Permutations and Combinations

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

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} $

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} $