Permutations and Combinations

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

$=2^{2 n+1}$

$ \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

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$

To make a no. divisible by 3 we can use the digits 1,2,5,6,7 or 1,2,3,5,7.

Using 1,2,5,6,7, number of even numbers is = 4 × 3 × 2 × 1 × 2 = 48

Using 1,2,3,5,7, number of even numbers is = 4 × 3 × 2 × 1 × 1 = 24

Required answer is 72.

Case 1 :

Among four bags from one bag 4 balls are taken. Number of ways of chosing one bag from 4 bags = ${}^4{C_1}$

From this bag number of ways of taking 4 balls are

(a) 3 Red and 1 Blue balls, which can be chosen in ${}^3{C_3} \times {}^2{C_1}$ ways.

(b) 2 Red and 2 Blue balls, which can be chosen in ${}^3{C_2} \times {}^2{C_2}$ ways.

$ \therefore $ Total number of ways of choosing this 4 balls from this bag = ${}^4{C_1}$$\left( {{}^3{C_3} \times {}^2{C_1} + {}^3{C_2} \times {}^2{C_2}} \right)$

Now, two balls are taken from remaining three bags.

From each bag two balls can be taken in ${{}^3{C_1} \times {}^2{C_1}}$ ways.

So, for three bags two balls can be taken in ${\left( {{}^3{C_1} \times {}^2{C_1}} \right)^3}$ ways.

$ \therefore $ Total number of ways of chosing 10 balls from these four bags

= ${}^4{C_1}$$\left( {{}^3{C_3} \times {}^2{C_1} + {}^3{C_2} \times {}^2{C_2}} \right)$ $ \times $ ${\left( {{}^3{C_1} \times {}^2{C_1}} \right)^3}$

Case 2 :

Among four bags from two bags 3 balls are taken. Number of ways of chosing two bags from 4 bags = ${}^4{C_2}$

From each bag, number of ways of taking 3 balls are

(a) 2 Red and 1 Blue balls, which can be chosen in ${}^3{C_2} \times {}^2{C_1}$ ways.

(b) 1 Red and 2 Blue balls, which can be chosen in ${}^3{C_1} \times {}^2{C_2}$ ways.

So, for two bags three balls can be taken in ${\left( {{}^3{C_2} \times {}^2{C_1} + {}^3{C_1} \times {}^2{C_2}} \right)^2}$ ways.

$ \therefore $ Total number of ways of choosing this 3 balls from this two bags = ${}^4{C_2}$${\left( {{}^3{C_2} \times {}^2{C_1} + {}^3{C_1} \times {}^2{C_2}} \right)^2}$

Now, two balls are taken from remaining two bags.

From each bag two balls can be taken in ${{}^3{C_1} \times {}^2{C_1}}$ ways.

So, for two bags two balls can be taken in ${\left( {{}^3{C_1} \times {}^2{C_1}} \right)^2}$ ways.

$ \therefore $ Total number of ways of chosing 10 balls from these four bags

= ${}^4{C_2}$${\left( {{}^3{C_2} \times {}^2{C_1} + {}^3{C_1} \times {}^2{C_2}} \right)^2}$ $ \times $ ${\left( {{}^3{C_1} \times {}^2{C_1}} \right)^2}$

From Case 1 and Case 2, total number of ways of chosing 10 balls from these 4 boxes so that from each box at least one red ball and one blue ball are chosen

= ${}^4{C_1}$$\left( {{}^3{C_3} \times {}^2{C_1} + {}^3{C_2} \times {}^2{C_2}} \right)$ $ \times $ ${\left( {{}^3{C_1} \times {}^2{C_1}} \right)^3}$

+ ${}^4{C_2}$${\left( {{}^3{C_2} \times {}^2{C_1} + {}^3{C_1} \times {}^2{C_2}} \right)^2}$ $ \times $ ${\left( {{}^3{C_1} \times {}^2{C_1}} \right)^2}$

$ \begin{aligned} & =4(5)(6)^3+6(3 \times 2+3)^2(6)^2 \\\\ & =4320+17496 \\\\ & =21816 \end{aligned} $

Given, ${ }^m P_r-{ }^{(m-1)} P_r=a \cdot{ }^{(m-1)} P_s$

$ \Rightarrow{ }^m P_r={ }^{(m-1)} P_r+a^{(m-1)} P_s \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Also, we know that

$ { }^n P_r={ }^{n-1} P_r+r \cdot{ }^{n-1} P_{r-1} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

On comparing Eqs. (i) and (ii), we get

$ \begin{aligned} & a=r \\ \text { and } s=r-1 & \\ \Rightarrow \quad & a-s=r-(r-1) \\ \Rightarrow \quad & a-s=1 \end{aligned} $

Given, word is TSEAMCET

In the word, T and E occur twice. So, we have S, A, M, C, TT, EE. We have 6 distinct letter.

