Question

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

Answer 1

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.
∴ 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. 
∴ 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 ⁸C₄ 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