Permutations and Combinations

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 $=421+1=422$

We need to find the sum of all the 4-digit numbers formed using the digits $\{0,3,6,9\}$ without repetition. Note that since the number is 4-digit, the thousands place cannot be $0$.

Step 1. Count the Valid Numbers

Thousands Place: Can be chosen from $\{3,6,9\}$ — $3$ possibilities.

Other Places: After choosing the thousands digit, the remaining three digits can be arranged in $3!$ ways (i.e. $6$ ways).

Thus, the total number of valid numbers is:

$ 3 \times 6 = 18. $

Step 2. Sum by Place Value Contribution

We calculate the contribution to the total sum from each digit place (thousands, hundreds, tens, and units).

(a) Thousands Place

Allowed digits: $3, 6, 9$.

Frequency: Each digit (3, 6, 9) appears in the thousands place $6$ times.

Contribution:

$ \text{Sum}_{\text{thousands}} = (3+6+9) \times 6 \times 1000 = 18 \times 6 \times 1000 = 108000. $

(b) Hundreds Place

For the hundreds place, consider the following:

Case 1: When thousands digit is $3$, the remaining digits are $\{0,6,9\}$. In these $6$ numbers, each of $0$, $6$, and $9$ appears exactly $2$ times in the hundreds place.

Case 2: When thousands digit is $6$, the remaining digits are $\{0,3,9\}$. Each appears $2$ times.

Case 3: When thousands digit is $9$, the remaining digits are $\{0,3,6\}$. Each appears $2$ times.

Thus, the total frequency in the hundreds place is:

$0:$ appears $2+2+2 = 6$ times.

$3:$ appears in cases 2 and 3: $2+2 = 4$ times.

$6:$ appears in cases 1 and 3: $2+2 = 4$ times.

$9:$ appears in cases 1 and 2: $2+2 = 4$ times.

Contribution:

$ \text{Sum}_{\text{hundreds}} = (0\cdot6 + 3\cdot4 + 6\cdot4 + 9\cdot4) \times 100. $

First calculate the inner sum:

$ 3\cdot4 + 6\cdot4 + 9\cdot4 = 4(3+6+9) = 4 \times 18 = 72. $

So,

$ \text{Sum}_{\text{hundreds}} = 72 \times 100 = 7200. $

(c) Tens Place

By symmetry (since after the thousands digit is fixed, the remaining three digits are uniformly permuted among the hundreds, tens, and units places), the distribution for the tens place is identical to that for the hundreds place.

Contribution:

$ \text{Sum}_{\text{tens}} = 72 \times 10 = 720. $

(d) Units Place

Again, by similar reasoning, the units place has the same digit frequency as the hundreds and tens places.

Contribution:

$ \text{Sum}_{\text{units}} = 72 \times 1 = 72. $

Step 3. Total Sum

Add the contributions from all four places:

$ \text{Total Sum} = 108000 + 7200 + 720 + 72. $

Calculate step-by-step:

$ 108000 + 7200 = 115200, $

$ 115200 + 720 = 115920, $

$ 115920 + 72 = 115992. $

Final Answer

The sum of all the 4-digit numbers formed is 115992.

Thus, the correct option is:

Option B – 115992.

Every things has two choices to go in to lst box and there are 6 such things there are two cases in which all the things go into lst or 2 nd box.

$\because$ Total number of ways $=2^6-2=62$

To find the number of ways in which the number 831600 can be divided into two factors that are relatively prime, we start by determining the prime factorization of 831600:

$ M = 831600 = 2^4 \times 3^3 \times 5^2 \times 7 \times 11 $

The number of different prime factors is 5: $2, 3, 5, 7, 11$.

When dividing a number into two relatively prime factors, each prime factor must entirely go to one factor or the other. The general formula to calculate the number of ways to partition these prime factors into two groups is given by:

$ 2^{n-1} $

where $n$ represents the number of distinct prime factors. Thus, for 831600:

$ 2^{5-1} = 2^4 = 16 $

Therefore, there are 16 different ways to split 831600 into two factors that are relatively prime.

