Permutations and Combinations

Given, $S=\{1,2,3,4,5,6,7,8,9\}$

1. Calculation for x

For $x$, we have to form a 9-digit number in which exactly one digit is repeated, and it is repeated exactly twice. So, one digit appears twice, and the remaining 7 digits appear once each.

First, select the digit that will be repeated: from $S$, this can be done in ${ }^9 C_1$ ways.

Now, from the remaining 8 digits, choose 7 digits that will appear once: this can be done in ${ }^8 C_7$ ways.

Now we have 9 digits in total, but one digit is repeated twice. So, the number of different arrangements is $=\frac{9!}{2!}$ ways.

$ x={ }^9 C_1 \times{ }^8 C_7 \times \frac{9!}{2!} $

2. Calculation for y

For $y$, we have to form a 9-digit number in which exactly two digits are repeated, and each of these is repeated exactly twice. So, two digits appear twice each (total 4 places), and the remaining 5 digits appear once each.

First, select the 2 digits that will be repeated from $S$: this can be done in ${ }^9 C_2$ ways.

Now, from the remaining 7 digits, choose 5 digits that will appear once: this can be done in ${ }^7 C_5$ ways.

Now we have 9 digits in total, with two digits repeated twice each. So, the number of different arrangements is $=\frac{9!}{2!2!}$ ways.

$ y={ }^9 C_2 \times{ }^7 C_5 \times \frac{9!}{2!2!} $

To find the relation between $x$ and $y$, divide $x$ by $y$:

To find the relation, divide x by y :

$ \frac{x}{y}=\frac{{ }^9 C_1 \times{ }^8 C_7 \times \frac{1}{2}}{{ }^9 C_2 \times{ }^7 C_5 \times \frac{1}{4}}=\frac{9 \times 8 \times \frac{1}{2}}{36 \times 21 \times \frac{1}{4}} $

So,

$\frac{x}{y}=\frac{36}{189}=\frac{4}{21}$

$\Rightarrow 21 x=4 y$

Arrang UDAYPUR in alphabetical order A, D, P, R, UU, Y

Number of words start with A. fix A and arrange DPRUUY $=\frac{6!}{2!}=\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1}=360$ ways.

Similarly, number words starts from D, P, R will be same as there is arrangement of 6 letters in which $U$ is repeating twice.

So, total number of words starting with A, D, P \& R are $=4 \times 360=1440$.

Now, number of words starts with UA

fix UA and arrange D, P, R, U, Y = 5! ways = 120 .

the next letter is D after A, so number of words starts with UDAP arranging $(R, U, Y)=3!$ ways $=6$

So, number of words start with UDAR and UDAU is same $=3!+3!=12$

next is UDAY matches with pattern(UDAYPUR)

number of words start with UDAYP arrang(RU) $=2!=2$

number of words start with UDAYPR is UDAYPRU $=1$

final pattern UDAYPUR = 1

So, rank of word UDAYPUR is =

$ 1440+120+6+12+1+1=1580 . $

To find largest value of $n$ for which 60 ! is divisible by $40^n$

This means we need to find exponent of 40 in 60 !

$ 40=2^3 \times 5 $

Key Concept: formula for exponent of prime $p$ in $k!$ is given as $\left[\frac{k}{p}\right]+\left[\frac{k}{p^2}\right]+\left[\frac{k}{p^3}\right]+\ldots$ where [] is greatest integer function.

so, exponent of 2 in 60 !

$ \begin{aligned} & {\left[\frac{60}{2}\right]+\left[\frac{60}{2^2}\right]+\left[\frac{60}{2^3}\right]+\left[\frac{60}{2^4}\right]+\left[\frac{60}{2^5}\right]+\left[\frac{60}{2^6}\right]+\ldots} \\ & 30+15+7+3+1=56-\text { (1) } \end{aligned} $

exponent of 5 in 60 !

$ \begin{aligned} & {\left[\frac{60}{5}\right]+\left[\frac{60}{5^2}\right]+\left[\frac{60}{5^3}\right]+\ldots} \\ & 12+2+0 \cdots=14-(2) \end{aligned} $

Combining (1) and (2)

60 ! is divisible by $2^{56} \times 5^{14} \times N$, where $N$ is natural number.

60 ! is divisible by $4 \times\left(2^3\right)^{18} \times 5^{14} \times N$

60 ! is divisible by $4 \times\left(2^3\right)^4 \times\left(2^3\right)^{14} \times 5^{14} \times N$

$60!$ is divisible by $2^{14} \times 40^{14} \times N$

So largest value of $n$ is 14

To distribute 16 oranges to 4 children such that each child gets atleast one orange.

First distribute one orange to each child so that each child get atleast one.

