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A committee of three people is to be chosen from four

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A committee of three people is to be chosen from four [#permalink] New post 22 May 2006, 04:06
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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?

1. 16
2. 24
3. 26
4. 30
5. 32
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Re: PS - Permutation (GMATPrep) [#permalink] New post 24 May 2006, 19:58
ffgmat wrote:
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?

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


3 members in a committe from four groups of 2 people each with two persons from the same group

No of different commites = (2*2*2) +(2*2*2) +(2*2*2) +(2*2*2)
= 4* (8)
= 32
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Re: PS - Permutation (GMATPrep) [#permalink] New post 24 May 2006, 21:10
Professor wrote:
ffgmat wrote:
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?

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


= 8c3 - 4x6
= 32

Prof please explain what is 4x6.I am stuck with this part
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Re: PS - Permutation (GMATPrep) [#permalink] New post 25 May 2006, 08:59
Yurik79 wrote:
Professor wrote:
ffgmat wrote:
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?

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


= 8c3 - 4x6
= 32

Prof please explain what is 4x6.I am stuck with this part

-----------------------

3 member committee can be formed only in two ways:

1. All three are unrelated. Say X
2. 1 couple + another person. Say Y

Total number of ways to form 3 member committee can be formed = 8c3

Hence 8c3 = X + Y

Y = (select one couple) * (select one person from remaining 6) = 4.6

So, X = 8c3 - 4.6

Hope this clarifies
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 [#permalink] New post 14 Jun 2006, 13:57
Hi, I have a quick question.

I understand how you got the answer however can you please explain to me how to calculate 8c3 ?

I know the answer is 768. Can anyone show me how to get to 768?

768/34 = 32

Thanks
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 [#permalink] New post 15 Jun 2006, 01:09
4 married couples = 8 people

# of ways to pick 3 people = 8C3 = 8!/3!5! = 56

# of ways that two peopel who are married are in the same committe:

4 * 6C1 = 24

Total # of ways to pick 3 people who are not married to each oterh = 56-24 = 32
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 [#permalink] New post 18 Jun 2006, 09:05
4 couples - 4 Men (M1,M2,M3,M4) and 4 Women (W1,W2,W3,W4)

Group of 3 can be
(1) 3 Men -> 4c3 = 4 ways
(2) 3 Women -> 4c3 = 4 ways
(3) 2M and 1W -> 4c2 * 2c1 = 12 (After selecting 2 Men you have to eliminate their corresponding female counterparts)
(4) 2W and 1M -> 4c2 *2c1 = 12

(1) +(2) +(3) +(4) = 32
  [#permalink] 18 Jun 2006, 09:05
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