Given, word Probability

Number of letter in probability

P,R,O,A,L,T,Y,BB,II

(i) 4 letter are selected alone different

i.e. $ { }^{9} C_{4}=\frac{9!}{4!5!}=126 $

(ii) Two letter or alike other are different

$ 2 C_{1} \times 8 C_{2}=\frac{2 \times 8!}{2!6!}=56 $

(iii) Two letter are alike

$ 2 C_{2}=1 $

Total number of sample space

$ =126+56+1=183 $

Favorable case in which at least one letter is repeated

$ \text { i.e., } \quad 56+1=57 $

$ \therefore \quad \text { Reqmend probability }=\frac{57}{183}=\frac{19}{61} $

To find the number of ways to arrange all the letters of the word 'COMBINATIONS' around a circle such that no two vowels are together, let's break it down step-by-step.

Identify Vowels and Consonants:

Vowels: O, O, I, I, A (total of 5 vowels)

Consonants: C, M, B, N, T, N, S (total of 7 consonants)

Arrangement of Consonants:

Since the arrangement is in a circle, we arrange the consonants in $(7-1)! = 6!$ ways because in circular permutations, we fix one position to break the circle and linearly order the rest.

Interleaving Vowels:

Arranging the consonants creates 7 gaps (before the first consonant, between each pair, and after the last consonant) to place the vowels. We need to select 5 gaps for the vowels, ensuring no two vowels are adjacent, which is crucial since we want them not together.

Arrangement of Vowels:

Arrange the vowels O, O, I, I, A in these gaps.

The arrangement of the vowels is affected by repetition. The formula for arranging 'n' items where there are repetitions is given by $\frac{n!}{k1! \cdot k2! \cdot \ldots \cdot kr!}$, where $k_i$ represents the number of times each repeated letter occurs.

The necessary calculation for the vowels: $\frac{5!}{2! \cdot 2!}$ because O and I each repeat twice.

Total Arrangements:

Combine the arrangements of consonants and vowels as follows:

$ \text{Total ways} = 6! \times \frac{5!}{2! \times 2!} $

By further expanding this calculation based on the repetitions:

$ \frac{7! \times 6!}{(2!)^4} $

This gives the total number of arrangements of the letters in 'COMBINATIONS' around a circle, ensuring that no two vowels are adjacent.

Explanation

To find the difference between the number of odd and even numbers formed by the digits 3, 5, 6, 7, and 8, where the numbers are greater than 6000 and less than 10000, we consider the following process:

Odd Numbers:

Set I: When the first digit (thousands place) is 6, 7, or 8, and the last digit (units place) is 3.

Choices:

3 choices for the first digit (6, 7, 8).

3 choices for the second digit.

2 choices for the third digit.

1 choice for the last digit (3).

Combinations:

$ 3 \times 3 \times 2 \times 1 = 18 $

Set II: Similar to Set I, but the last digit (units place) is 5.

Choices:

3 choices for the first digit (6, 7, 8).

3 choices for the second digit.

2 choices for the third digit.

1 choice for the last digit (5).

Combinations:

$ 3 \times 3 \times 2 \times 1 = 18 $

Set III: The first digit (thousands place) is 6 or 8, and the last digit is 7.

Choices:

2 choices for the first digit (6, 8).

3 choices for the second digit.

2 choices for the third digit.

1 choice for the last digit (7).

Combinations:

$ 2 \times 3 \times 2 \times 1 = 12 $

Total Odd Numbers:

$ 18 + 18 + 12 = 48 $

Even Numbers:

Set I: The first digit (thousands place) is 7 or 8, and the last digit (units place) is 6.

Choices:

2 choices for the first digit (7, 8).

3 choices for the second digit.

2 choices for the third digit.

1 choice for the last digit (6).

Combinations:

$ 2 \times 3 \times 2 \times 1 = 12 $

Set II: The first digit (thousands place) is 6 or 7, and the last digit (units place) is 8.