Now, distribute remaining ( $\mathrm{n}=12$ ) oranges to ( $\mathrm{r}=4$ ) children using beggar's method $={ }^{n+r-1} C_{r-1}$

$ \begin{aligned} & ={ }^{12+4-1} C_{4-1}={ }^{15} C_3 \\\\ & =\frac{15 \times 14 \times 13}{3 \times 2 \times 1}=5 \times 7 \times 13=455 \end{aligned} $

$ \text { Exponent of } 7 \text { in } 101!=\left[\frac{101}{7}\right]+\left[\frac{101}{7^2}\right]+\ldots . . \text { (Where [.] is G.I.F.) }=14+12=16 $

Case (i) If $f(1)=2$ then ${ }^7 C_5$ functions

Case (ii) $f(1)=3$ then ${ }^6 C_5$

Case (iii) $f(1)=4$ then ${ }^5 C_5$

$ \Rightarrow 21+6+1=28 $

Number of ways ( $\boldsymbol{n}$ ) $=\left({ }^4 C_2\right) \cdot\left({ }^3 C_1\right) \cdot(2) \cdot(1)=36$

First, decide which $4$ out of the $9$ persons will get down together. The remaining $5$ persons are then fixed automatically.

Number of ways to form the two groups of sizes $4$ and $5$ is $\frac{9!}{4!5!}=126$

Now, the lift does not stop at the $1^{\text{st}}$ and $2^{\text{nd}}$ floors. So people can get down only on floors $3$ to $10$, i.e., $8$ possible floors.

Choose any $2$ different floors out of these $8$ floors for the two groups to exit: ${ }^8 \mathrm{C}_2=28$

After choosing the two floors, we must decide which floor is for the group of $4$ and which floor is for the group of $5$. This can be done in $2$ ways.

So, total number of ways is $126 \times 28 \times 2=7056$

From the word INCONSEQUENTIAL, let us first identify the vowels and consonants available.

The letters are:

$ I, N, C, O, N, S, E, Q, U, E, N, T, I, A, L $

We have to form $4$-letter words containing:

• exactly $2$ vowels

• exactly $2$ consonants

• no letter repeated

Since letters of the word may repeat in the original word, we count only distinct available letters for selection.

Step 1: Distinct vowels

The vowels present are:

$ A, E, I, O, U $

So, number of distinct vowels $= 5$.

Step 2: Distinct consonants

The consonants present are:

$ N, C, S, Q, T, L $

So, number of distinct consonants $= 6$.

Step 3: Choose 2 vowels and 2 consonants

Number of ways to choose $2$ vowels from $5$:

$ {}^5C_2 = 10 $

Number of ways to choose $2$ consonants from $6$:

$ {}^6C_2 = 15 $

So, total selections of letters:

$ {}^5C_2 \times {}^6C_2 = 10 \times 15 = 150 $

Step 4: Arrange the 4 chosen letters

All 4 letters are distinct, so number of arrangements:

$ 4! = 24 $

Hence, total number of such 4-letter words is:

$ 150 \times 24 = 3600 $

Therefore, the correct answer is:

$ \boxed{3600} $

So, Option D is correct.

Let us first understand the problem carefully. We are given a box containing:

Total number of blue balls = $5$

Total number of yellow balls = $6$

Total number of red balls = $4$

We need to draw a total of $8$ balls such that there are at least two balls of each colour. Let the number of blue, yellow, and red balls drawn be denoted by $b$, $y$, and $r$ respectively.

According to the question, we have the following equation:

$b + y + r = 8$

Since we need at least $2$ balls of each colour, we must satisfy the conditions:

$b \ge 2, y \ge 2, \text{ and } r \ge 2$

Since $2+2+2 = 6$, we only need to decide how to distribute the remaining $2$ balls among the three colours. Let us list all the possible combinations for $(b, y, r)$:

Case 1: $4$ blue, $2$ yellow, $2$ red ($b=4, y=2, r=2$)

Case 2: $2$ blue, $4$ yellow, $2$ red ($b=2, y=4, r=2$)

Case 3: $2$ blue, $2$ yellow, $4$ red ($b=2, y=2, r=4$)

Case 4: $3$ blue, $3$ yellow, $2$ red ($b=3, y=3, r=2$)

Case 5: $3$ blue, $2$ yellow, $3$ red ($b=3, y=2, r=3$)

Case 6: $2$ blue, $3$ yellow, $3$ red ($b=2, y=3, r=3$)

Now, we will find the number of ways for each case using the combinations formula $^nC_r = \frac{n!}{r!(n-r)!}$.

1. Number of ways for Case 1:

$^5C_4 \times ^6C_2 \times ^4C_2 = 5 \times 15 \times 6 = 450$

2. Number of ways for Case 2:

$^5C_2 \times ^6C_4 \times ^4C_2 = 10 \times 15 \times 6 = 900$

3. Number of ways for Case 3:

$^5C_2 \times ^6C_2 \times ^4C_4 = 10 \times 15 \times 1 = 150$

4. Number of ways for Case 4:

$^5C_3 \times ^6C_3 \times ^4C_2 = 10 \times 20 \times 6 = 1200$

5. Number of ways for Case 5:

$^5C_3 \times ^6C_2 \times ^4C_3 = 10 \times 15 \times 4 = 600$

6. Number of ways for Case 6:

$^5C_2 \times ^6C_3 \times ^4C_3 = 10 \times 20 \times 4 = 800$

To find the total number of ways to draw the $8$ balls satisfying the given conditions, we simply add the number of ways from all the above cases:

$\text{Total ways} = 450 + 900 + 150 + 1200 + 600 + 800$

$\text{Total ways} = 4100$

Therefore, the total number of ways of drawing $8$ balls containing at least two balls of each colour is $4100$.

Option A is the correct answer.

