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A Committee of 6 is chosen from 8 men and 5 women, so as to

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Joined: 20 Dec 2004
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A Committee of 6 is chosen from 8 men and 5 women, so as to [#permalink] New post 25 Dec 2007, 12:56
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A Committee of 6 is chosen from 8 men and 5 women, so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?

A- 3510
B- 2620
C- 1404
D- 700
E- 635
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 [#permalink] New post 25 Dec 2007, 13:45
My guess would be E, by POE.

the committee can be 2 men, 4 women or 3 men and 3 women.

2 men out of 8 = C(8,2) = 28
4 women out of 5 = C(5,4) = 5

28 x 5 = 140

3 men out of 8 = C(8,3) = 56
3 women out of 5 = C(5,3) = 10

56 x 10 = 560

Add the possible combos
560 + 140 = 700

But we need to eliminate the combos that the 2 men who refused to server together.

Therefore, less than 700

of the 28 combos of 2 men, 1 combo contains the 2 men who don't serve together ---> 1 x 5 (4 women combo) = 5

of the 56 combos of 3 men, if 2 men don't serve together, then C(6,1) = 6
6 x 10 (3 women combo) = 60

60 + 5 = 65

700-65=635
  [#permalink] 25 Dec 2007, 13:45
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