Permutations and Combinations

Here, n = 11 and subset have atmost 5 elements

$\therefore$ Total subset = (Subset with 0 element) + (Subset with 1 element) + (Subset with 2 element) + (Subset with 3 element) + (Subset with 4 element) + (Subset with 5 element)

$\begin{aligned} & =\left({ }^{11} C_0+{ }^{11} C_1\right)+\left({ }^{11} C_2+{ }^{11} C_3\right)+\left({ }^{11} C_4+{ }^{11} C_5\right) \\ & ={ }^{12} C_1+{ }^{12} C_3+{ }^{12} C_5 \quad\left[\because{ }^n C_r+{ }^n C_{r-1}={ }^{n+1} C_r\right]\end{aligned}$

$\begin{aligned} &^6 P_4+4 \cdot{ }^6 P_3 \\ &\left(\frac{6 !}{2 !}+4 \cdot \frac{6 !}{3 !}\right)=360+480 \quad\left[\because{ }^n P_r=\frac{n !}{(n-r) !}\right] \\ &=840\end{aligned}$

$\underline{B_1} G_1 \underline{B_2} G_2 \underline{B_3}$

$G_1$ and $G_2$ can be arranged in 2! ways. $B_1, B_2$ and $B_3$ can be arranged in 3! ways.

$\therefore$ Required number of arrangements

$=2 ! \times 3!$

$=12$

5 balls $\to$ 4 Tins

Each ball can be placed in 4 ways.

So, number of ways $=4\times4\times4\times4\times4=4^5$

$\begin{aligned} & \frac{1}{r+1}\left\{{ }^n P_{r+1}-{ }^{n-1} P_{r+1}\right\} \\ & \quad=\frac{1}{r+1}\left[\frac{n !}{(n-r-1) !}-\frac{(n-1) !}{(n-r-2) !}\right] \\ & \quad=\frac{1}{(r+1)} \cdot \frac{(n-1) !}{(n-r-2) !}\left(\frac{n}{n-r-1}-1\right)\end{aligned}$

$\begin{aligned} & =\frac{1}{(r+1)} \frac{(n-1) !}{(n-r-2) !} \frac{(r+1)}{(n-r-1)} \\ & =\frac{(n-1) !}{(n-r-1) !}={ }^{n-1} P_r\end{aligned}$

Total number of balls $=10$

Total number of ways of selecting four balls $={ }^{10} C_4=210$

Number of ways of selecting all green balls $={ }^4 C_4=1$

Number of ways of selecting 4 balls in which at least one black ball is selected $=210-1=209$

Total number of arrangements = 9!

Number of arrangements of best and worst paper together = 2! $\times$ 8!

Number of arrangements when best and worst are never together = 9! $-$ 2! $\times$ 8!

Total number of coins = 3

The number of different sums

= ³C₁ + ³C₂ + ³C₃

= 3 + 3 + 1 = 7

Number of white balls = 7

Number of black balls = 3 $\times$ W $\times$ W $\times$ W $\times$ W $\times$ W $\times$ W $\times$ W $\times$

$\times$ are the places in row where black balls can be placed, which are 8 in number for 3 black balls.

Number of arrangements $ = {}^8{C_3} = {{8 \times 7 \times 6} \over {1 \times 2 \times 3}} = 56$

Total number of ways of dividing n identical rings to r different girls, so that every girl gets at least one ring $={}^{n - 1}{C_{r - 1}}$

Here, we have to distribute 8 identical rings to 3 different girls.

This can be done in ${}^{8 - 1}{C_{3 - 1}} = {}^7{C_2} = {{7 \times 6} \over {1 \times 2}} = 21$ ways

Number of letters in the word REGULATIONS = 11

Number of possible arrangements = 11!

Except R and E₁ there are remaining 9 letters

Favourable number of arrangements

$ = {}^9{C_4} \times 4! \times 2! \times 6!$

Required probability $ = {{{}^9{C_4} \times 4! \times 2! \times 6!} \over {11!}} = {6 \over {55}}$

Two circles intersect maximum at two distinct points.

