Find the number of combinations and permutations of 4 letters taken from the word ‘EXAMINATION’
Question:

Find the number of combinations and permutations of 4 letters taken from the word ‘EXAMINATION’

Solution:

There are 11 letters in the word EXAMINATION, namely AA, NN, II, E, X, M, T and O.

The four-letter word may consist of

(i) 2 alike letters of one kind and 2 alike letters of the second kind

(ii) 2 alike letters and 2 distinct letters

(iii) all different letters

Now, we shall discuss the three cases one by one.

(i) 2 alike letters of one kind and 2 alike letters of the second kind:

There are three sets of 2 alike letters, namely AA, NN and II.

Out of these three sets, two can be selected in 3C2 ways.

So, there are 3C2 groups, each containing 4 letters out of which two are alike letters of one kind and two 2 are alike letters of the second kind.

Now, 4 letters in each group can be arranged in $\frac{4 !}{2 ! 2 !}$ ways.

$\therefore$ Total number of words that consists of 2 alike letters of one kind and 2 alike letters of the second kind $={ }^{3} C_{2} \times \frac{4 !}{2 ! 2 !}=3 \times 6=18$

(ii) 2 alike and 2 different letters:

Out of three sets of two alike letters, one set can be chosen in 3C1 ways.

Now, from the remaining 7 letters, 2 letters can be chosen in 7C2 ways.

Thus, 2 alike letters and 2 distinct letters can be chosen in $\left({ }^{3} C_{1} \times{ }^{7} C_{2}\right)$ ways.

So, there are $\left({ }^{3} C_{1} \times{ }^{7} C_{2}\right)$ groups of 4 letters each.

Now, the letters in each group can be arranged in $\frac{4 !}{2 !}$ ways.

$\therefore$ Total number of words consisting of 2 alike and 2 distinct letters $=\left({ }^{3} C_{1} \times{ }^{7} C_{2}\right) \times \frac{4 !}{2 !}=756$

(iii) All different letters:

There are 8 different letters, namely A, N, I, E, X, M, T and O. Out of them, 4 can be selected in 8C4 ways.

So, there are ${ }^{8} C_{4}$ groups of 4 letters each. The letters in each group can be arranged in $4 !$ ways.

$\therefore$ Total number of four-letter words in which all the letters are distinct $={ }^{8} C_{4} \times 4 !=1680$

$\therefore$ Total number of four-letter words $=18+756+1680=2454$