18,19,20)$).
The number of ways to select $4$ non-consecutive numbers is the total number of selections minus the number of consecutive selections.
$\text{Required selections} = {}^{20}C_{4} - 17$
$\text{Required selections} = \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} - 17$
$\text{Required selections} = 4845 - 17 = 4828$
Calculating the options: $284 \times 17 = 4828$.
Thus,the correct option is $A$.

(D) Let the number of balls be $n = 5$ and the number of boxes be $k = 3$.
Since the balls are distinct and the boxes are distinct,we use the Principle of Inclusion-Exclusion to ensure no box is empty.
The total number of ways to place $5$ distinct balls into $3$ distinct boxes is $3^5 = 243$.
Let $S$ be the set of all possible distributions. Let $A_i$ be the property that box $i$ is empty.
We want to find the number of ways such that no box is empty,which is $|S| - |A_1 \cup A_2 \cup A_3|$.
By the Principle of Inclusion-Exclusion:
$|A_1 \cup A_2 \cup A_3| = \sum |A_i| - \sum |A_i \cap A_j| + |A_1 \cap A_2 \cap A_3|$.
$|A_i| = 2^5 = 32$. There are $\binom{3}{1} = 3$ such cases.
$|A_i \cap A_j| = 1^5 = 1$. There are $\binom{3}{2} = 3$ such cases.
$|A_1 \cap A_2 \cap A_3| = 0^5 = 0$.
Number of ways = $3^5 - \binom{3}{1} 2^5 + \binom{3}{2} 1^5 = 243 - 3(32) + 3(1) = 243 - 96 + 3 = 150$.

(D) The total number of ways to distribute $n$ distinct objects among $n$ distinct persons is $n^n$.
Each person can receive any number of objects.
The number of ways in which each person gets exactly one object is given by the number of permutations of $n$ objects taken $n$ at a time,which is $n!$.
If each person gets exactly one object,then no person is left without an object.
Therefore,the number of ways in which at least one person does not get any object is the total number of ways minus the number of ways where everyone gets exactly one object.
Required number of ways $= n^n - n!$.

(C) We use the identity ${ }^{n} C_{r} + { }^{n} C_{r-1} = { }^{n+1} C_{r}$.
Expanding the summation,the expression is:
${ }^{47} C_4 + { }^{51} C_3 + { }^{50} C_3 + { }^{49} C_3 + { }^{48} C_3 + { }^{47} C_3$
Using the identity ${ }^{47} C_4 + { }^{47} C_3 = { }^{48} C_4$,the expression becomes:
${ }^{48} C_4 + { }^{48} C_3 + { }^{49} C_3 + { }^{50} C_3 + { }^{51} C_3$
Applying the identity repeatedly:
${ }^{48} C_4 + { }^{48} C_3 = { }^{49} C_4$
${ }^{49} C_4 + { }^{49} C_3 = { }^{50} C_4$
${ }^{50} C_4 + { }^{50} C_3 = { }^{51} C_4$
${ }^{51} C_4 + { }^{51} C_3 = { }^{52} C_4$
Thus,the final value is ${ }^{52} C_4$.

(C) We know the property of binomial coefficients: $\frac{{}^n C_r}{{}^n C_{r-1}} = \frac{n-r+1}{r}$.
Using this,we have:
$1) \frac{{}^n C_r}{{}^n C_{r-1}} = \frac{84}{36} = \frac{7}{3}$ $\Rightarrow \frac{n-r+1}{r} = \frac{7}{3}$ $\Rightarrow 3n - 3r + 3 = 7r$ $\Rightarrow 3n - 10r = -3$ (Eq. $1$)
$2) \frac{{}^n C_{r+1}}{{}^n C_r} = \frac{126}{84} = \frac{3}{2}$ $\Rightarrow \frac{n-(r+1)+1}{r+1} = \frac{3}{2}$ $\Rightarrow \frac{n-r}{r+1} = \frac{3}{2}$ $\Rightarrow 2n - 2r = 3r + 3$ $\Rightarrow 2n - 5r = 3$ (Eq. $2$)
Multiplying Eq. $2$ by $2$,we get $4n - 10r = 6$ (Eq. $3$).
Subtracting Eq. $1$ from Eq. $3$: $(4n - 10r) - (3n - 10r) = 6 - (-3) \Rightarrow n = 9$.
Substituting $n=9$ into Eq. $2$: $2(9) - 5r = 3$ $\Rightarrow 18 - 5r = 3$ $\Rightarrow 5r = 15$ $\Rightarrow r = 3$.
Thus,${}^n C_8 = {}^9 C_8 = {}^9 C_{9-8} = {}^9 C_1 = 9$.

(A) Let $S = 1! + 2! + 3! + 4! + \ldots + 11!$.
We know that for any $n \ge 4$,$n!$ contains $4 \times 3 = 12$ as a factor.
Therefore,$4!, 5!, \ldots, 11!$ are all divisible by $12$.
Thus,the remainder of $S$ when divided by $12$ is the same as the remainder of $(1! + 2! + 3!)$ when divided by $12$.
$1! + 2! + 3! = 1 + 2 + 6 = 9$.
Since $9 < 12$,the remainder is $9$.

