Find the number of ways of selecting 4 letters from the word examination.
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245
EXAMINATION has 11 letters, and in which 'A', 'I' and 'N', all occur twice. 11C4, would have been fine if all letters were distinct.
So, we have E, X, M, T, O, (AA), (II), (NN). 8 distinct letters.
1. 4 letters selected, which are all distinct: 8C4 = 70
2. 2 letters alike, and 2 distinct (eg: AAEX) = 3C1 x 7C2 = 63
3. 2 letters alike, and 2 letters alike (eg: AAII) = 3C2 = 3
So answer is, 70 + 63 + 3 = 136.
@skb
So, we have E, X, M, T, O, (AA), (II), (NN). 8 distinct letters.
1. 4 letters selected, which are all distinct: 8C4 = 70
2. 2 letters alike, and 2 distinct (eg: AAEX) = 3C1 x 7C2 = 63
3. 2 letters alike, and 2 letters alike (eg: AAII) = 3C2 = 3
So answer is, 70 + 63 + 3 = 136.
@skb
Answered by
38
3C2 + 8C4 + 3C1×7C2
3 + 70 + 63 =136
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