Question

The number of permutations of $4$ letters that can be made out of the letters of the word 'EXAMINATION' is

• A. $2454$
• B. $2452$
• C. $2450$
• D. $1806$

Answer 1

The correct option is A

$2454$

Explanation for the correct option.

Finding the number of permutations of $4$ letters that can be made out of the letters of the word 'EXAMINATION' is

Given word: EXAMINATION

There are 11 letters out of which there are three letters that are repeated twice and rest five are single .

Repeated letters are two I's two A's and two N's and rest letters are E,X,M,T,O

Now we need to choose 4 letters out of the 11

($1$)All four letters are different.

We have $8$ different types of letters i.e. A, E, I, M, N, O, T, X.
Out of these $8$ letters $4$ can be arranged in :
$${}^8{p}_4={\displaystyle \frac{8!}{4!}}=8\times 7\times 6\times 5=1680$$

($2$)Two of them are alike and two are different.

Two alike letters can be chosen from one of the $3$ pairs (A,A), (I,I) and (N,N)
So total number of ways to choose one pair$=3$
To choose $2$ different letters we have $7$ options so total number of ways :
${}^7{C}_2={\displaystyle \frac{7!}{5!2!}}={\displaystyle \frac{7\times 6}{2}}=21$
Hence, total number of groups with $2$ alike and $2$ different $=63$
Each of the group have $4$ letter in which $2$ are same and $2$ are different and they can be arranged in themselves in ${\displaystyle \frac{4!}{2!}}=12$
Hence, total number of words is$=63\times 12$
$=756$

$\left(3\right)$Two alike of one kind and two of another kind.

Out of three pair of letter, we have to choose two of them.
This can be done in ${}^3{C}_2=3$ ways.
For example NNAA.
There will be arranged within the word also and they are arranged in:
${\displaystyle \frac{4!}{2!2!}}=6$
Hence we have $6\times 3=18$ words of this type.

Therefore $1680+756+18=2454$

Hence the correct answer is option (A)