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

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A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]

29 Aug 2004, 19:22

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?

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29 Aug 2004, 20:12

two options - 3Men 3 Women
or 2Men 4 Women

first one, (8C3 - 6) * (5C3)

second (8C2 - 1) * (5C4)

add 'em up, comes out too

50 * 10 + 27 * 5 = 635

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29 Aug 2004, 20:31

635/1716
Total # of ways to select a committee of 3 women 3 men, 2 of which do not serve together: 5C3 * (8C3 - 6C1) = 10*50 = 500
Total # of ways to select a committee of 4 women 2 men, 2 of which do not serve together: 5C4 * (8C2 - 6C0) = 5*27 = 135
Total # ways to pick 6 out of 13 people: 13C6 = 1716
Total favorable: (500+135) / 1716 = 635/1716
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29 Aug 2004, 20:38

Oh well, I calculated the probability of ...
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29 Aug 2004, 22:13

Contains at least 2 men and 3 women, so can be 2m4w or 3m3w.

Consider 2m4w case:

8C2 combinations of men - 28
Each pair of men, can be paired upw ith 5C4 = 5 combinatins of women
So 2m4w cases can be 140
but 2 men refuse to be on the same team. number of combinations these men can be in the same team is 5 (assuming m1,m2 cannot be on the same team, but if they are, they will be paired with on of the 5 combinations of women)
therefore, 2m4w = 140-5 = 135

Now consider 3m3w
8C3 combinatins of men - 56
Each combination can be paired up with 5C3= 10 combinations of women
So 3m3w case can be 560
2 men refuse to be on the same team, number of combinations is 6*10 = 60
therefore, 3m3w = 500

So total number of different combinations = 500+135 = 635

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29 Aug 2004, 22:30

Yes, You all are right and Paul is right + :wink:

[#permalink] 29 Aug 2004, 22:30

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

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

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