Now, two circles can be selected in ¹⁰C₂ ways.

The total number of points of intersection are ¹⁰C₂ $\times$ 2

$ = {{10 \times 9} \over {1 \times 2}} \times 2 = 90$

Given,

$ \begin{aligned} A & =\left\{a_n / a_n=\left[\begin{array}{cc} (-1)^{n / 2}\left(\frac{n}{2}\right), & n \text { is even } \\ (-1)^{\frac{n-1}{2}}\left(\frac{n-1}{2}\right), & n \text { is odd } \end{array}\right\}\right. \\ & \therefore A=\{0,-1,2,-3,4,-5, \ldots-25\} \\ & \therefore B=\{0,-1,2,-3,4,-5, \ldots, 24,-25\} \end{aligned} $

Here given integer $=12$, total integer $=26$

∴ Required permutation $=\frac{26!}{12!}$

Given all men are together

∴ Linear arrangement of $p$ men and $q$ women $=p!(q+1)$ ! and circular arrangement of $p$ men and $q$ women $=p!q!$

$ \Rightarrow \quad \frac{\alpha}{\beta}=\frac{p!(q+1)!}{p!q \mid!}=(q+1): 1 $

The number of positive integral solution of $x_1+x_2+x_3+x_4=10$ is ${ }^9 C_3=\frac{9 \times 8 \times 7}{1 \times 2 \times 3}=84$

∴ Statement I is false.

The exponent of 2 in 25! is

$ \begin{aligned} {\left[\frac{25}{2}\right]+\left[\frac{25}{4}\right] } & +\left[\frac{25}{8}\right]+\left[\frac{25}{16}\right] \\ & =12+6+3+1 \\ & =22 \end{aligned} $

The exponent of 5 in (25)! is

$ \left[\frac{25}{5}\right]+\left[\frac{25}{25}\right]=5+1=6 $

∴ Exponent of 10 in $25!=6$

Hence, (25)! $=10^6 \times k$

$ \Rightarrow \quad k \in \mathbf{N} $

∴ Statement II is true.

The total number of collection of atleast $(n+1)$ books from $(2 n+1)$ books is,

$ { }^{2 n+1} C_{n+1}+{ }^{2 n+1} C_{n+2}+\ldots{ }^{2 n+1} C_{2 n+1}=255 $

$ \begin{aligned} &\begin{aligned} \Rightarrow & & \frac{2^{2 n+1}}{2}-1 & =255 \\ \Rightarrow & & 2^{2 n} & =256 \\ \Rightarrow & & 2^{2 n} & =2^8 \\ \Rightarrow & & n & =4 \end{aligned}\\ &\text { ∴ Total number of books }\\ &\begin{aligned} & =2 n+1 \\ & =2 \times 4+1=9 \end{aligned} \end{aligned} $

Given, word ATRAPATRAM

Here, $\mathrm{A} \rightarrow 4$ times

$ \begin{aligned} & \mathrm{T} \rightarrow 2 \text { times } \\ & \mathrm{R} \rightarrow 2 \text { times } \\ & \mathrm{P} \rightarrow 1 \text { time } \\ & \mathrm{M} \rightarrow 1 \text { time } \end{aligned} $

(i) All A's are together,

Then it will be consider as one letter and remaining 6 letters and $\mathrm{l}^{\prime} \mathrm{A}^{\prime} \mathrm{s}$ (including 4 A's) will be consider as 7 letters.