Way 14 letters selected and all are distinct.

$ { }^6 C_4={ }^6 C_2=\frac{6 \times 5}{2 \times 1}=15 $

Way 22 letters are alike and 2 are distinct

$ { }^2 C_1 \times{ }^5 C_2=2 \times \frac{5 \times 4}{2 \times 1}=20 $

Way 32 letters are alike and rest 2 are also alike $\quad{ }^2 C_2=1$

So, the total number of ways

$ =15+20+1=36 $

Given, $a+b+c=5$ and $a, b, c \in N L$ is least value of $2^a 3^b 5^c$ and $M$ is the greatest value of $2^a 3^b 5^5$.

Possible pairs of $(a, b, c)$

$ \begin{array}{r} =(1,1,3),(1,3,1),(3,1,1),(1,2,2) \\ (2,1,2),(2,2,1) \end{array} $

Value for $(1,1,3)=\left(2 \cdot 3 \cdot 5^3\right)=750$

Value for $(1,3,1)=\left(2 \cdot 3^3 \cdot 5\right)=270$

Value for $(3,1,1)=\left(2^3 \cdot 3 \cdot 5\right)=120$

Value for $(1,2,2)=\left(2^1 \cdot 3^2 \cdot 5^2\right)=450$

Value for $(2,1,2)=\left(2^2 \cdot 3^1 \cdot 5^2\right)=300$

Value for $(2,2,1)=\left(2^2 \cdot 3^2 \cdot 5^1\right)=180$

$ \begin{aligned} L=2^3 \cdot 3 \cdot 5 & =120, M=2 \cdot 3 \cdot 5^3=750 \\ \therefore \quad M-L & =2 \cdot 3 \cdot 5^3-2^3 \cdot 3 \cdot 5 \\ & =2 \cdot 3 \cdot 5(25-4) \\ & =2 \cdot 3 \cdot 5 \cdot 21=2 \cdot 3^2 \cdot 5 \cdot 7 \end{aligned} $

Since, $360=2^3 \cdot 3^2 \cdot 5^1$

The number of $(+\mathrm{ve})$ divisors which are multiple of 3

$ =(3+1)(2+1-1)(1+1)=4 \times 2 \times 2=16 $

The total possible arrangement of the word 'LINEAR' is $6!=720$.

Number of possible arrangement of the word 'LINEAR' in which $E$ and $A$ come together is $5!2!=120 \times 2=240$

Number of possible arrangement in the word 'LINEAR' in which $E$ and $A$ come together and $N$ and $R$ come together $=2!2!4!=2 \times 2 \times 24=96$

Number of possible arrangement in which $E$ and $A$ come together but $N$ and $R$ do not come together $=240-96=144$

Given, 15 lines are concurrent at a point $P$.

Now, if a line is not passing through a point $P$, then the triangle can be formed if those 15 times, concurrent at 2 points, so the required number of ways of forming a triangle $={ }^{15} C_2$

$ =\frac{15 \times 14}{1 \times 2}=15 \times 7=105 $

Case I When $x=1$, then $y+z=11$

There are 4 triplets in this case.

i.e. $(1,2,9),(1,3,8),(1,4,7),(1,5,6)$

Case II When $x=2$, then $y+z=10$

There are 2 triplets in this case i.e. $(2,3,7)$ and $(2,4,6)$.

Case III When $x=3$, then $y+z=9$

There is only one triplet in this case i.e. $(3,4,5)$.

∴ Total number of triplets $=4+2+1=7$

There are total 12 questions.

