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A committee of three people is to be chosen from four marrie

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A committee of three people is to be chosen from four marrie [#permalink] New post 13 Nov 2007, 06:31
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E

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Question Stats:

59% (02:15) correct 40% (01:42) wrong based on 92 sessions
A committee of three people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?

A. 16
B. 24
C. 26
D. 30
E. 32

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-committee-of-3-people-is-to-be-chosen-from-four-married-94068.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 01 Oct 2013, 00:29, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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 [#permalink] New post 13 Nov 2007, 06:41
Answer E 32

The committee could have

0 men and 3 women = 4 possibilities ( 4c3 )
1 man and 2 women = 12 possibilities ( no couple should be in the team : 4 * 3c2 )
2 men and 1 woman = 12 possibilities ( no couple should be in the team : 4c2 *2c1 )
3 men and 0 women = 4 possibilities ( 4c3 )

Therefore 4+12+12+4=32 possible teams.
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 [#permalink] New post 09 Dec 2007, 06:08
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E.

N=4C3*2C1*2C1*2C1=4*2*2*2=32

4C3 - 3 couples from 4 couples
2C1 - a man or woman from a couple.
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 [#permalink] New post 09 Dec 2007, 11:29
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Another way to look at this is the Manhattan GMAT technique. For a restriction problem like this you want to look at it in two parts:

Total Possible Combos - Restricted Combos.

So total possible without any restrictions is:

8!/(3!)(5!) = 56 total possibilities

Now you look at the restrictions. What is the restriction -- a couple can't be together. How many are restricted? When looking at a restriction it is best to look at them as a total unit. So 1 couple is a unit. We have four couples. How can we arrange this 4 units? 4*3*2*1 = 24 combos restricted.

So 56 - 24 = 32. Your answer.
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Re: PS...Probability ... prep [#permalink] New post 07 Sep 2009, 09:36
The same as here ps-combinatorics-m02q05-55472.html
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Re: PS...Probability ... prep [#permalink] New post 28 Sep 2009, 03:11
A committee of three people is to be chosen from four couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee

1. 16
2. 24
3. 26
4. 30
5. 32


Ans: 8 * 6 * 4/3!

32 ways..

Ans is option 5
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Re: PS...Probability ... prep [#permalink] New post 02 May 2011, 22:05
8*6*4 = total ways
3! is the number of permutations of 3 people.

hence 24*8/6 = 32
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Re: A committee of three people is to be chosen from four [#permalink] New post 30 Sep 2013, 11:52
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Re: A committee of three people is to be chosen from four marrie [#permalink] New post 01 Oct 2013, 00:31
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A committee of three people is to be chosen from four married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?

A. 16
B. 24
C. 26
D. 30
E. 32

Each couple can send only one "representative" to the committee. Let's see in how many ways we can choose 3 couples (as there should be 3 members) out of 4 to send only one "representatives" to the committee: 4C3=4.

But each of these 3 couples can send two persons (husband or wife): 2*2*2=2^3=8.

Total # of ways: 4C3*2^3=32.

Answer: E.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-committee-of-3-people-is-to-be-chosen-from-four-married-94068.html

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Re: A committee of three people is to be chosen from four marrie   [#permalink] 01 Oct 2013, 00:31
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