Question

Find the number of combinations and permutations of $4$ letters taken from the word 'EXAMINATION'.

Answer 1

The word EXAMINATION has letters E, X, A, M, I, N, T, O where A, I, N repeat twice.
So, four letter word may consist of
(i) 2 alike letters of one kind and 2 alike of letter of the second kind.
There are three sets of alike letters, namely AA, NN, II.
Out of these three sets, two can be selected in ${}^3{C}_2$ ways.
So, there are ${}^3{C}_2$ groups, each containing $4$ letters out of which two are alike letters of one kind and two are like letters of the second kind.
Now, $4$ letters in each group can be arranged in $\frac{4!}{2!\times 2!}$
Therefore, the total number of four letters words are ${}^3{C}_2\times \frac{4!}{2!\times 2!}=3\times 6=18$
Since, The total number of letter $=11$
Similar way,
(ii) All different ${}^8{C}_4\times 4!=\frac{8!}{4!}$

As selecting $4$ out of $8$ different letters is ${}^8{C}_4$
Arrangement of these $4$ are done by $4!$

Thus, $\frac{8!}{4!}=\frac{8\times 7\times 6\times 5\times 4!}{4!}$

$=56\times 30$

$=1680$

(iii) $2$ alike and $2$ distinct letter:

${}^3{C}_1{\times }^7{C}_2=3\times \frac{7\times 6\times 5!}{5!\times 2}=\frac{3\times 7\times 6}{2}=378$ [ Since, ${}^n{C}_r{=}^n{C}_{n-r}$]

$\therefore$ Total number of ways in which $4$ letter words are formed $=1680+378+18+378=2454$ ways.