Choices:

2 choices for the first digit (6, 7).

3 choices for the second digit.

2 choices for the third digit.

1 choice for the last digit (8).

Combinations:

$ 2 \times 3 \times 2 \times 1 = 12 $

Total Even Numbers:

$ 12 + 12 = 24 $

Calculating the Difference:

The required difference between the number of odd and even numbers is:

$ 48 - 24 = 24 $

This result can be expressed as $ {}^4 P_3 $.

There are 4 possibilities for inviting the relatives to the party:

3 ladies from the husband's side and 3 gentlemen from the wife's side:

Total number of ways:

$ = { }^{4} C_{3} \times { }^{4} C_{3} = 16 $

3 gentlemen from the husband's side and 3 ladies from the wife's side:

Total number of ways:

$ = { }^{3} C_{3} \times { }^{3} C_{3} = 1 $

2 ladies and 1 gentleman from the husband's side and 2 gentlemen and 1 lady from the wife's side:

Total number of ways:

$ = \left( { }^{4} C_{2} \times { }^{3} C_{1} \right) \times \left( { }^{3} C_{1} \times { }^{4} C_{2} \right) = 324 $

1 lady and 2 gentlemen from the husband's side, and 2 ladies and 1 gentleman from the wife's side:

Total number of ways:

$ = \left( { }^{4} C_{1} \times { }^{3} C_{2} \right) \times \left( { }^{3} C_{2} \times { }^{4} C_{1} \right) = 144 $

The total number of ways to invite 3 of the man's relatives and 3 of the wife's relatives is:

$ = 16 + 1 + 324 + 144 = 485 $

Alphabets present in the word 'COLLEGE' are C, O, L, E, G

Alphabets in ascending order are

C, E, G, L, O

Number of words starts with $ \mathrm{CE}=\frac{5!}{2!} $

Number of words starts with CG $ =\frac{5!}{2!2!} $

Number of words starts with $ \mathrm{CL}=\frac{5!}{2!} $

Number of words starts with $ \mathrm{COE}=\frac{4!}{2!} $

Number of words starts with COG $ =\frac{4!}{2!2!} $

Number of words starts with COLE $ =3 $ !

Number of words starts with COLG $ =\frac{3!}{2!} $

Number of words starts with

COLLEE $ =1 $ !

Next words in COLLEGE

$ \therefore $ Required number of ways

$ \begin{array}{l} =\frac{5!}{2!}+\frac{5!}{2!2!}+\frac{5!}{2!}+\frac{4!}{2!}+\frac{4!}{2!2!}+3!+\frac{3!}{2!}+1+1 \\\\ =60+30+60+12+6+6+3+2 \\\\ =179 \end{array} $

Case I Let one digit be 1 or 7 , then sum of the other two digits be $ 2,5,8 $, 11 or 14 or $ 3 x+2 $ Therefore, the 2 digits can be $ (3,5),(5,9) $.

$ \therefore $ The possible number are 8 .

Case II Let one digit be 3 or 9 the sum of other two digits can be multiple of 3 . The other 2 digit can be $ (1,5) $ or $ (7,5) $ $ \therefore $ The possible numbers are 8 .

Case III Let one digit can be 5 , then sum of other digit can be $ 1,4,7,10,13 $ or 16.

$ \therefore $ The other 2 digit numbers can be $ (1,3),(1,9),(3,7),(7,6) $

$ \therefore $ The possible number can be 8 in $ ^{-} $ number.

Hence, the required number are

$ 8+8+8=24 $

We have, part A has 7 questions, Part B has 5 questions and part $ C $ has 3 questions.

Case I 4 questions from part $ A $, then 3 questions from part $ B $ and $ C $

Number of ways

$ ={ }^{7} C_{4} \times\left[{ }^{5} C_{3}+{ }^{5} C_{2} \times{ }^{3} C_{1}+{ }^{5} C_{1} \times{ }^{3} C_{2}\right] $

$ =\frac{7 \times 6 \times 5}{1 \times 2 \times 3}[10+30+15] $

$ =35 \times 55=1925 $

Case II 3 questions from Part $ A $, then 4 questions from part $ B $ and $ C $.

