Permutations and Combinations

$a+b=23\lambda+2$

$\lambda=0,1,2,$ ...., but $\lambda$ cannot be even as $a+b$ is odd

$\lambda=1$ $(a, b)\to12$ pairs

$\lambda=3$ $(a,b)\to35$ pairs

$\lambda=5$ $(a,b)\to42$ pairs

$\lambda=7$ $(a,b)\to19$ pairs

$\lambda=9$ $(a,b)\to0$ pairs

$ \vdots $

Total $=12+35+42+19=108$

Lets arrange the letters of OUGHT in alphabetical order.

G, H, O, T, U

Number divisible by 3 = 300

Number divisible by 4 = 225

Number divisible by 12 = 75

Number divisible by 48 = 18

Total required number = 300 + 225$-$ 75 $-$ 18 = 432

$\therefore$ Option (1) is correct.

$\because$ In every row and every column there would be exactly one 1 and four zeroes.

Number of matrices $={ }^{5} C_{1} \cdot{ }^{4} C_{1} \cdot{ }^{3} C_{1} \cdot{ }^{2} C_{1} \cdot{ }^{1} C_{1}=$ 120

$ \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} $

To determine the number of ways to arrange the letters in "SUNITHA" such that the vowels occupy the first, middle, and last positions, consider the following:

The vowels in "SUNITHA" are A, U, and I.

Arrangement of Vowels:

There are 3 vowels that need to be placed in specific positions: the first, middle, and last slots.

These 3 vowels can be arranged in these 3 slots in $3!$ (factorial of 3) ways.

Arrangement of Consonants:

The remaining letters, S, N, T, and H, which are consonants, can occupy the remaining 4 slots.

These 4 consonants can be arranged among themselves in $4!$ (factorial of 4) ways.

Calculating the Total Number of Arrangements:

Multiply the number of arrangements for vowels by the number of arrangements for consonants:

$ 3! \times 4! = 6 \times 24 = 144 $

Thus, there are a total of 144 ways to arrange the letters in "SUNITHA" with vowels occupying the first, middle, and last positions.

To determine the number of four-digit numbers that can be formed using the digits 0, 1, 2, 3, 4, and 5 without repeating any digits, consider the following:

Thousand's Place: This digit cannot be 0 (as it would not be a four-digit number), leaving us with 5 options: 1, 2, 3, 4, or 5.

Hundred's Place: After choosing a digit for the thousand's place, there are still 5 remaining digits (including 0) that can be used, since repetition is not allowed.

Ten's Place: Once the thousand's and hundred's places are filled, we are left with 4 digits to choose from.

Unit's Place: Finally, for the unit's place, there are 3 remaining digits available.

Thus, the total number of four-digit numbers that can be formed is calculated by multiplying the number of choices for each place:

$ 5 \times 5 \times 4 \times 3 = 300 $

For a number to be divisible by 4, last 2 digits must be divisible by 4 . Possible cases $12,16,24,32,36,52,56$, 64, 72, 76

To make 6-digit numbers from given digits $1,3,5, 7$ and 9 , we must repeat only one digit and we can done the same in ${ }^5 C_1$ ways.

Now, the arrangement of these 6 -digits in which two are identical is 6!/ 2.

So, required numbers of different 6-digit numbers

$={ }^5 C_1 \frac{6!}{2!}=5 \times \frac{720}{2}=1800$

Given word is LETTER, having vowels E, E and consonants L, T, T, R. Now, the number of ways to arrange the consonants are $\frac{4!}{2!}=12$

Now, we have five place to put vowel E, E.

So, number of ways to arrange vowels is ${ }^5 C_2=10$

So, number of required words $=12 \times 10=120$

By multinomial theorem, no. of ways to distribute 30 identical candies among four children C₁, C₂ and C₃, C₄

= Coefficient of x³⁰ in (x⁴ + x⁵ + .... + x⁷) (x² + x³ + .... + x⁶) (1 + x + x² ....)²

= Coefficient of x²⁴ in ${{(1 - {x^4})} \over {1 - x}}{{(1 - {x^5})} \over {1 - x}}{{{{(1 - {x^{31}})}^2}} \over {{{(1 - x)}^2}}}$

= Coefficient of x²⁴ in $(1 - {x^4} - {x^5} + {x^9}){(1 - x)^{ - 4}}$

$ = {}^{27}{C_{24}} - {}^{23}{C_{20}} - {}^{22}{C_{19}} + {}^{18}{C_{15}} = 430$