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 $

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} $

Given word, MATHEMATICS

$ \begin{array}{ll} 2 \mathrm{M} & 1 \mathrm{H} \\ 2 \mathrm{~A} & 1 \mathrm{E} \\ 2 \mathrm{~T} & 1 \mathrm{I} \\ & 1 \mathrm{C} \\ & 1 \mathrm{~S} \end{array} $

Number of words formed using 4 letters in which two are of same kind and 2 are different

$ \begin{aligned} & ={ }^3 C_1 \times{ }^7 C_2 \times \frac{4!}{2!} \\ & =3 \times 7 \times 3 \times 4 \times 3 \\ & =27 \times 28 \\ & =756 \end{aligned} $

Digits $-0,5,6,7,8$ and 9

No. greater than 6000

Case I 4 digit numbers greater than 6000

$ \begin{aligned} & \bar{\uparrow} \bar{\uparrow} \bar{\uparrow} \bar{\uparrow}=240 \\ & 4543 \end{aligned} $

$ \begin{aligned} &\text { Case II 5-digit numbers }\\ &\begin{aligned} & \bar{\uparrow} \bar{\uparrow} \bar{\uparrow} \bar{\uparrow} \bar{\uparrow}=600 \\ & 55432 \end{aligned} \end{aligned} $

Case III 6-digit numbers

$ \begin{aligned} & \bar{\uparrow} \bar{\uparrow} \bar{\uparrow} \bar{\uparrow} \bar{\uparrow} \bar{\uparrow}=600 \\ & 554321 \end{aligned} $

∴ Total $=1440$

∴ Required case

$=(2$ particular are in Ist group $)+$ (2 particular are in 3rd group) + (one is in Ist group and other is in 3rd group)

$ \begin{aligned} & =\frac{13!}{1!5!7!}+\frac{13!}{3!5!5!}+\frac{13!}{2!5!6!}(24) \\ & =\frac{13!}{7!5!}+\frac{13!}{6!5!}+\frac{13!}{5!6!}=\frac{13!(1+7+7)}{7!5!} \\ & =\frac{13 \times 12 \times 11!\times 15}{7!\times 5 \times 4 \times 3!}=\frac{117(11!)}{7!3!} \end{aligned} $

The digit are 1 to 9 . Each such number is strictly increasing sequence of digits, i.e., digits don't repeat and are in order.

e.g., 12, 135, 1239 and soon.

So, total such numbers in the number of increasing digit sequence of length 2 to 4.

We choose subsets of digits from 1 to 9 of lengths 2,3 and 4 in increasing order.

$\therefore 2$-digits numbers

$ ={ }^9 C_2=\frac{9!}{2!7!}=\frac{9 \times 8}{2}=36 $

3-digits numbers

$ ={ }^9 C_3=\frac{9!}{3!6!}=\frac{9 \times 8 \times 7 \times 6!}{6 \times 6!}=84 $

$ \begin{aligned} &\begin{aligned} 4 \text {-digits numbers } & ={ }^9 C_4=\frac{9!}{4!5!} \\ & =\frac{9 \times 8 \times 7 \times 6 \times 5!}{4 \times 3 \times 2 \times 5!}=126 \end{aligned}\\ &\text { So, total such digits }\\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=36+84+126=246 \end{aligned} $

Given, word is AGAIN,

So the letters in alphabatical order : A, A, G, I, N

Words starting with $A=\frac{4!}{2!}=12$

Words starting with $G=\frac{4!}{2!}=12$

Words starting with $I=\frac{4!}{2!}=12$

Words starting with $N=\frac{4!}{2!}=12$

So, total words $=12+12+12+12=48$

Thus, the 50th word starts with $N$.

49th word = NAAGI

$\therefore 50$ th word $=$ NAAIG.

We select exactly I wicket keeper from 4 .

So, number of ways $={ }^4 C_1=4$

Now, we have 10 players left to select from batsmen, bowlers and all-rounders with the condition.

Batsmen $\geq 4$, Bowlers $\geq 3$, All-rounders $\geq 2$

So, total of these three $=10$

So, number of batsmen, $B$ from 4 to 6 , number of bowlers, $L$ (say)from 3 to 6 and number of all-rounder, $A$ (say) from 2 to 4 .

$\therefore B+L+A=10$ (after 1 wicket keeper)

Now, the possible combinations for, (B, L, A) are

$(4,4,2),(4,3,3),(5,3,2)$

Total number of ways for 10 players

$ \begin{aligned} & =\left({ }^6 C_4{ }^6 C_4{ }^4 C_2+{ }^6 C_4{ }^6 C_3{ }^4 C_3+{ }^6 C_5{ }^6 C_3{ }^4 C_2\right) \\ & =(15 \times 15 \times 6+15 \times 20 \times 4+6 \times 20 \times 6) \\ & =(1350+1200+720)=3270 \end{aligned} $

$\therefore$ Total number of ways for 11 players

$ \begin{aligned} & =4 \times 3270 \\ & =13080 \end{aligned} $

Number of combinations of choosing 4 digits out of 4

$ ={ }^5 C_4=5 $

Each 4-digit set can be arranged in 4! ways $=24$ ways

So, total number of 4 -digit numbers

$ =5 \times 24=120 $

In each group, each digit appears

$ \frac{24}{4}=6 \text { times } $

So, contribution per group

$ =\text { sum of digits } \times 6 \times(1000+100+10+1) $

Now, compute total sum of digits for all 5 groups

$ \begin{array}{r} \begin{array}{r} (2+3+5+8)+(1+3+5+8) \\ +(1+2+5+8)+(1+2+3+8) \\ +(1+2+3+5) \end{array} \\ =18+17+16+14+11=76 \end{array} $

$ \text { Hence, total sum } \begin{aligned} & =76 \times 6666 \\ & =506616 \end{aligned} $

Total possible ordered pairs $(p, q)$ of first 50 natural numbers

$ =50 \times 50=2500 $

Out of these, exactly half will satisfy $p>q$ and there are 50 pairs where $p=q \{(1,1),(2,2) \ldots(50,50)\}$

So, number of pairs with $p \neq q$

$ =2500-50=2450 $

and for $p>q$

$ =\frac{2450}{2}=1225 $

Fathers $(F)=5$, Teachers $(T)=6$, students $(S)=4$

for, $T+F+S=7, T \geq 0$, thus $F \geq 1, S \geq 1$ total combination ( $T, F, S$ )

$ (4,1,2),(4,2,1),(5,1,1),(3,2,2) $

Case $\mathbf{I}(T=4, F=1, S=2)$

$ \begin{aligned} & ={ }^6 C_4 \cdot{ }^5 C_1 \cdot{ }^4 C_2 \\ & =15 \times 5 \times 6=450 \end{aligned} $

$ \text { } \begin{aligned} \mathbf{Case \,I I}\,\,(T & =4, F=2, S=1) \\ & ={ }^6 C_4 \cdot{ }^5 C_2 \cdot{ }^4 C_1 \\ & =15 \times 10 \times 4=600 \end{aligned} $

Case III ( $T=5, F=1, S=1$ )

$ \begin{aligned} & ={ }^6 C_5 \cdot{ }^5 C_1 \cdot{ }^4 C_1 \\ & =6 \times 5 \times 4=120 \end{aligned} $

Case IV ( $T=3, F=2, S=2$ )

$ \begin{aligned} & ={ }^6 C_3 \cdot{ }^5 C_2 \cdot{ }^4 C_2 \\ & =20 \times 10 \times 6=1200 \end{aligned} $

Hence, total number of ways

$ \begin{aligned} & =450+600+120+1200 \\ & =2370 \end{aligned} $