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PS- Committee [#permalink] New post 07 Dec 2008, 03:04
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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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Re: PS- Committee [#permalink] New post 07 Dec 2008, 15:33
vivektripathi wrote:
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

D.
men = 8
women = 5
total = 13
required combinations = (2 men and 4 women) + (3 men and 3 women)
required combinations = (8c2 x 5c4) + (8c3 x 5c3) = (28x5) + (56 x 10) = 700
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Re: PS- Committee [#permalink] New post 07 Dec 2008, 17:20
i think it should be 635, since 2 men dont want to serve together..

700 without such restriction
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Re: PS- Committee [#permalink] New post 07 Dec 2008, 18:04
Indeed it is 700 - (5c4+5c3) = 700 - 15 = 685...

Is there a mistype or am I missing sthg?
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Re: PS- Committee [#permalink] New post 07 Dec 2008, 19:08
agree with 635.
I missed the major part - to exclude 2 men who do not share the same comittees.

= 700 - [(1 x 5c4) + (6c1 x 5c3)]
= 700 - 65
= 635
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Re: PS- Committee [#permalink] New post 07 Dec 2008, 21:40
I am clear till getting 700. But why did we subtract 5C4+5C3. Can you please explain.
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Re: PS- Committee [#permalink] New post 08 Dec 2008, 03:16
Somehow, I am getting a different answer. What is wrong here?
2*(7C2*5C4) + 2*(7C3*5C3) = 210 + 700 = 910 :(
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Re: PS- Committee [#permalink] New post 08 Dec 2008, 05:06
echizen wrote:
Indeed it is 700 - (5c4+5c3) = 700 - 15 = 685...

Is there a mistype or am I missing sthg?


[(8c2 -1) x 5c4] + [(8c3 -6) x 5c3] = 635

Blue: minus 1 case

Red:
Total men: A1, A2, A3, A4, A5, A6, A7, A8
If A1 & A2 don't want to serve together, as these 2 men have to be combined with one of other 6 men, so the number of the cases to be subtracted is 6
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Re: PS- Committee [#permalink] New post 08 Dec 2008, 11:40
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scthakur wrote:
Somehow, I am getting a different answer. What is wrong here?
2*(7C2*5C4) + 2*(7C3*5C3) = 210 + 700 = 910 :(


Thats a good question and I understand what you are saying.
what you are doing is taking out one man at a time but doing so repeats some of the team that you form.
suppose m1, m2, m3, m4, m5, m6, m7, and m8 are the members of 8 men. Of them, m1 and m2 cannot be in the team. Lets consider each condition.

Condition 1: Taking out m1

Remaining members are: m2, m3, m4, m5, m6, m7, and m8, we need to combine 2 out of these 7 members. the possibilities is 7c2. fine.

Condition 2: Taking out m2

Remaining members are: m1, m3, m4, m5, m6, m7, and m8. Again we need to combine 2 men out of 7. The possibilities are same i.e. 7c2. Here the problem arises. In condition 1 you have, for example, a team of 2 consisting m3 and m4 that is repeted here as well. This way any combination from "m3, m4, m5, m6, m7, and m8" is repeted.

Therefore we cannot do => "2*(7C2*5C4) + 2*(7C3*5C3) = 210 + 700 = 910".
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Re: PS- Committee [#permalink] New post 08 Dec 2008, 23:40
Thanks GMATTIGER for a detailed explanation. I guess, in the beginning, I could not comprehend the repeat part. +1
Re: PS- Committee   [#permalink] 08 Dec 2008, 23:40
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