We need to count the number of ordered triples $(a,b,c)$ of non-negative integers satisfying

$ a+b+2c=22 $

Since $a,b,c \ge 0$, let us fix $c$ first.

For a fixed value of $c$, the equation becomes

$ a+b=22-2c $

Now, the number of non-negative integer solutions of

$ a+b=n $

is

$ n+1 $

So, for fixed $c$, number of solutions is

$ (22-2c)+1=23-2c $

Now find all possible values of $c$.

Since $2c \le 22$,

$ c=0,1,2,\dots,11 $

Hence total number of solutions is

$ \sum_{c=0}^{11} (23-2c) $

This is an arithmetic series:

• first term $=23$

• last term $=23-2(11)=1$

• number of terms $=12$

So,

$ n(A)=\frac{12}{2}(23+1) $

$ n(A)=6 \cdot 24=144 $

Therefore,

$ n(A)=144 $

So the correct option is:

$ \boxed{144} $

Hence, $\boxed{\text{Option C}}$.

We have a total of 4 boys and 3 girls.

Total number of children = $4 + 3 = 7$.

The total number of ways to arrange 7 children in a queue is given by the permutation of 7 distinct objects taken all at a time, which is $7!$.

$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$

Number of arrangements where all the girls are together

To find this, let us treat all 3 girls as a single unit or block, since they must remain together.

Now, we have 4 boys and 1 unit of girls. This gives us a total of $4 + 1 = 5$ entities to arrange in the queue.

These 5 entities can be arranged among themselves in $5!$ ways.

Additionally, the 3 girls within their single unit can rearrange themselves in $3!$ ways.

By the Fundamental Principle of Counting (Multiplication Principle), the total number of ways in which all the girls are together is:

$5! \times 3!$

We know that $5! = 120$ and $3! = 6$, so:

$120 \times 6 = 720$

Number of arrangements where all the girls are NOT together

To find the required number of ways, we subtract the arrangements where all girls are together from the total possible arrangements.

$\text{Required ways} = \text{Total arrangements} - \text{Arrangements with all girls together}$

$\text{Required ways} = 5040 - 720 = 4320$

Therefore, the number of ways of forming a queue such that all the girls are not together is 4320.

The correct choice is Option D.

$P_n = {}^nC_3$

In an $n$-sided polygon, any $3$ vertices form a triangle, so the total number of triangles is $P_n = {}^nC_3$.

$P_{n+1} - P_n = 66$

This means when we increase the number of sides from $n$ to $n+1$, the number of triangles increases by $66$.

${}^{n+1}C_3 - {}^nC_3 = 66$

Now we write both combinations using the formula ${}^nC_3 = \frac{n(n-1)(n-2)}{6}$.

$\Rightarrow \frac{(n+1)n(n-1)}{6} - \frac{n(n-1)(n-2)}{6} = 66$

Take the common factor $\frac{n(n-1)}{6}$ from both terms.

$\Rightarrow \frac{n(n-1)}{6}[n+1-n+2] = 66$

Inside the bracket, simplify: $n+1-n+2 = 3$.

$\Rightarrow n(n-1) = 132$

So we need two consecutive integers whose product is $132$. That gives $n=12$ because $12 \cdot 11 = 132$.

$n = 12$

Prime divisors of 12 are 2 & 3

The distinct prime divisors of $12$ are $2$ and $3$.

Sum $= 2 + 3 = 5$

We have to form a $7$-digit number using the digits $1,2,3,5,7$ such that each digit is used at least once.

Since there are $5$ distinct digits and the number has $7$ places, after using each digit once, there are $2$ extra places.

So, the repetition can happen in the following ways:

Case 1: One digit is repeated $3$ times

Then the distribution of digits is:

$3,1,1,1,1$

Choose the digit which appears $3$ times:

$^5C_1 = 5$

Now arrange the $7$ digits:

$\frac{7!}{3!}$

So, total numbers in this case:

$5 \times \frac{7!}{3!} = 5 \times \frac{5040}{6} = 5 \times 840 = 4200$

Case 2: Two digits are repeated $2$ times each

Then the distribution is:

$2,2,1,1,1$

Choose the two digits which are repeated:

$^5C_2 = 10$

Now arrange the $7$ digits:

$\frac{7!}{2!2!}$

So, total numbers in this case:

$10 \times \frac{7!}{2!2!} = 10 \times \frac{5040}{4} = 10 \times 1260 = 12600$

Total number of such $7$-digit numbers

$4200 + 12600 = 16800$

Hence, the required number is

$\boxed{16800}$

So, the correct option is Option C.

We need to find the number of ordered pairs $(r,k)$ satisfying

$ {}^{36}C_{r+1}=\frac{6\left({}^{35}C_r\right)}{k^2-3}, $

where $k \in \mathbb{Z}$.

Let us simplify the relation using the standard NCERT combination formula.

