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 ^{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 2 are alike letters of the second kind.

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

∴ 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 ^{3}*C*_{1} ways.

Now, from the remaining 7 letters, 2 letters can be chosen in ^{7}*C*_{2} ways.

Thus, 2 alike letters and 2 distinct letters can be chosen in

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

∴ Total number of words consisting of 2 alike and 2 distinct letters =\[\left( {}^3 C_1 \times {}^7 C_2 \right) \times \frac{4!}{2!} = 756\]

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

^{8}

*C*

_{4}ways.

So, there are

^{8}

*C*

_{4}groups of 4 letters each. The letters in each group can be arranged in \[4!\]ways.

∴ Total number of four-letter words in which all the letters are distinct =\[{}^8 C_4 \times 4! = 1680\]

∴ Total number of four-letter words = 18 + 756 + 1680 = 2454