Permutation and Combination Questions in English

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

(A) We know the property of combinations: ${ }^{n} C_{a} = { }^{n} C_{b}$ implies either $a = b$ or $a + b = n$.
Given the equation: ${ }^{2 n} C_{r} = { }^{2 n} C_{r+2}$.
Here,$r \neq r+2$,so we must have the sum of the lower indices equal to the upper index:
$r + (r + 2) = 2n$
$2r + 2 = 2n$
$2r = 2n - 2$
Dividing by $2$,we get:
$r = n - 1$.

(A) Given that ${ }^{18} C_{r}={ }^{18} C_{r+2}$.
We know the property ${ }^{n} C_{a}={ }^{n} C_{b}$ implies either $a=b$ or $a+b=n$.
Here,$r \neq r+2$,so we must have $r + (r+2) = 18$.
$2r + 2 = 18$
$2r = 16$
$r = 8$.
Now,we need to find ${ }^{r} C_{5} = { }^{8} C_{5}$.
${ }^{8} C_{5} = { }^{8} C_{8-5} = { }^{8} C_{3}$.
${ }^{8} C_{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 8 \times 7 = 56$.

(B) Given equation: $12 \cdot {}^{n}C_{2} = {}^{2n}C_{3}$.
Using the formula ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$,we have:
$12 \cdot \frac{n!}{2!(n-2)!} = \frac{(2n)!}{3!(2n-3)!}$
$12 \cdot \frac{n(n-1)}{2} = \frac{(2n)(2n-1)(2n-2)}{6}$
$6n(n-1) = \frac{2n(2n-1) \cdot 2(n-1)}{6}$
$6n(n-1) = \frac{4n(2n-1)(n-1)}{6}$
Dividing both sides by $2n(n-1)$ (assuming $n > 1$):
$3 = \frac{2(2n-1)}{6}$
$18 = 4n - 2$
$20 = 4n$
$n = 5$.

(B) The expression is $\sum_{r=1}^{5} C(5, r) = C(5, 1) + C(5, 2) + C(5, 3) + C(5, 4) + C(5, 5)$.
We know that the sum of binomial coefficients is $\sum_{r=0}^{n} C(n, r) = 2^n$.
Here,$n = 5$,so $\sum_{r=0}^{5} C(5, r) = 2^5 = 32$.
The given sum is $\sum_{r=1}^{5} C(5, r) = \left( \sum_{r=0}^{5} C(5, r) \right) - C(5, 0)$.
Since $C(5, 0) = 1$,we have $32 - 1 = 31$.
Thus,the correct option is $B$.

(C) The number of ways to select $r$ items from a set of $n$ items is given by the combination formula: $C(n, r) = \frac{n!}{r!(n-r)!}$.
Here,$n = 10$ and $r = 5$.
Therefore,the number of ways $= C(10, 5) = \frac{10!}{5!5!}$.
$= \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}$.
$= \frac{30240}{120} = 252$.

(D) To select a team of $11$ players from $16$ players,where $2$ specific players must always be included,we follow these steps:
$1$. Since $2$ players are already fixed,we only need to select the remaining $11 - 2 = 9$ players.
$2$. The number of available players remaining is $16 - 2 = 14$.
$3$. The number of ways to select $9$ players from $14$ is given by the combination formula ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$.
$4$. Substituting the values: ${}^{14}C_{9} = \frac{14!}{9!5!} = \frac{14 \times 13 \times 12 \times 11 \times 10}{5 \times 4 \times 3 \times 2 \times 1} = 2002$ ways.

(B) To select a team of $11$ players from $16$ players where $1$ particular player must always be excluded,we effectively need to choose $11$ players from the remaining $15$ players $(16 - 1 = 15)$.
This is calculated using the combination formula ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$.
Here,$n = 15$ and $r = 11$.
Number of ways = ${}^{15}C_{11} = {}^{15}C_{15-11} = {}^{15}C_{4}$.
${}^{15}C_{4} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 15 \times 7 \times 13 = 1365$ ways.

(A) Total players available = $16$.
We need to select a team of $11$ players.
Given conditions:
$1$. $2$ particular players must be included.
$2$. $1$ particular player must be rejected.
Since $2$ players are already included,we need to select $11 - 2 = 9$ more players.
Since $1$ player is rejected and $2$ are already included,the number of players remaining to choose from is $16 - 2 - 1 = 13$.
Therefore,the number of ways to select the remaining $9$ players from $13$ players is given by the combination formula ${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$.
Number of ways = ${}^{13}C_{9} = {}^{13}C_{13-9} = {}^{13}C_{4}$.
${}^{13}C_{4} = \frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1} = 13 \times 11 \times 5 = 715$.

(D) The number of ways to choose $8$ questions from $10$ in part $A$ is given by $C(10, 8)$.
The number of ways to choose $5$ questions from $10$ in part $B$ is given by $C(10, 5)$.
Since these are independent selections,the total number of ways is the product of the two combinations:
Total ways $= C(10, 8) \times C(10, 5)$
$= \frac{10!}{8!2!} \times \frac{10!}{5!5!}$
$= \frac{10 \times 9}{2 \times 1} \times \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}$
$= 45 \times 252 = 11340$.

