find no. of combinations and permutations of the 4 letter taken from the word "EXAMINATION"

The word examination consists of 11 letters -
  (AA), (II), (NN), E, X, M, T, O.
The following combinations are possible:
(a) 2 alike, 2 alike:³C2 = 3 ways
(b) 2 alike, 2 different:³C1×7C2 = 63 ways
(c) all 4 different: 8C4 = 70 ways
Hence required number of combinations = 3 +63 +70 = 136.

Number of permutations:
In (a), the number of permutations = 3×4!/[2!2!] = 18
In (b), the number of permutations = 63×4!/2! = 756
In (c), the number of permutations = 70×4!= 1680
Hence required number of permutations = 18 +756 +1680 = 2454.

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