We know that

$ {}^{n}C_{r+1}=\frac{n!}{(r+1)!(n-r-1)!} $

and

$ {}^{n-1}C_r=\frac{(n-1)!}{r!(n-r-1)!}. $

So,

$ \frac{{}^{36}C_{r+1}}{{}^{35}C_r} = \frac{\dfrac{36!}{(r+1)!(35-r)!}}{\dfrac{35!}{r!(35-r)!}} = \frac{36!}{35!}\cdot \frac{r!}{(r+1)!} = 36\cdot \frac{1}{r+1} = \frac{36}{r+1}. $

Hence,

$ {}^{36}C_{r+1} = {}^{35}C_r \cdot \frac{36}{r+1}. $

Given,

$ {}^{36}C_{r+1}=\frac{6\left({}^{35}C_r\right)}{k^2-3}. $

Comparing both expressions,

$ {}^{35}C_r \cdot \frac{36}{r+1}=\frac{6\left({}^{35}C_r\right)}{k^2-3}. $

Since ${}^{35}C_r \neq 0$, cancel it:

$ \frac{36}{r+1}=\frac{6}{k^2-3}. $

Cross-multiplying,

$ 36(k^2-3)=6(r+1). $

Divide by $6$:

$ 6(k^2-3)=r+1. $

So,

$ r = 6k^2-19. $

Now $r$ must be valid for the combinations ${}^{35}C_r$ and ${}^{36}C_{r+1}$.

Thus,

$ 0 \le r \le 35. $

Substitute $r=6k^2-19$:

$ 0 \le 6k^2-19 \le 35. $

Add $19$ throughout:

$ 19 \le 6k^2 \le 54. $

Divide by $6$:

$ \frac{19}{6} \le k^2 \le 9. $

Now possible integer square values are

$ k^2=4 \quad \text{or} \quad k^2=9. $

So the integer values of $k$ are:

• If $k^2=4$, then $k=\pm 2$

• If $k^2=9$, then $k=\pm 3$

Now find corresponding $r$:

For $k^2=4$,

$ r=6(4)-19=24-19=5. $

So pairs are

$ (5,2),\ (5,-2). $

For $k^2=9$,

$ r=6(9)-19=54-19=35. $

So pairs are

$ (35,3),\ (35,-3). $

Hence total number of elements in the set $S$ is

$ 4. $

Therefore, the correct answer is:

$ \boxed{4} $

So, Option B is correct.

To count the number of distinct triangles:

Start with all possible triples of points.

From 12 points, the number of ways to choose any 3 is

$ \binom{12}{3}= \frac{12\cdot11\cdot10}{3\cdot2\cdot1}=220. $

Subtract the “invalid” triples that are collinear.

The only collinear sets of three points arise from the single line containing the 5 collinear points.

Number of ways to pick 3 points from those 5 is

$ \binom{5}{3}=10. $

Valid triangles = total triples − collinear triples.

$ 220-10 = 210. $

Hence, the total number of triangles that can be formed is 210.

Correct option: B

1 Captain, 1 vice-captain are already present

$\Rightarrow$ We need to select 8 players such that atleast 3 batsman and bowler must be there

Total = 155 ways

Total triangles

(2 points as $y=\frac{x}{2}, 1$ point on $y=\frac{x}{2}$)

$+2\left(\right.$ points and $y=\frac{x}{2}, 1$ point on $y=2 x$)

+(1 point on $y=2 x, 1)$ point on $y=\frac{x}{2}$ and origin)

$={ }^9 C_2,{ }^{12} C_1+{ }^9 C_1{ }^{12} C_2+{ }^9 C_1 \cdot{ }^{12} C_1 \cdot{ }^1 C_1$

$=1134$

Let $x, y, z$ be the number of box which are filled

$\Rightarrow 1 \leq x \leq 3,1 \leq y \leq 3,1 \leq z \leq 2$

Total ways $=(48)$ to fill boxes

Now to arrange $a, b, c, d$ and $e$

Number of ways will be 48.5! = 5760

To find the number of sequences of ten terms, each being either 0, 1, or 2, containing exactly five 1s and exactly three 2s, follow these steps:

Determine the Remaining Terms: Since there are 5 ones and 3 twos, you will need 2 zeros to fill the sequence (because $5 + 3 + 2 = 10$).

Calculate the Number of Arrangements: You have a total of 10 positions to fill with these numbers (5 ones, 3 twos, 2 zeros). The formula for calculating permutations of a multiset is:

$ \frac{10!}{5! \times 3! \times 2!} $

Where:

$10!$ is the factorial of the total number of terms.

$5!$ is the factorial for the number of 1s.

$3!$ is the factorial for the number of 2s.

$2!$ is the factorial for the number of 0s.

Result: Simplifying the calculation gives:

$ \frac{10!}{5! \times 3! \times 2!} = 2520 $

Thus, there are 2520 possible sequences with the given conditions.

(i) number of numbers created using

$1111133=\frac{7!}{5!2!} \Rightarrow 21$

(ii) number of numbers created using

$1111223=\frac{7!}{4!2!} \Rightarrow 105$

(iii) number of numbers created using

$\begin{aligned} & 1112222=\frac{7!}{4!3!} \Rightarrow 35 \\ & \text { Total }=161 \end{aligned}$

$\begin{aligned} & { }^n C_{r-1}=28,{ }^n C_r=56 \\ & \frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}-1}}{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}=\frac{28}{56} \\ & \frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{n!}{r!(n-r)!}}=\frac{1}{2} \\ & \frac{\mathrm{r}}{(\mathrm{n}-\mathrm{r}+1)}=\frac{1}{2} \\ & 3 \mathrm{r}=\mathrm{n}+1\quad\text{..... (i)} \end{aligned}$

