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# GMAT Prep - combinations

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Senior Manager
Joined: 15 Jul 2006
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25 Sep 2006, 19:36
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Sorry if this was already posted but I could not find it....

A committee of 3 people is to be choosen from 4 married couples. What is the # of different committees that can be formed if 2 people married to eachother can't both serve on the same commeettee?

My approach was :
1st place = 8 people
2nd place = 6 people
3rd place = 4 people

answer: 8x6x4. why is this wrong?
Manager
Joined: 13 Sep 2006
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25 Sep 2006, 22:42
This is a combination problem... use the combo formula..

picking 3 from 4 married couples (8 total) yields a potential of 56 combinations...

now you need to subtract the number of possibilities where the married couples could be in the same group, which is 24 (i think..i had to diagram it out to conceptualize it)

Im getting 32...
Senior Manager
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26 Sep 2006, 02:56
committee can consist of

(1) 3 women = 4C3 = 4 ways
(2) 3 Men = 4C3 = 4 ways
(3) 1 woman 2 Men = 4C1 * 3C2 = 12
(4) 1 Man 2 women = 4C1 * 3C2 = 12

# of different committees = 4 +4+12+12= 32
GMAT Instructor
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Re: GMAT Prep - combinations [#permalink]

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26 Sep 2006, 03:30
EconGirl wrote:
Sorry if this was already posted but I could not find it....

A committee of 3 people is to be choosen from 4 married couples. What is the # of different committees that can be formed if 2 people married to eachother can't both serve on the same commeettee?

My approach was :
1st place = 8 people
2nd place = 6 people
3rd place = 4 people

answer: 8x6x4. why is this wrong?

Only one of the four couples will not have a member of the committee, so there are 4 ways to choose which couple will be excluded. Then each of the 3 couples represented will choose which member to put on the committee: (2*2*2) ways. So in total, we have 4*8=32 different committees.
Re: GMAT Prep - combinations   [#permalink] 26 Sep 2006, 03:30
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