(C) Part $1$: Number of ways to select $11$ players out of $15$ is given by $C(15, 11)$.
$C(15, 11) = C(15, 4) = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365$.
Part $2$: If a particular player must be included,we have already selected $1$ player. Now,we need to select the remaining $10$ players from the remaining $14$ players.
This is given by $C(14, 10)$.
$C(14, 10) = C(14, 4) = \frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1} = 1001$.
Thus,the number of ways a particular player is included is $1001$.

(C) The word '$INVOLUTE$' contains $8$ distinct letters.
The vowels in the word are $\{I, O, U, E\}$,which are $4$ in total.
The consonants in the word are $\{N, V, L, T\}$,which are $4$ in total.
We need to select $3$ vowels out of $4$ and $2$ consonants out of $4$.
The number of ways to select the letters is $C(4, 3) \times C(4, 2) = 4 \times 6 = 24$.
These $5$ selected letters can be arranged among themselves in $5!$ ways.
$5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$.
Therefore,the total number of words that can be formed is $24 \times 120 = 2880$.

(B) To form a line,we need to select $2$ distinct points from the given set of points.
Since no $3$ points are collinear (as they lie on a circle),every pair of points determines a unique line.
The number of lines that can be drawn through $n$ points is given by the combination formula $C(n, 2) = \frac{n(n-1)}{2}$.
Here,$n = 21$.
Therefore,the number of lines $p = C(21, 2) = \frac{21 \times 20}{2} = 21 \times 10 = 210$ lines.

(C) To select $9$ balls such that there are $3$ balls of each colour,we need to choose $3$ red balls from $6$,$3$ white balls from $5$,and $3$ blue balls from $5$.
The number of ways to choose $3$ red balls from $6$ is given by $C(6,3) = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
The number of ways to choose $3$ white balls from $5$ is given by $C(5,3) = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10$.
The number of ways to choose $3$ blue balls from $5$ is given by $C(5,3) = \frac{5!}{3!2!} = \frac{5 \times 4}{2 \times 1} = 10$.
By the fundamental principle of counting,the total number of ways is the product of these individual selections:
$p = 20 \times 10 \times 10 = 2000$.

(B) Out of the available $9$ courses,$2$ are compulsory.
Since the student must choose a total of $5$ courses and $2$ are already fixed,the student needs to select the remaining $5 - 2 = 3$ courses from the remaining $9 - 2 = 7$ available courses.
The number of ways to select $3$ courses out of $7$ is given by the combination formula $C(n, r) = \frac{n!}{r!(n-r)!}$.
Here,$n = 7$ and $r = 3$.
$C(7, 3) = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.
Therefore,there are $35$ ways to choose the programme.

(C) To select $4$ questions from each part,we use the combination formula $C(n, r) = \frac{n!}{r!(n-r)!}$.
Number of ways to select $4$ questions from Part-$I$ ($6$ questions): $C(6, 4) = \frac{6!}{4!2!} = \frac{6 \times 5}{2 \times 1} = 15$.
Number of ways to select $4$ questions from Part-$II$ ($7$ questions): $C(7, 4) = \frac{7!}{4!3!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.
Number of ways to select $4$ questions from Part-$III$ ($8$ questions): $C(8, 4) = \frac{8!}{4!4!} = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$.
Total number of combinations = $C(6, 4) \times C(7, 4) \times C(8, 4) = 15 \times 35 \times 70 = 36750$.

(NONE) The total number of persons is $8,$ and we need to select $6$ persons.
Case $1$: Person $A$ is not selected.
If $A$ is not selected,we must choose $6$ persons from the remaining $7$ persons $(8 - 1 = 7)$. Since $A$ is not selected,the condition regarding $B$ is satisfied (as $B$ can be selected or not). The number of ways is $C(7, 6) = 7$.
Case $2$: Person $A$ is selected.
If $A$ is selected,then $B$ must also be selected. We have already selected $2$ persons ($A$ and $B$). We need to select $4$ more persons from the remaining $6$ persons $(8 - 2 = 6)$. The number of ways is $C(6, 4) = \frac{6!}{4!2!} = \frac{6 \times 5}{2 \times 1} = 15$.
However,we must check if the condition is violated. The condition is: if $A$ is chosen,$B$ must be chosen. In Case $1$,$A$ is not chosen,so the condition is not violated. In Case $2$,$A$ is chosen and $B$ is chosen,so the condition is satisfied.
Total ways $= 7 + 15 = 22$.

(D) The word '$THERAPY$' consists of $7$ distinct letters: $T, H, E, R, A, P, Y$.
There are $2$ vowels $(E, A)$ and $5$ consonants $(T, H, R, P, Y)$.
First,calculate the total number of arrangements of all $7$ letters,which is $7! = 5040$.
Next,calculate the number of arrangements where the $2$ vowels come together. Treat the $2$ vowels as a single unit. Now we have $5$ consonants + $1$ unit = $6$ entities,which can be arranged in $6!$ ways.
The $2$ vowels within the unit can be arranged in $2!$ ways.
So,the number of ways the vowels come together is $6! \times 2! = 720 \times 2 = 1440$.
The number of ways the vowels never come together is the total arrangements minus the arrangements where they are together:
$5040 - 1440 = 3600$.

