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If a committee of 3 people is to be selected from among 5

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Manager
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If a committee of 3 people is to be selected from among 5 [#permalink] New post 22 Jul 2006, 15:44
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If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

A. 20
B. 40
C. 50
D. 80
E. 120
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 [#permalink] New post 22 Jul 2006, 17:26
D

No of ways of choosing 3 of 10 people = 10C3 = 120
No of ways of choosing 3 people with a married couple in it = 5 X 8C1=40

No of ways of choosing 3 without a married couple = 120-40 =80
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 [#permalink] New post 23 Jul 2006, 11:49
Can someone explain the second part of the question?

This sound complex: choosing 3 people with a married couple


No of ways of choosing 3 people with a married couple in it = 5 X 8C1=40
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 [#permalink] New post 24 Jul 2006, 01:17
TOUGH GUY wrote:
Can someone explain the second part of the question?

This sound complex: choosing 3 people with a married couple


No of ways of choosing 3 people with a married couple in it = 5 X 8C1=40


You have to choose 3 people in which there is a couple. There are 5 couples in all, so 2 out of the 3 people must be one out of these 5 couples. Since there are exactly 5 couples, there are exactly 5 ways in which you could choose a couple.

Now after we chose a couple there are 8 people left, from which we must choose exactly 1 more person to make the total number of people in our group equal to 3. Since there are 8 people, and we need to choose 1, there are 8C1 ways of choosing one person.

This gives us 5 x 8C1. PM me if you want to discuss this further.
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 [#permalink] New post 25 Jul 2006, 03:11
Number of ways 3 people can be chosen from 10 people = 10C3 = 120

Number of ways 3 people with married couple = 5 * 8C1 = 40

Total number of ways there are no married couple = 120 -40 = 80
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 [#permalink] New post 25 Jul 2006, 11:11
D. 80

Total number = 10
Committee size = 3

# of ways of selecting first member = 10

# of ways of selecting second member = 8 (excluding first member's spouse)

# of ways of selecting third member = 6 (excluding 1st and 2nd member's spouses)

Total # of ways = 10x8x6/3! = 80

(3! because the order is not relevant, {1,2,3} = {2,1,3} committee.
  [#permalink] 25 Jul 2006, 11:11
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If a committee of 3 people is to be selected from among 5 smily_buddy 1 12 Aug 2007, 11:53
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If a committee of 3 people is to be selected from among 5 TOUGH GUY 5 26 Dec 2005, 08:53
If a committee of 3 people is to be selected from among 5 pb_india 3 21 May 2005, 19:43
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