(B) We need to find the remainder of $(1! + 2! + 95!) \pmod{15}$.
First,calculate the factorials:
$1! = 1$
$2! = 2$
$3! = 6$
$4! = 24$
$5! = 120$
Since $5! = 120$ and $120 = 15 \times 8$,$5!$ is divisible by $15$.
Consequently,all factorials $n!$ for $n \geq 5$ are divisible by $15$.
Thus,$95! \equiv 0 \pmod{15}$.
The expression becomes $1! + 2! + 95! \equiv 1 + 2 + 0 \pmod{15} = 3 \pmod{15}$.
Therefore,the remainder is $3$.

(C) The set $S$ contains $9$ distinct digits.
For $x$: We choose $1$ digit to be repeated twice,which can be done in ${}^{9}C_{1}$ ways. The remaining $7$ digits are chosen from the remaining $8$ digits in ${}^{8}C_{7}$ ways. The total number of arrangements is $\frac{9!}{2!}$. Thus,$x = {}^{9}C_{1} \times {}^{8}C_{7} \times \frac{9!}{2!} = 9 \times 8 \times \frac{9!}{2} = 36 \times 9!$.
For $y$: We choose $2$ digits to be repeated twice,which can be done in ${}^{9}C_{2}$ ways. The remaining $5$ digits are chosen from the remaining $7$ digits in ${}^{7}C_{5}$ ways. The total number of arrangements is $\frac{9!}{2! \times 2!}$. Thus,$y = {}^{9}C_{2} \times {}^{7}C_{5} \times \frac{9!}{2! \times 2!} = \frac{9 \times 8}{2} \times \frac{7 \times 6}{2} \times \frac{9!}{4} = 36 \times 21 \times \frac{9!}{4} = 189 \times 9!$.
Calculating the ratio: $\frac{x}{y} = \frac{36 \times 9!}{189 \times 9!} = \frac{36}{189} = \frac{4}{21}$.
Therefore,$21x = 4y$.

(B) Let the $4$-digit number be $d_1 d_2 d_3 d_4$. Since the number is between $5000$ and $9000$,the first digit $d_1$ can be $5$ or $9$.
Case $1$: $d_1 = 5$.
The sum of digits $S = 5 + d_2 + d_3 + d_4$ must be divisible by $3$.
Possible values for $d_2, d_3, d_4 \in \{0, 1, 2, 5, 9\}$.
There are $5 \times 5 = 25$ possible combinations for $(d_2, d_3)$. For each pair,$d_4$ is determined such that $5 + d_2 + d_3 + d_4 \equiv 0 \pmod{3}$.
If $5 + d_2 + d_3 \equiv 0 \pmod{3}$,then $d_4 \in \{0, 9\}$ ($2$ choices).
If $5 + d_2 + d_3 \equiv 1 \pmod{3}$,then $d_4 \in \{2, 5\}$ ($2$ choices).
If $5 + d_2 + d_3 \equiv 2 \pmod{3}$,then $d_4 \in \{1\}$ ($1$ choice).
Counting the pairs $(d_2, d_3)$ by their sum modulo $3$:
Sum $\equiv 0$: $(0,0), (0,9), (1,2), (1,5), (2,1), (2,4) \text{ (not in set)}, (5,1), (5,4) \dots$
Actually,checking all $25$ pairs:
Sum $\equiv 0 \pmod{3}$: $9$ pairs. $9 \times 2 = 18$ numbers.
Sum $\equiv 1 \pmod{3}$: $8$ pairs. $8 \times 2 = 16$ numbers.
Sum $\equiv 2 \pmod{3}$: $8$ pairs. $8 \times 1 = 8$ numbers.
Total for $d_1=5$ is $18 + 16 + 8 = 42$.
Wait,the condition is $d_1 < 9$. So $d_1$ can only be $5$.
Total numbers = $42$.

(A) To find the largest $n$ such that $7^n$ divides $101!$,we use Legendre's Formula for the exponent of a prime $p$ in $m!$,which is given by $E_p(m!) = \sum_{k=1}^{\infty} [\frac{m}{p^k}]$.
Here,$m = 101$ and $p = 7$.
$E_7(101!) = [\frac{101}{7}] + [\frac{101}{7^2}] + [\frac{101}{7^3}] + \dots$
$E_7(101!) = [14.428] + [2.061] + [0.294] + \dots$
$E_7(101!) = 14 + 2 + 0 = 16$.
Thus,the largest $n$ is $16$.