∴ Number of arrangement

$ =\frac{7!}{2!2!}=x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

(ii) no two A's are together. such that-T-R-P-T-R-M-

∴ Number of arrangement

$ \begin{aligned} & ={ }^7 C_4 \times \frac{6!}{2!2!}=y \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ \therefore & \quad x+y=\frac{7!}{2!2!}+{ }^7 C_4 \times \frac{6!}{2!2!} \\ = & \frac{7 \times 6!}{2!2!}+{ }^7 C_3 \times \frac{6!}{2!2!}=\frac{6!}{2!2!}\left(7+{ }^7 C_3\right) \\ = & \frac{6!}{2!2!}(7+35)=\frac{6!}{2!2!}(42) \end{aligned} $

Number between 1 and 10000 are formed using digit 2 and 3 only one and digit 4 twice

$ \text { one- } \text { digit }=\{2,3,4\}=3 $

Two digit

$ =\{23,24,32,34,42,43,44\}=7 $

Three digit number

$ \begin{aligned} & =\{234,243,244,324,342,344 \\ & 423,432,424,434,442,443\}=12 \end{aligned} $

Four digit number

$ \begin{array}{r} =\{2344,2434,2443,3244,3424, \\ 3442,4234,4324 \ldots\} \end{array} $

Rank of $324=14$

Rank of $4324=3+7+12+9=31$

$ \therefore \quad x-y=31-14=17 $

Total number of three digit numbers $=5 \times 5 \times 4=100$

Total number of five digit numbers

$ =5 \times 5 \times 4 \times 3 \times 2=600 $

∴ Total number of required numbers

$ =100+600=700 $

$ \begin{aligned} &\text { Number of ways of selections }\\ &={ }^{12} C_1+{ }^{12} C_2+{ }^{12} C_3+{ }^{12} C_4 \end{aligned} $

$ \begin{aligned} & =12+\frac{12 \times 11}{2 \times 1}+\frac{12 \times 11 \times 10}{3 \times 2 \times 1} +\frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} \\ & =12+66+220+495=793 \end{aligned} $

Possible even digits are 2, 4, 6, 8, 0.

Case I Number has digits 4, 6, 8, 0.

(Here, sum of digits 18, divisible by 3)

Number of arrangements = 3 $\times$ 3!

[1st place can be filled using 4, 6, 8]

= 3 $\times$ 6 = 18

Case II Number has digits 2, 4, 6, 0

(Here, sum of digits is 12, divisible by 3)

1st place cannot be filled by 0.

Number of arrangements = 3 $\times$ 3! = 18

$\therefore$ Number of numbers = 18 + 18 = 36

Let's denote the three digits of the number as $a $, $b$, and $ c $, and thus the number we're looking at is $100a + 10b + c$. According to the property described, the middle digit $b $ is the arithmetic mean of $a $ and $c$, which means:

$ b = \frac{a+c}{2} $

In order for $b$ to be an integer, $a + c$ must be an even number since the sum of two numbers divided by 2 can only result in an integer if the sum is even.

Since $a$, $b$, and $c$ are digits, they can only take values from 0 to 9, inclusive. However, for $a$, it cannot be 0 because we are looking for a three-digit number, which means $a$ must be from 1 to 9.

Let's consider all possible even $a + c$ sums:

$a = 1$ allows $c = 1, 3, 5, 7, 9$ (since $1 + c$ must be even and $c$ can be from 0 to 9).

$a = 2$ allows $c = 0, 2, 4, 6, 8$ (for the same reason).

$a = 3$ allows $c = 1, 3, 5, 7, 9$.

...

$a = 9$ allows $c = 1, 3, 5, 7, 9$.

So for each odd $a$, there are 5 possible values for $c$, and for each even $a$, there are also 5 possible values for $c$, totaling 5 choices for $c$ for each value of $a$.

Since $a$ can be from 1 to 9, there are, in total:

$ 9 \times 5 = 45 $

possible three-digit numbers where the middle digit is the arithmetic mean of the first and last digits. Therefore, the correct answer is:

Option B: 45

With four 6-sided dics, there are possible combinations that can be rolled. Out of these, only 80 sum to 10 as the table shows,

$ 6 \times 6 \times 6 \times 6=6^4=1296 $