$ \begin{array}{l} \therefore \text { Number of ways } \\\\ ={ }^{7} C_{3} \times\left[{ }^{5} C_{3} \times{ }^{3} C_{1}+{ }^{5} C_{2} \times{ }^{3} C_{2}\right] \\\\ =35 \times[30+30]=35 \times 60=2100 \end{array} $

Case III 2 questions from part $ A $, then 5 question from part $ A $ and $ B $

Number of ways $ ={ }^{7} C_{2}\left[{ }^{5} C_{3} \times{ }^{3} C_{2}\right] $

$ =21 \times 10 \times 3=630 $

$ \therefore $ Required number of ways

$ =1925+2100+630=4655 $

The numbers will be divisible by 6 if and only if the unit digit of numbers is an even number and the sum of digits is divisible by 3 .

Collection of 4 -digits whose sum is divisible by 3 .

(1236) $\rightarrow \ldots \ldots \ldots . .2$ or $6 \rightarrow 3 \times 2 \times 1 \times 2=12$

$(1245) \rightarrow \ldots \ldots \ldots 2$ or $4 \rightarrow 3 \times 2 \times 1 \times 2=12$

$(1356) \rightarrow \ldots \ldots \ldots 6 \rightarrow 3 \times 2 \times 1 \times 1=6$

$(2346) \rightarrow \ldots \ldots \ldots .2$ or 4 or $6 \rightarrow$

$ 3 \times 2 \times 1 \times 3=18 $

(3456) $\rightarrow \ldots \ldots . .4$ or $6 \rightarrow 3 \times 2 \times 1 \times 2=12$

$ \text { Total }=12+12+6+18+12=60 $

$\therefore$ The number of numbers which are divisible by 4 are 60.

Given that the number of circular permutation of 9 distinct things taken 5 at a time is $n_1$.

$ \begin{aligned} \therefore \quad n_1 & ={ }^9 C_5(5-1)! \\\\ & =\frac{9!}{5!(9-5)!} \times 4!=9 \times 8 \times 7 \times 6 \end{aligned} $

Also, given that the number of linear permutation of 8 distinct things taken 4 at a time is $n_2$.

$ \begin{aligned} \therefore n_2={ }^8 C_4(4!) & =\frac{8!}{4!(8-4)!} \times 4! \\\\ & =8 \times 7 \times 6 \times 5 \end{aligned} $

Now, $\frac{n_1}{n_2}=\frac{9 \times 8 \times 7 \times 6}{8 \times 7 \times 6 \times 5}=\frac{9}{5}$

We know that

The number of ways in which $n$ different things can be distributed to $m$ person so that no person gets all the things is same as the number of functions (whose range is not a singleton set) from set $A$ (containing $n$ elements) to set $B$ (containing $m$ elements).

$\therefore$ The number of ways in which 4 different things can distributed to 6 person so that no person gets all the things

$ \begin{aligned} & =6^4-6=6(216-1) \\\\ & =6(215)=1290 \end{aligned} $

We have,

$ 2,3,5,7 $

No repetition

Number of given digits $=n=4$

Sum of digits = 17

Sum of all four digits formed

$ \begin{aligned} & =(n-1)!(\text { Sum of digits) (11111 ......... n times) } \\\\ & =3!(17)(1111) \\\\ & =6 \times 17 \times 1111 \\\\ & =113322 \end{aligned} $

Given that there are 15 identical gold coins to be distributed among 3 persons.

Also, each person gets at least 3 coins $\Rightarrow$ total $=3 \times 3=9$ coins

$\therefore 9$ coins are already distributed in 1 ways. we now have 6 coins to be distributed between 3 people.