$\begin{aligned} & \frac{{ }^n C_r}{{ }^n C_{r+1}}=\frac{56}{70} \\ & \frac{(r+1)}{(n-r)}=\frac{56}{70} \Rightarrow 9 r=4 n-5\quad\text{.... (ii)} \end{aligned}$

$\begin{aligned} & \text { By (i) & (ii) } \\ & \begin{array}{l} (r=3),(n=8) \\ A(4 \cos t, 4 \sin t) \quad B(2 \sin t,-2 \cos t) C\left(3 r-n, r^2-n-1\right) \\ A(4 \cos t, 4 \sin t) \quad B(2 \sin t,-2 \operatorname{cost}) C(1,0) \\ (3 x-1)^2+(3 y)^2=(4 \operatorname{cost}+2 \sin t)^2+(4 \sin t-\operatorname{cost})^2 \\ (3 x-1)^2+(3 y)^2=20 \quad \therefore \text { option }(1) \end{array} \end{aligned}$

Case I $5{ }_{---} 0$

Case II $5{ }_{---} 1$

$ \begin{array}{ll} 5 & 2 \\ 5 & 3 \\ 6 & 0 \\ 6 & 1 \\ 6 & 2 \\ 7 & 0 \end{array}$

Case IX $7{ }_{---} 1$

$9 \times(8 \times 8 \times 8)=4608$ but 50000 is not included, so total numbers $4608-1=4607$

$\begin{array}{ll} \text { C-I } & (3 \mathrm{G} \& 2 \mathrm{~B}) \&(1 \mathrm{G} \& 2 \mathrm{~B}) \\ \text { C-II } & (2 \mathrm{G} \& 3 \mathrm{~B}) \&(2 \mathrm{G} \& 1 \mathrm{~B}) \\ \text { C-III } & (1 \mathrm{G} \& 4 \mathrm{~B}) \&(3 \mathrm{G} \& 0 \mathrm{~B}) \end{array}$

$\begin{aligned} & \text { Total }=\text { C-I }+ \text { C-II }+ \text { C-III } \\ & ={ }^7 \mathrm{C}_2 \cdot{ }^3 \mathrm{C}_3 \cdot{ }^6 \mathrm{C}_2 \cdot{ }^5 \mathrm{C}_1+{ }^7 \mathrm{C}_3 \cdot{ }^3 \mathrm{C}_2 \cdot{ }^6 \mathrm{C}_1{ }^5 \mathrm{C}_2+{ }^7 \mathrm{C}_4 \cdot{ }^3 \mathrm{C}_1 \cdot{ }^6 \mathrm{C}_0 \cdot{ }^5 \mathrm{C}_3 \\ & =8925 \end{aligned}$

DAUGHTER

Total words $=8$ !

Total words in which vowels are together $=6!\times 3!$ words in which all vowels are not together

$\begin{aligned} & =8!-6!\times 3! \\ & =6![56-6] \\ & =720 \times 50 \\ & =36000 \end{aligned}$

Let's break the problem down step by step:

There are two blocks because all girls must stand together and all boys must stand together. The two blocks can be arranged in:

$2 \text{ ways} \quad \text{(i.e., girls first then boys, or boys first then girls).}$

The girls can be arranged among themselves in:

$3! = 6 \text{ ways.}$

For the boys (4 in total), they must be arranged such that the specific boys $B_1$ and $B_2$ are not adjacent.

First, calculate the total number of arrangements of 4 boys:

$4! = 24.$

Next, count the arrangements where $B_1$ and $B_2$ are adjacent. Think of $B_1$ and $B_2$ as a single unit. This unit can be arranged in:

$2! = 2 \text{ ways (since }B_1\text{ and }B_2\text{ can swap positions).}$

Now, with this new unit, we have 3 units in total (the $B_1B_2$ unit and the other 2 boys), which can be arranged in:

$3! = 6 \text{ ways.}$

So, the number of arrangements where $B_1$ and $B_2$ are adjacent is:

$2! \times 3! = 2 \times 6 = 12.$

Therefore, the number of valid arrangements for the boys where $B_1$ and $B_2$ are not adjacent is:

$24 - 12 = 12.$

Finally, multiply all the factors together:

$\text{Total ways} = 2 \times 6 \times 12 = 144.$

Thus, the number of ways in which the girls and boys can stand in the queue under the given conditions is $\boxed{144}.$

This corresponds to Option D.

First, note that we are choosing 5 distinct letters (in strictly increasing alphabetical order) such that the middle (third) letter is ‘M’. Symbolically, if we denote the chosen letters as:

$ L_1 < L_2 < L_3 < L_4 < L_5, $

we want $L_3 = \text{M}$. The English alphabet has 26 letters, and M is the $13^\text{th}$.

Step 1: Letters before M

The letters before M are $\{A, B, C, \ldots, L\}$.

There are 12 letters here ($A$ through $L$).

We need to pick 2 of these 12 letters to occupy $L_1$ and $L_2$.

The number of ways to choose 2 letters out of 12 is ${ }^{12} \mathrm{C}_2$.

Step 2: Letters after M

The letters after M are $\{N, O, P, \ldots, Z\}$.

There are 13 letters here ($N$ through $Z$).

We need to pick 2 of these 13 letters to occupy $L_4$ and $L_5$.

