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

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Senior Manager
Joined: 10 Dec 2004
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If a committee of 3 people is to be selected from among 5 [#permalink]  21 May 2005, 19:43
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?

=> I know this is wrong, I want to know why.
Why Can't I use 10(ways of selecting 1st person) * 8*(2nd person) * 6 (third one) = 480?
SVP
Joined: 03 Jan 2005
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I'd say you first choose three couples from the five, then one person from each chosen couple:

C(5,3)*2*2*2=80

The way you are doing is ok, just remember you have got some duplicates, since you were doing them by order (permutation instead of combination). For example, you could get A C E, and you might also get C A E, and you'd count them as two outcomes while in fact it is the same combination of people in different orders. To get to the correct answer from your approach, you would divide your answer by the total possible ways to put three people in order.

10*8*6/P(3,3)=80
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Director
Joined: 18 Feb 2005
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HongHu wrote:
I'd say you first choose three couples from the five, then one person from each chosen couple:

C(5,3)*2*2*2=80

The way you are doing is ok, just remember you have got some duplicates, since you were doing them by order (permutation instead of combination). For example, you could get A C E, and you might also get C A E, and you'd count them as two outcomes while in fact it is the same combination of people in different orders. To get to the correct answer from your approach, you would divide your answer by the total possible ways to put three people in order.

10*8*6/P(3,3)=80

Impeccable Solution!!
Senior Manager
Joined: 10 Dec 2004
Posts: 274
Followers: 1

Kudos [?]: 68 [0], given: 0

Thanks Hong Hu!
That was neat explanation.

Also we can do this way -

10C3 - 5*(2C2*8C1) = 80
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