$ \begin{aligned} & \Rightarrow x_1+x_2+x_3=6, x_i \geq 0 \\\\ & \text { Number of } \end{aligned} $

Number of solutions is given by

$ \begin{aligned} & { }^{n+r-1} C_{r-1} \\\\ & \Rightarrow \quad n=6, r=3 \\\\ & \therefore \text { Reanired solution } \end{aligned} $

$\therefore$ Required Number $={ }^{6+3-1} C_{3-1}$

$ ={ }^8 C_2=28 $

4 letter words from

ACCOMMODATION

There are different letters

"ACCOMMODATION"

namely,

A, C, D, I, M, N, O, T

$\begin{array}{llllllll}2 & 2 & 1 & 1 & 2 & 1 & 3 & 1\end{array}$

Case (I) all 4 letters of the word are different numbers of combinations

$ ={ }^8 c_4=70 $

Case (II) 2 letters are same and 2 are different.

$ ={ }^4 c_1 \cdot{ }^7 c_2-6=78 $

( $\because$ when we select 2 same letter from 0 and 3rd letter also from, 0 then this need to be subtracted and such cases are 6)

Case (III) When 3 letters are same

$ =1 .{ }^7 c_1=7 $

Case (IV) When 2 letters are same and other 2 are also same.

$ ={ }^4 c_1 \times{ }^3 c_1=12 $

Hence, total number of combination

$ =70+78+7+12=167 $

We have,

$ \begin{aligned} & 2 \frac{c_1}{c_0}+4 \frac{c_2}{c_1}+6 \frac{c_3}{c_2}+\ldots .+2 n \frac{c_n}{c_{n-1}}=650 \\\\ & \Rightarrow \quad \sum_{r=1}^n 2 r \frac{{ }^n c_r}{{ }^n c_{r-1}}=650 \end{aligned} $

$\begin{aligned} & \Rightarrow \quad 2 \sum_{r=1}^n r\left[\frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r-1)!(n-r+1)!}}\right]=650 \\\\ & \Rightarrow 2 \sum_{r=1}^n r\left[\begin{array}{l}\frac{n!}{r \cdot(r-1)!(n-r)!} \\\\ \times \frac{(r-1)!(n-r+1) \cdot(n-r)!}{n!}\end{array}\right]\end{aligned}$

$ \begin{array}{ll} =650 \\\\ \Rightarrow & 2 \Sigma(n-r+1)=650 \\\\ \Rightarrow & 2\left[n^2+n-\frac{n(n+1)}{2}\right]=650 \\\\ \Rightarrow & n^2+n-650=0 \\\\ \Rightarrow & (n+26)(n-25)=0 \\\\ \Rightarrow & n=25 \end{array} $

Hence, ${ }^n c_2={ }^{25} c_2=300$

$ \begin{aligned} &\\ &\begin{aligned} & \text { } 0,3,6,7,8 \text { and } 9 \\ & \frac{6,7,8,9}{4}-\frac{}{4!}-\frac{3}{1}=96 \\ & \frac{6,8,9}{3}-\frac{}{4!}-\frac{7}{1}=72 \\ & \frac{6,7,8}{3} \times 4!\times-\frac{9}{1}=72 \\ & \text { Total ways }=96+72+72 \\ & =240 \end{aligned} \end{aligned} $

6 men +4 women

Arrangement of 5 men and 3 women around at table

$ (5+3-1)!\times{ }^8 C_2 \times 2! $

{2 gaps needed out of available 8 gaps }

$ =7!\times 8 \times 7=7 \times 8! $

If polygon has $n$ sides, then number of diagonals is

$ \begin{aligned} & & { }^n C_2-n & =35 \\ \Rightarrow & & \frac{n(n-1)}{2}-n & =35 \\ \Rightarrow & & n^2-n-2 n & =70 \\ \Rightarrow & & n^2-3 n-70 & =0 \\ \Rightarrow & & n^2-10 n+7 n-70 & =0 \\ \Rightarrow & & n(n-10)+7(n-10) & =0 \\ & \Rightarrow & n=10 \text { or } n=-7 \Rightarrow n & =10 \end{aligned} $

Thus, there are 10 vertices.

