The number of words of four letters that can be formed from the letters of the word EXAMINATION is

There are 11 letters A, A; I, I; N, N; E, X, M, T, O. For the selection of 4 letters we have the following possibilities:

(i) 2 alike, 2 alike
(ii) 2 alike, 2 different
(iii) All four different

(i) There are 3 pairs of 2 letters. So, the number of ways of selection of 2 pairs is 3 C, and permutation of these 4 letters is 4! / 2! 2! Therefore, the number of words in this case is ${ }^3{C}_2\times 4!/2!2!=18$. 

(ii) We have to select one pair from 3 pairs and 2 distinct letters from remaining 7 distinct letters. For illustration, let us select both A, A; then we have I, N, E, X, M, T, O, i.e., 7 as remaining distinct letters. Hence, the number of selections is ${ }^3{C}_1{\times }^7{C}_2$ and these 4(2 same, 2 distinct) can be permuted in 4!/2! ways. Therefore, number of words is ${ }^3{C}_1{\times }^7{C}_2\times 4!/2!=3\times 21\times 12=756$. 

(iii) There are 8 distinct letters so number of words of 4 letters is ${ }^8{C}_4\times 4!=1680$. By sum rule, the total number of words is 18 + 756 + 1680 = 2454.