The number of ways to choose 2 letters out of 13 is ${ }^{13} \mathrm{C}_2$.

Step 3: Multiply the choices

Since these choices are independent (picking the two letters before M and two letters after M), the total number of ways is:

$ { }^{12} \mathrm{C}_2 \;\times\;{ }^{13} \mathrm{C}_2 $

Calculate each combination:

$ { }^{12} \mathrm{C}_2 = \frac{12 \times 11}{2} = 66, \quad { }^{13} \mathrm{C}_2 = \frac{13 \times 12}{2} = 78. $

So,

$ { }^{12} \mathrm{C}_2 \times { }^{13} \mathrm{C}_2 = 66 \times 78 = 5148. $

Answer: 5148 (Option B)

Total number of words formed by using the letter's of word M,O,T,H,E,R when repetition is not allowed $=6!=720$

$ \begin{array}{cccccc} 3 & 4 & 6 & 2 & 1 & 5 \\ \mathrm{M} & \mathrm{O} & \mathrm{~T} & \mathrm{H} & \mathrm{E} & \mathrm{R} \\ 2 & 2 & 3 & 1 & 0 & 0 \\ \downarrow & \downarrow & \downarrow & \downarrow & \downarrow & \downarrow \\ 5! & 4! & 3! & 2! & 1! & 0! \end{array} $

Rank

$ \begin{aligned} 2 \times 5!+ & 2 \times 4!+2 \times 3!+1 \times 2!+1 \\ & =240+48+18+2+1 \\ & =309 \end{aligned} $

Number of words that appear after word MOTHER $=720-309=411$

$ \begin{aligned} & \text { } \frac{1}{x}+\frac{1}{y}=\frac{1}{2025} \\ & 2025(x+y)=x y \\ & 2025 x+2025 y-x y=0 \\ & x(2025-y)-2025(-y+2025)=-(2025)^2 \\ & (2025-y)(2025-x)=(2025)^2 \\ & \underbrace{(2025-y)}_{5^2 \times 3^4} \underbrace{(2025-x)}_{5^2 \times 3^4}=5^4 \times 3^8 \\ & \text { Or }(x-2025)(y-2025)=5^4 \times 3^8 \end{aligned} $

Total number of positive integral solution

$=$ (ways of distribute 4 coins among 2 beggars) $\times$ (ways to distribute 8 coins among 2 beggars)

$ ={ }^5 C_1 \times{ }^9 C_1=45 $

$x y z=60$

$ x y z=2^2 \times 3^1 \times 5^1 $

Number of positive integral solutions

$ \begin{aligned} & =\left({ }^{2+3-1} C_{3-1}\right) \times\left({ }^{1+3-1} C_{3-1}\right) \times\left({ }^{1+3-1} C_{3-1}\right) \\ & ={ }^4 C_2 \times{ }^3 C_2 \times{ }^3 C_2 \end{aligned} $

Number of ways to arrange 5 boys in a circle is $4!

$ \begin{aligned} & =4!\times{ }^5 C_5 \times 5! \\ & =24 \times 12 \times 10 \\ & =2880 \end{aligned} $

Remaining words excluding GENTLE are IINCEL.

$ \begin{aligned} \therefore \text { Number of ways } & =\frac{{ }^1 C_1 \times{ }^4 C_1 \times 1 \times 6!}{2!} \\ & =\frac{4 \times 720}{2}=1440 \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \frac{20!}{15!}-\frac{19!}{14!} \\ & \quad=\frac{20 \times 19!}{15 \times 14!}-\frac{19!}{14!} \\ & \quad=\frac{19!}{14!}\left(\frac{20-15}{15}\right) \\ & \quad=\frac{19 \times 18 \times 17 \times 16 \times 15 \times 5}{15} \\ & \quad=19 \times 17 \times 2 \times 9 \times 80 \\ & \quad=6!\times 646 \end{aligned} \end{aligned} $

$ \text { Arrangement }=\frac{4!}{2!} $

Selection of 3 gaps out of 5 for $3 A^{\prime} 5={ }^5 C_3$

∴ Required arrangements

$ \begin{aligned} & =\frac{4!}{2!} \cdot{ }^5 C_3 \cdot 1 \times \frac{5!}{2!} \\ & =\frac{4!}{2!} \times \frac{5 \times 4}{2 \times 1} \times \frac{5!}{2!} \\ & =12 \times 5 \times 120=7200 \end{aligned} $

Selecting 1 vowel 2 consonants (arrangement in which vowel is at middle place)

$ \begin{aligned} & =\left({ }^3 C_1 \cdot{ }^4 C_2\right) 2!+\left({ }^3 C_1 \cdot{ }^1 C_1\right) \cdot 1 \\ & =3 \times 6 \times 2+3=39 \end{aligned} $

Similarly,

Selecting, 2 vowels 1 consonant

$ ={ }^3 C_2 \cdot{ }^4 C_1(2!) \times 2=3 \times 4 \times 4=48 $

Selecting 3 vowels $=3!=6$

Total words $=39+48+6=93$

6 Boys and 4 girls

Number of arrangements in a row such that between any 2 girls there must be exactly 2 boys is