Now, if $A$ and $B$ are two vertices, then the number of triangle formed by $A B$ as one of its side is equal to 8 .

We know that, any triangle can be obtained by joining any 3 points not in the same straight line. Thus, number of triangles obtained from 10 different point, number 3 of which are collinear $={ }^{10} C_3=120$

Triangles obtained from 4 points $={ }^4 C_3=4$ and triangle obtained from remaining 6 points $={ }^6 C_3=20$

So, required number of triangles $=120-20-4=96$

Case I A student chooses 4 out of first five questions and six out of the remaining 8 questions.

$ \text { ∴ Total number of ways }={ }^5 C_4 \times{ }^8 C_6 $

Case II A student chooses 5 out of first five questions and 5 out of remaining 8 questions.

$ \text { ∴ Total number of ways }={ }^5 C_5 \times{ }^8 C_5 $

∴ Total number of ways

$ \begin{aligned} & =\left({ }^5 C_4 \times{ }^8 C_6\right)+\left({ }^5 C_5 \times{ }^8 C_5\right) \\ & =(5 \times 28)+(1 \times 56)=140+56=196 \end{aligned} $

The number of words start with

$ \mathrm{D}=\frac{5!}{2!}=60 $

The number of words start with

$ E=\frac{5!}{2!}=60 $

The number of words start with

$ I=\frac{5!}{2!2!}=30 $

The number of words start with

$ \mathrm{ND}=\frac{4!}{2!}=12 $

The number of words start with

$ \mathrm{NE}=\frac{4!}{2!}=12 $

Now, next word be 'NIDDEE'.

Thus, the rank of NIDDEE be

$ 60+60+30+12+12+1=175 $

The sum of digits $1,2,3,5,6$ and 8 is 25 .

Since, required numbers are divisible by 3 but not by 6 . So, the sum of digits must be divisible by 3 and the numbers must be odd.

The sum of digits will be divisible by 3 only, when digits be $2,3,5,6$ and 8 .

Now,

$ \begin{aligned} &\text { Thus, the number of required number be }\\ &4 \times 3 \times 2 \times 1 \times 2=48 \end{aligned} $

$ \begin{aligned} &\text { Required number of ways }\\ &\begin{array}{r} ={ }^7 C_2{ }^6 C_1{ }^5 C_1{ }^4 C_1+{ }^7 C_1{ }^6 C_2{ }^5 C_1{ }^4 C_1+ \\ { }^7 C_1{ }^6 C_1{ }^5 C_2{ }^4 C_1+{ }^7 C_1{ }^6 C_1{ }^5 C_1{ }^4 C_2 \end{array} \end{aligned} $

$ \begin{aligned} &=21 \times 6 \times 5 \times 4+7 \times 15 \times 5 \times 4+ \\ & 7 \times 6 \times 10 \times 4+7 \times 6 \times 5 \times 6 \\ &=2520+2100+1680+1260=7560 \end{aligned} $

We know that

$ { }^{n-1} P_r+r \cdot{ }^{n-1} P_{r-1}={ }^n P_r $

Then,

$ \begin{gathered} { }^n P_{r+1}+r^2{ }^{n-1} P_{r-1}+(r+1){ }^{n-1} P_r+r \cdot{ }^{n-1} P_{r-1} \\ ={ }^n P_{r+1}+r\left(r{ }^{n-1} P_{r-1}+{ }^{n-1} P_r\right)+\left({ }^{n-1} P_r+r{ }^{n-1} P_{r-1}\right) \\ ={ }^n P_{r+1}+r \cdot{ }^n P_r+{ }^n P_r \\ ={ }^n P_{r+1}+(r+1){ }^n P_r={ }^{n+1} P_{r+1} \end{gathered} $

The number of ways of arranging $n$ boys and $n$ girls in a row $=n!\times n!\times 2!$ Also, the number of ways $n$ boys and $n$ girls can be arranged such that no two girls are together and no two boys are together $=n!\times n!\times 2!$