A candidate can choose 8 questions from 12 in the following pattern

$ \begin{array}{ccc} \hline \text { Ist } & \text { Ilnd } & \text { Illrd } \\ \hline 2 & 2 & 4 \\ \hline 2 & 3 & 3 \\ \hline 2 & 4 & 2 \\ \hline 3 & 2 & 3 \\ \hline 3 & 3 & 2 \\ \hline 4 & 2 & 2 \\ \hline \end{array} $

$ \begin{aligned} &\text { ∴ Total number of ways }\\ &\begin{aligned} = & \left({ }^4 C_2 \times{ }^4 C_2 \times{ }^4 C_4\right)+\left({ }^4 C_2 \times{ }^4 C_3 \times{ }^4 C_3\right) \\ & +\left({ }^4 C_2 \times{ }^4 C_4 \times{ }^4 C_2\right)+\left({ }^4 C_3 \times{ }^4 C_2 \times{ }^4 C_3\right) \\ & +\left({ }^4 C_3 \times{ }^4 C_3 \times{ }^4 C_2\right)+\left({ }^4 C_4 \times{ }^4 C_2 \times{ }^4 C_2\right) \\ = & 6 \times 6 \times 1+6 \times 4 \times 4+6 \times 1 \times 6+4 \times \\ & 6 \times 4+4 \times 4 \times 6+1 \times 6 \times 6 \\ = & 36+96+36+96+96+36=396 \end{aligned} \end{aligned} $

Total ways $=10$ !

Number of ways all three are together

$ =3!\times 8! $

∴ Required number of ways

$ =10!-3!\times 8!=84 \times 8! $

$6=2 \times 3$

Highest power of 2 in 72!

$ \begin{aligned} =\left[\frac{72}{2}\right]+\left[\frac{72}{2^2}\right]+\left[\frac{72}{2^3}\right]+\left[\frac{72}{2^4}\right] +\left[\frac{72}{2^5}\right]+\left[\frac{72}{2^6}\right] \end{aligned} $

$ \begin{aligned} &=36+18+9+4+2+1=70\\ &\text { Highest power of } 3 \text { in } 72 \text { ! }\\ &\begin{aligned} & =\left[\frac{72}{3}\right]+\left[\frac{72}{3^2}\right]+\left[\frac{72}{3^3}\right]=24+8+2=34 \\ & \therefore \text { Exponent of } 6 \text { in } 72!=34 \\ & \qquad\left(\because 2^{34} \times 3^{34}=6^{34}\right) \end{aligned} \end{aligned} $

$ \text { Taking } 1 \text { at one's place, } $

So, two numbers can be formed using the above combination.

So, two numbers can be formed using the above combination.

Total numbers that can be formed when 1 is at the unit's place

$ =2+2+2+2=8 $

Now, taking 3 at one's place,

Total numbers that can be formed when 3 is at the unit's place $=2+2+2+2=8$

Now, taking 5 at one's place,

Total numbers that can be formed when 5 is at the unit's place $=2+2+2+2=8$ ∴ Total number of 3 -digit odd numbers divisible by 3 that can be formed using the digits $1,2,3,4,5,6$ when repetition is not allowed $=8+8+8=24$

(A) The number of ways of not selecting $(n-r)$ things from $n$ different things $={ }^n C_{n-r}$

(B) $(n-r+1) \cdot{ }^n C_{r-1}$

$ \begin{aligned} & =(n-r+1) \times \frac{n!}{(r-1)!(n-r+1)!} \\ & =\frac{(n-r+1) \times n!}{(r-1)!(n-r+1)(n-r)!} \\ & =\frac{n!}{(r-1)!(n-r)!}=r \frac{n!}{r!(n-r)!}=r \cdot{ }^n C_r \end{aligned} $

(C) The number of ways of selecting atleast $(n-r)$ things from $n$ different things

$ \begin{aligned} & ={ }^n C_{n-r}+{ }^n C_{n-r+1}+{ }^n C_{n-r+2} +\ldots+{ }^n C_n \\ & ={ }^n C_n+{ }^n C_{n-1}+{ }^n C_{n-2}+\ldots \quad+{ }^n C_{n-r+1}+{ }^n C_{n-r} \end{aligned} $

$ \begin{aligned} & =1+{ }^n C_1+{ }^n C_2+\ldots+{ }^n C_r \\ & \text { (D) }(n-r)\left({ }^{n-1} C_{r-1}+{ }^{n-1} C_r\right) \\ & =(n-r)\left({ }^n C_r\right) \\ & {\left[\because{ }^n C_r+{ }^n C_{r+1}={ }^{n+1} C_{r+1}\right]} \\ & =(n-r) \times \frac{n!}{(n-r)!r!}=\frac{n!}{(n-r-1)!r!} \\ & =(r+1) \frac{n!}{(r+1)!(n-r-1)!}=(r+1){ }^n C_{r+1} \end{aligned} $