$ \begin{aligned} &=\xrightarrow{G_1}{} \uparrow \xrightarrow{G_2} \uparrow \xrightarrow{G_3} \uparrow{ }\xrightarrow{G_4}\\ & =4!\left(\frac{6!}{(2!)^3 \cdot 3!}\right) \cdot 3!(2!)^3 \\ & =6!\times 24=144(5!) \end{aligned} $

∵ Total number of ways in which train has stopped at exactly 5 stations $={ }^{15} C_5$

And total number of ways in which train has not stopped in consecutive stations

$ x_1+x_2+x_3+x_4+x_5+L=15-4=11 $

$x_i=$ number of station left before $i^{\text {th }}$ station

$ x_1 \geq 0 x_2, x_3, x_4, x_5 \geq 1 \text {, and } L=\text { station } $

left after 4 stations

$ \begin{aligned} x_1+\left(x_2-1\right)+ & \left(x_3-1\right)+\left(x_4-1\right) +\left(x_5-1\right)+L \\ = & 11-4=7 \end{aligned} $

$ \begin{aligned} \therefore \text { Total number of solution } & ={ }^{7+5-1} C_{6-1} \\ & ={ }^{11} C_5 \end{aligned} $

$ \text { ∴ Required ways }={ }^{15} C_5-{ }^{11} C_5 $

Number of all possible ways of distributing eight identical apples among 3 persons is number of non-negative integral solution of equation is

$ \begin{aligned} & x+y+z=8 \\ & \therefore{ }^{8+3-1} C_{3-1} \\ & \quad={ }^{10} C_2 \\ & \quad=\frac{10 \times 9}{2 \times 1}=45 \end{aligned} $

The number of words in CAB and NET both appears $=$ CAB NET $I=3$ !

∴ Required number of ways

$ \begin{aligned} & =7!-(5!+5!-3!) \\ & =5040-(120+120-6) \\ & =5040-234 \\ & =4806 \end{aligned} $

We have, $x+y+z+t=10\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

When $x \geq 2, z \geq 5$

let $\quad x^{\prime}=x-2$

$\Rightarrow \quad x=x^{\prime}+2$

$ z^{\prime}=z-5 $

$\Rightarrow \quad z=z^{\prime}+5$

Here, $x^{\prime}, z^{\prime}, y \geq 0$ and $t \geq 0$ are non-negative integers.

$\therefore$ Substitude these value in Eq. (i), we get

$ \begin{aligned} & \left(x^{\prime}+2\right)+y+\left(z^{\prime}+5\right)+t=10 \\ \Rightarrow \quad & x^{\prime}+y+z^{\prime}+t=3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

$\therefore$ Number of non-negative integral solution for Eq. (ii)

$ \begin{aligned} & ={ }^{3+4-1} C_{4-1}={ }^6 C_3 \\ & =\frac{6 \times 5 \times 4}{3 \times 2 \times 1}=20 \end{aligned} $

Integers lying between 1000 and 10000 such that the sum of all the digits in each of those numbers becomes 30 .

$\therefore$ Coefficient of $x^{30}$ in the expansion of

$ \begin{aligned} \left(x^1+x^2\right. & \left.+\ldots+x^9\right)\left(x^0+x^1+\ldots+x^9\right)^3 \\ & =x\left(\frac{1-x^9}{1-x}\right)\left(\frac{1-x^{10}}{1-x}\right)^3 \\ & =x\left(1-x^9\right)\left(1-x^{10}\right)^3(1-x)^{-4} \end{aligned} $

$\Rightarrow$ Coefficient of $x^{29}$ in

$ \left(1-x^9\right)\left(1-x^{10}\right)^3(1-x)^{-4} $

$ \begin{aligned} & =\left(1-x^9\right)\left(1-3 x^{10}+3 x^{20}-x^{30}\right)(1-x)^{-4} \\ & =\left(1-x^9-3 x^{10}+3 x^{19}+3 x^{20}-3 x^{29}\right. \\ & \left.\quad \quad-x^{30}+x^{39}\right)(1-x)^{-4} \end{aligned} $

Here, $(1-x)^{-4}=\sum_{k=0}^{\infty} k+{ }^3 C_3 x^k$

$\therefore$ Coefficient of $x^{29}$ from $1 \cdot(1-x)^{-4}$

$ ={ }^{29+3} C_3={ }^{32} C_3=4960 $

Coefficient of $x^{20}$ from $-x^9 \cdot(1-x)^{-4}$

$ =-{ }^{20+3} C_3=-{ }^{23} C_3=-1771 $

Coefficient of $x^{19}$ from $-3 x^{10} \cdot(1-x)^{-4}$

$ \begin{aligned} & =-3\left({ }^{19+3} C_3\right) \\ & =-3(1540)=-4620 \end{aligned} $

Coefficient of $x^{10}$ from $3 x^{19}(1-x)^{-4}$

(Means form is $3 x^{19} \Sigma^{k+3} C_3 x^k$ need

$ \begin{aligned} x^{29}, k & =10) \\ & =3^{10+{ }^3} C_3=3\left({ }^{13} C_3\right)=858 \end{aligned} $

Coefficient of $x^9$ from $3 x^{20} \cdot(1-x)^{-4}$

$ =3\left({ }^{9+3} C_3\right)=3(220)=660 $

Coefficient of $x^0$ from $-3 x^{29} \cdot(1-x)^{-4}$

$ =-3\left({ }^{0+3} C_3\right)=-3(1)=-3 $

$\therefore$ Total numbers

$ \begin{aligned} & =4960-1771-4620+858+660-3 \\ & =84 \end{aligned} $

Given word in most

Words start with $M=3!=6$

Words start with $\mathrm{O}=3!=6$

Words start with $\mathrm{SM}=2!=2$

Words start with $\mathrm{SO}=2!=2$

Words start with STM = 1 ! = 1

Words start with STO is STOM

Rank of STOM is $(6+6+2+2+1)+1 =18$

The total number of ways to choose any subset of the 5 options $=2^5=32$

These 32 subsets include choosing 2, 3, 4 or 5 options.