(A) The word '$PRAISE$' consists of $6$ distinct letters: $P, R, A, I, S, E$.
Since all letters are distinct,the number of ways to arrange them is given by the factorial of the number of letters.
Number of arrangements $= 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$ ways.

(A) The word '$PRAISE$' consists of $6$ distinct letters: $P, R, A, I, S, E$.
The number of ways to arrange $n$ distinct objects is given by $n!$.
Here,$n = 6$.
Therefore,the number of arrangements $= 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$.

(D) The word '$BANKING$' consists of $7$ letters,where the letter '$N$' repeats $2$ times.
The number of ways to arrange $n$ objects where $p$ objects are of one kind is given by the formula $\frac{n!}{p!}$.
Here,$n = 7$ and $p = 2$.
Therefore,the required number of arrangements $= \frac{7!}{2!} = \frac{5040}{2} = 2520$.

(D) The word $ATTEND$ consists of $6$ letters in total.
In this word,the letter $T$ appears $2$ times,while all other letters $(A, E, N, D)$ appear once.
The number of ways to arrange $n$ objects where $p$ objects are of one kind,$q$ objects are of another kind,etc.,is given by the formula $\frac{n!}{p!q!...}$.
Here,$n = 6$ and the frequency of $T$ is $2$.
Therefore,the required number of arrangements $= \frac{6!}{2!} = \frac{720}{2} = 360$.

(D) The word '$CYCLE$' contains $5$ letters in total.
In this word,the letter '$C$' appears $2$ times,while the letters '$Y$','$L$',and '$E$' appear $1$ time each.
The number of ways to arrange $n$ objects where $p$ objects are of one kind and $q$ objects are of another kind is given by $\frac{n!}{p!q!...}$.
Here,$n = 5$ and the frequency of '$C$' is $2$.
Therefore,the total number of arrangements $= \frac{5!}{2!} = \frac{120}{2} = 60$ ways.

(C) Total number of students $= 54 \times 30 = 1620$.
When these students are arranged in rows of $45,$ the number of rows formed is calculated by dividing the total number of students by the number of students per row.
Number of rows $= \frac{1620}{45} = 36$.
Therefore,$36$ rows will be formed.

(D) The word $ATTEND$ consists of $6$ letters in total.
In this word,the letter $T$ appears $2$ times,and all other letters $(A, E, N, D)$ appear $1$ time each.
The number of ways to arrange $n$ objects where $p$ objects are of one kind and $q$ objects are of another kind is given by $\frac{n!}{p!q!}$.
Here,$n = 6$ and $p = 2$ (for the letter $T$).
Therefore,the number of arrangements $= \frac{6!}{2!} = \frac{720}{2} = 360$.

(A) To form a committee of $5$ members,we consider the two given cases:
Case $1$: The committee has all $4$ professors and $1$ research associate.
The number of ways to choose $4$ professors out of $4$ is ${}^{4}C_{4} = 1$.
The number of ways to choose $1$ research associate out of $6$ is ${}^{6}C_{1} = 6$.
Total ways for Case $1 = 1 \times 6 = 6$.
Case $2$: The committee has all $3$ trainees and $2$ professors.
The number of ways to choose $3$ trainees out of $3$ is ${}^{3}C_{3} = 1$.
The number of ways to choose $2$ professors out of $4$ is ${}^{4}C_{2} = \frac{4 \times 3}{2 \times 1} = 6$.
Total ways for Case $2 = 1 \times 6 = 6$.
Since these cases are mutually exclusive,the total number of ways is the sum of the ways from both cases:
Total ways $= 6 + 6 = 12$.

(C) To form the committee,we need to select $2$ trainees from $3$ available trainees and $3$ research associates from $6$ available research associates.
The number of ways to select $2$ trainees out of $3$ is given by the combination formula ${ }^{n} C_{r} = \frac{n!}{r!(n-r)!}$.
${ }^{3} C_{2} = \frac{3!}{2!(3-2)!} = \frac{3 \times 2!}{2! \times 1!} = 3$.
The number of ways to select $3$ research associates out of $6$ is:
${ }^{6} C_{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.
Since these selections are independent,the total number of ways to form the committee is the product of these two values:
Total ways $= { }^{3} C_{2} \times { }^{6} C_{3} = 3 \times 20 = 60$.

(A) The word '$OFFICES$' contains $7$ letters in total.
In this word,the letter '$F$' is repeated $2$ times,while all other letters $(O, I, C, E, S)$ appear once.
The number of ways to arrange $n$ objects where $p$ objects are of one kind and $q$ objects are of another kind is given by the formula $\frac{n!}{p!q!...}$.
Here,$n = 7$ and $p = 2$ (for the letter '$F$').
Therefore,the required number of arrangements $= \frac{7!}{2!} = \frac{5040}{2} = 2520$.