Jitendra hegde (Student) (50 Points)
25 July 2012
Deepak Gupta
(CA Student)
(15922 Points)
Replied 25 July 2012
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.
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I don't know how the solution has been done. I have just googled the answer and pasted the same here for your reference.
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