But choosing options 0 and 1 are not allowed.

Now, subsets with 0 options

$ ={ }^5 C_0=\frac{5!}{0!5!}=1 $

Subsets with 1 options

$ ={ }^5 C_1=\frac{5!}{1!4!}=5 $

So, total subsets with options 0 and 1

$ =1+5=6 $

Thus, the number of ways in which the student can answer that questions

$ =32-6=26 $

Let the triangle be $A B C$ with point $(x, y)$,

where $0 \leq x \leq 4$ and $0 \leq y \leq 4$

So, the points $(x, y)$ are $(0,0),(0,1)$, $(0,2),(0,3),(0,4),(1,0),(1,1),(1,2)$, $(1,3),(1,4),(2,0),(2,1),(2,2),(2,3)$, $(2,4),(3,0),(3,1),(3,2),(3,3),(3,4)$, $(4,0),(4,1),(4,2),(4,3),(4,4)$.

Total coordinates are 25 , but we need to select only 3 .

Now, the number of triangles with sets of 3 non-collinear points. Total number of ways to choose 3 points $={ }^{25} C_3$

But, we must subtract the number of collinear triplets.

Case I Horizontal lines

Number of ways to choose 3 collinear points

$ \Rightarrow \quad{ }^5 C_3=10 $

So, in 5 rows $=5 \times 10=50$

Case II Vertical lines

Same as above $\rightarrow 5$ columns

$ =5 \times 10=50 $

Now, diagonal from bottom-left to top-right

$=$ Length $3: 2$ such diagonals + Length

4 : 2 diagonals + Length 5:1 diagonal

$ \begin{aligned} = & 2\left({ }^3 C_3\right)+2\left({ }^4 C_3\right)+{ }^5 C_3 \\ & 2(1)+2(4)+10 \\ = & 2+8+10=20 \end{aligned} $

And, diagonals from top-left to bottom-right $=20$

Total collinear triplets

$ =50+50+20+20=140 $

So, total number of triangles $={ }^{25} C_3-140$

$ \Rightarrow \quad 2300-140=2160 $

Total Alphabets are A, D, E, H, L, N Letters before H are A, D, E.

So, the number of letters before $\mathrm{H}=3 \times 5$ !

$ =3 \times 120=360 $

Letters before $E$ are A, D.

So, the number of letters before $E=2 \times 4$ !

$ =2 \times 24=48 $

Letters before L are $\mathrm{A}, \mathrm{D}$.

So, the number of letters before $L$

$ =2 \times 3!=12 $

Letters before N is D

So, the number of letters with N only

$ \Rightarrow \quad 1 \times 1!=1 $

So, total words $=360+48+12+1=421$

Now, only 1 letter remaining, that is D .

So, Rank of HELAND $=421+1=422$

Calculate the number excluding the digit 5 .

(a) 1-digit number

$ =8(1,2,3,4,6,7,8,9) $

Similarly,

(c) 3-digit numbers $=8 \times 9 \times 9=648$

(d) 4-digit numbers $=8 \times 9 \times 9 \times 9=5832$

Total numbers between 1 and 10000 excluding the digit 5

$ =8+72+648+5832=6560 $

Hence, required numbers $=9999-6560$

$ =3439 $

Total number of people $=5+4=9$

So, total arrangements $=9$ !

To find $\alpha$ Let the particular man and woman be treated as a single unit or block when they are together.

Remaining 7 peoples +1 block $=8$ units

$ \therefore \quad \alpha=2 \times 8! $

and $\beta=9$ ! $-\alpha=9$ ! $-2 \times 8$ !

Hence, $\alpha: \beta=\frac{2 \times 8!}{8}: \frac{9!-2 \times 8!}{8!}$

$ =2: 7 $

Number of ways to choose 4 couples out of $4={ }^4 C_4=1$

From 4 couples, pick 1 person per couple $=2^4=16$

Team formula married couples $=2^2=4$

Hence, required ways $=16-4=12$

Total 8-digit numbers formed by $1,2,3,4,5,6,7,8$ is 8 !

Total 8 -digit numbers formed by $0,2,3$, $4,5,6,7,9$ is $7 \times 7$ !

Total 8 -digit numbers formed by $1,0,3$, $4,5,6,8,9$ is $7 \times 7$ !

Total 8 -digit numbers formed by $1,2,0$, $4,5,7,8,9$ is $7 \times 7$ !

Total 8-digit numbers formed by $0,1,2$, $3,5,6,7,8$ is $7 \times 7$ !

$ \begin{aligned} \text { Total } & =8!+28 \times 7! \\ & =(8+28) 7! \\ & =36 \times 7! \end{aligned} $