(B) For $n(S)$,we need to choose $4$ distinct digits from the set ${0, 1, 2, 3, 4, 5, 6, 7, 8, 9}$. Since the condition $a > b > c > d$ is given,once $4$ digits are chosen,there is only $1$ way to arrange them in descending order. Thus,$n(S) = {}^{10}C_4 = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210$.
For $n(P)$,we need $5$-digit numbers whose product of digits is $20$. The prime factorization of $20$ is $2^2 \times 5$. The possible sets of $5$ digits are:
$1)$ ${5, 4, 1, 1, 1}$: The number of arrangements is $\frac{5!}{3!} = 20$.
$2)$ ${5, 2, 2, 1, 1}$: The number of arrangements is $\frac{5!}{2!2!} = 30$.
Thus,$n(P) = 20 + 30 = 50$.
Therefore,$n(S) + n(P) = 210 + 50 = 260$.

(C) We have $5$ distinct digits to fill $7$ positions,and each digit must appear at least once.
This implies that two digits must be repeated.
The possible partitions of $7$ into $5$ parts are $(3, 1, 1, 1, 1)$ and $(2, 2, 1, 1, 1)$.
Case $1$: Partition $(3, 1, 1, 1, 1)$
First,choose the digit that appears $3$ times: $\binom{5}{1} = 5$ ways.
Then,arrange these $7$ digits: $\frac{7!}{3!1!1!1!1!} = \frac{5040}{6} = 840$ ways.
Total for Case $1 = 5 \times 840 = 4200$.
Case $2$: Partition $(2, 2, 1, 1, 1)$
First,choose the $2$ digits that appear twice each: $\binom{5}{2} = 10$ ways.
Then,arrange these $7$ digits: $\frac{7!}{2!2!1!1!1!} = \frac{5040}{4} = 1260$ ways.
Total for Case $2 = 10 \times 1260 = 12600$.
Grand Total = $4200 + 12600 = 16800$.

(C) The given equation is $a + b + 2c = 22$,where $a, b, c \ge 0$.
We can rewrite this as $a + b = 22 - 2c$.
Since $a, b \ge 0$,we must have $22 - 2c \ge 0$,which implies $0 \le c \le 11$.
For a fixed $c$,the number of non-negative integer solutions to $a + b = 22 - 2c$ is given by the formula $(n+r-1)C(r-1)$,which simplifies to $(22 - 2c + 1) = 23 - 2c$.
Thus,the total number of solutions is $n(A) = \sum_{c=0}^{11} (23 - 2c)$.
This is an arithmetic progression: $n(A) = 23 + 21 + 19 + \dots + 1$.
The number of terms is $12$.
The sum is $\frac{12}{2} \times (23 + 1) = 6 \times 24 = 144$.

(A) Let $b, y, r$ be the number of blue,yellow,and red balls selected,respectively. We need to find the number of ways to select $8$ balls such that $b+y+r=8$ with $b \geq 2, y \geq 2, r \geq 2$.
Let $b=2+b', y=2+y', r=2+r'$,where $b', y', r' \geq 0$.
Substituting these into the sum: $(2+b') + (2+y') + (2+r') = 8 \implies b'+y'+r' = 2$.
Given the constraints $b \leq 5, y \leq 6, r \leq 4$,we have $b' \leq 3, y' \leq 4, r' \leq 2$.
The possible non-negative integer solutions for $(b', y', r')$ are:
$(2,0,0), (0,2,0), (0,0,2), (1,1,0), (1,0,1), (0,1,1)$.
These correspond to the following $(b, y, r)$ combinations:
$(4,2,2), (2,4,2), (2,2,4), (3,3,2), (3,2,3), (2,3,3)$.
Calculating the number of ways for each case:
$1. (4,2,2): C(5,4) \times C(6,2) \times C(4,2) = 5 \times 15 \times 6 = 450$
$2. (2,4,2): C(5,2) \times C(6,4) \times C(4,2) = 10 \times 15 \times 6 = 900$
$3. (2,2,4): C(5,2) \times C(6,2) \times C(4,4) = 10 \times 15 \times 1 = 150$
$4. (3,3,2): C(5,3) \times C(6,3) \times C(4,2) = 10 \times 20 \times 6 = 1200$
$5. (3,2,3): C(5,3) \times C(6,2) \times C(4,3) = 10 \times 15 \times 4 = 600$
$6. (2,3,3): C(5,2) \times C(6,3) \times C(4,3) = 10 \times 20 \times 4 = 800$
Total ways = $450 + 900 + 150 + 1200 + 600 + 800 = 4100$.

(C) The building has $10$ floors above the ground floor,but the lift does not stop at the $1$st and $2$nd floors. Thus,there are $10 - 2 = 8$ floors available for exiting.
Nine persons enter the lift. We need to select $4$ persons to exit at one floor and the remaining $5$ persons to exit at a different floor.
The number of ways to partition $9$ persons into a group of $4$ and $5$ is $\binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$.
The number of ways to choose $2$ distinct floors from the $8$ available floors is $P(8, 2) = 8 \times 7 = 56$ (since the order of floors matters,i.e.,which group exits at which floor).
The total number of ways is $126 \times 56 = 7056$.