$12600=2^3 \times 3^2 \times 5^2 \times 7$

For multiples of 3

$ \begin{aligned} 12600 & =3\left[2^3 \times 3 \times 5^2 \times 7\right] \\ n_1 & =(3+1)(1+1)(2+1)(1+1) \\ & =4 \times 2 \times 3 \times 2 \\ & =48 \end{aligned} $

For multiples of 14

$ \begin{aligned} 12600 & =14\left[2^2 \times 3^2 \times 5^2\right] \\ n_2 & =(2+1)(2+1)(2+1) \\ & =3 \times 3 \times 3=27 \\ \therefore \quad n_1+n_2 & =48+27=75 \end{aligned} $

Let three digits are $a, b$ and $c$ such that

$ a+b+c=10 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

where, $a \geq 1, b \geq 0, c \geq 0$

Let $d=a-1 \Rightarrow a=d+1$, where $d \geq 0$

$\Rightarrow d+b+c=9$, where $0 \leq d \leq 8$,

$ 0 \leq b \leq 9,0 \leq c \leq 9 $

Thus, total number of 3 -digits number is

$ \begin{aligned} { }^{9+3-1} C_{3-1}-1 & ={ }^{11} C_2-1 \\ & =\frac{11 \times 10 \times 9!}{9!\times 2}-1 \\ & =55-1=54 \end{aligned} $

The number of words start with E , $\mathrm{H}, \mathrm{M}, \mathrm{O}$ and $\mathrm{R}=5 \times 5!=5 \times 120=600$

The number of words start with TE $=4!=24$

The number of words start with

THE, THM and THO $=3 \times 3!=3 \times 6=18$

The number of words start with

THRE, THRM $=2 \times 2!=2 \times 2=4$

Next word will be "THROEM".

Hence, the position of word "THROEM" is $600+24+18+4+1=647$

A student is allowed to select at most $n$ books from a collection of $(2 n+1)$ books.

The number of ways, he select at least one book is 255.

This selection is given by combination concepts, we get

$ \begin{aligned} & { }^{2 n+1} C_1+{ }^{2 n+1} C_2+{ }^{2 n+1} C_3 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ & +\ldots+{ }^{2 n+1} C_n=255 \end{aligned} $

We know that, ${ }^n C_r={ }^n C_{n-r}$ and

$ \begin{aligned} & { }^{2 n+1} C_0+{ }^{2 n+1} C_1+{ }^{2 n+1} C_2+\ldots+ \\ & { }^{2 n+1} C_n=\frac{2^{2 n+1}}{2} \end{aligned} $

Thus, $2\left({ }^{2 n+1} C_0+{ }^{2 n+1} C_1+\ldots+{ }^{2 n+1} C_n\right)$

$ \begin{aligned} & \Rightarrow\left(^{2 n+1} C_0+{ }^{2 n+1} C_1+\ldots+{ }^{2 n+1} C_n\right) \\ & =\frac{2^{2 n} \cdot 2}{2}=2^{2 n} \end{aligned} $

$ \begin{aligned} \Rightarrow & \left(1+{ }^{2 n+1} C_1+{ }^{2 n+1} C_2+\ldots+{ }^{2 n+1} C_n\right) \\ & =2^{2 n} \\ & {\left[\because{ }^{2 n+1} C_0=1\right] } \end{aligned} $

$ \begin{aligned} \Rightarrow & { }^{2 n+1} C_1+{ }^{2 n+1} C_2+\ldots+{ }^{2 n+1} C_n \\ & =2^{2 n}-1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

Now, comparing the Eqs. (i) and (ii), we get

$ \begin{aligned} &\begin{aligned} 2^{2 \pi}-1 & =255 \\ \Rightarrow \quad 2^{2 \pi} & =256=2^8 \Rightarrow 2 n=8 \end{aligned}\\ &\left(\because a^m=a^n \Rightarrow a m=n, \quad \text { where } a>0\right)\\ &\Rightarrow \quad n=4\\ &\text { Then, the value of } n \text { is } 4 \text {. } \end{aligned} $