Permutations and Combinations

$10^n C_2=3^{n+1} C_3$

$\begin{aligned} & \frac{10 n!}{(n-2)!2!}=\frac{3(n+1)!}{(n+1-3)!\times 3!} \\ & \Rightarrow \quad \frac{10 n!}{2}=\frac{3(n+1) n!}{3 \times 2} \\ & \Rightarrow \quad 10=n+1 \\ & \Rightarrow \quad n=9 \\ \end{aligned}$

Here's how to solve this problem:

First, let's understand the concepts involved:

• Collinear points: Points that lie on the same straight line.
• Triangle: A polygon with three sides.

To form a triangle, we need to choose 3 points out of the 10. We can do this using combinations:

Total number of ways to choose 3 points out of 10 = $^{10}C_3 = \frac{10!}{3!7!} = 120$

However, this count includes triangles formed using the 6 collinear points. We need to subtract these invalid triangles.

Number of ways to choose 3 points out of 6 collinear points = $^6C_3 = \frac{6!}{3!3!} = 20$

Therefore, the total number of valid triangles (those not formed by the collinear points) is:

$N = 120 - 20 = 100$

So the answer is Option C: 100.

Total marks $=$ Marks for first 3 papers + Marks for fourth paper $=3 n+2 n=5 n$

Candidate needs to get $3 n$ marks.

Let $x_1, x_2, x_3, x_4$ be the marks of candidate in I, II, III and IV paper, respectively.

Then, $x_1+x_2+x_3+x_4 =3 n\left(0 \leq x_1, x_2, x_3, \leq n\right.$ and $\left.0 \leq x_4 \leq 2 n\right)$

Using multinomial theorem,

$\begin{aligned} & \begin{array}{l} \text { Number of ways }=\text { Coefficient of } x^{3 n} \text { in } \\ \underbrace{\left(1+x+\ldots+x^n\right)^3}_{\text {GP series }} \underbrace{\left(1+x+\ldots x^{2 n}\right)}_{\text {GP Series }} \end{array} \\ & \Rightarrow\left(\frac{1-x^{n+1}}{1-x}\right)^3\left(\frac{1-x^{2 n+1}}{1-x}\right) \\ & \Rightarrow 1-x^{3(n+1)}-3 x^{n+1}\left(1-x^{n+1}\right)\left(1-x^{2 n+1}\right)(1-x)^{-4} \\ & \Rightarrow\left(1-x^{3(n+1)}-3 x^{n+1}+3 x^{2 n+2}\right)\left(1-x^{2 n+1}\right)(1-x)^{-4} \\ & \Rightarrow\left(1-x^{3(n+1)}-3 x^{n+1}+3 x^{2 n+2}-x^{2 n+1}+x^{5 n+4}\right. \\ & \left.\quad+3 x^{3 n+2}-3 x^{4 n+3}(1-x)^{-4}\right) \end{aligned}$

Finding the coefficient of $x^{3 n}$ using binomial expansion, we get

$\begin{gathered} { }^{3 n+3} C_3-3^{2 n+2} C_3+3^{n+2} C_3-{ }^{n+3} C_3 \\ =\frac{1}{6}(n+1)\left(5 n^2+10 n+6\right) \end{gathered}$

If set $ A $ contains $ m $ elements and set $ B $ comprises $ n $ elements, the number of injections from $ A $ to $ B $ can be defined as follows:

$ \left\{ \begin{array}{cc} 0, & \text{if } n < m \\ {}^n P_m, & \text{if } n \geq m \end{array} \right. $

This means that if the number of elements in set $ B $ is less than the number in set $ A $ ($ n < m $), there are no injections possible. Conversely, if $ n \geq m $, the number of injections is given by the permutation notation $ {}^n P_m $.

MULTIPLE has 8 letters, with 5 consonant (L repeated twice) and 3 vowels.

The vowels in 2nd, 5th and 8th places, these can be arranged in 3 ! ways $=6$ ways

The remaining 5 consonants can be arranged in $\frac{5!}{2!}$ ways $=\frac{120}{2}=60$ ways

Thus, the total number of ways $=6 \times 60=360$

Number of zeroes at the end of 4000!

$\begin{aligned} & =\left[\frac{4000}{5}\right]+\left[\frac{4000}{5^2}\right]+\left[\frac{4000}{5^{-3}}\right]+\left[\frac{4000}{5^4}\right]+\left[\frac{4000}{5^5}\right] \\ & =800+160+32+6+1=999 \end{aligned}$

Now, we need one more pair of 5 and 2 so that the number ends in exactly 1000 zeroes.

$4005 \leq N<4009$

$\therefore$ From the option (c), $N=4009$

Total number of things $=n$

Repetition of things is allowed and atmost $r$ can be taken at a time.

The number of ways of taking 1 thing at a time $=n$

The number of ways of taking 2 things at a time $=n^2$

The number of ways of taking 3 things at a time $=n^3$ ..........

The number of ways of taking $r$ things at a time $=n^3$

Total number of permutations of $n$ different things taken not more than $r$ at a time i.e. number of permutations of taking $1,2, \ldots ., r$ at a time.

$\begin{aligned} & =n+n^2+n^3+\ldots .+n^{\prime} \\ & =\frac{n\left(n^2-1\right)}{n-1} \quad [\because ~\text{sum of GP}] \end{aligned}$

There are 21 points on the circle, denoted as $ n = 21 $. A chord is formed by connecting any two points on the circle.

Therefore, the number of possible chords is given by the combination formula $ { }^n C_2 $:

$ { }^{21} C_2 = \frac{21 \times 20}{2 \times 1} = 210 $

Hence, 210 chords can be drawn through the 21 points on the circle.

$\begin{aligned} & \text { Number of sides }=n \\ & \text { Number of diagonals }=560 \\ & \because \text { Number of diagonals }=\frac{n(n-3)}{2} \\ & \Rightarrow \quad 560=\frac{n(n-3)}{2} \\ & \Rightarrow \quad 1120=n^2-3 n \\ & \Rightarrow \quad n^2-3 n-1120=0 \\ & \Rightarrow \quad n^2-35 n+32 n-1120=0 \\ & \Rightarrow \quad n(n-35)+32(n-35)=0 \\ & \Rightarrow \quad(n-35)(n+32)=0 \\ & \Rightarrow \quad n=35,-32 \\ & \qquad n \neq-32 \\ & \therefore \quad n=35. \end{aligned}$

If the total number of letters is $n$, then the number of ways in which $r$ letters goes into wrong envelopes $={ }^n C_{n-r} D_r$

The number of ways in which atleast two of the letters are in wrong envelopes

$\begin{aligned} & ={ }^6 C_{6-2} D_2+{ }^6 C_{6-3} D_3+{ }^6 C_{6-4} D_4 \\ & +{ }^6 C_{6-5} D_5+{ }^6 C_{6-6} D_6 \quad[\because r=2,3,4,5,6] \\ & =\sum_{r=2}^6{ }^6 C_{6-}, D_r \quad\left[\text { where, } D^r=r!\left(\sum_{i=0}^r \frac{(-1)^r}{i!}\right)\right] \\ \end{aligned}$

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}}$