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

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Intern
Joined: 10 Aug 2007
Posts: 22

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

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13 Oct 2007, 08:15
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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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VP
Joined: 08 Jun 2005
Posts: 1144

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13 Oct 2007, 08:28
10C3 = 120 ---> choosing any three people from ten.

5C1= 5 ---> choosing a married couple out of five

8C1 = 8 ---> choosing the third person on the committee.

120 - 5*8 = 80

read more ---> http://www.themathpage.com/aPreCalc/per ... ations.htm

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Intern
Joined: 10 Aug 2007
Posts: 22

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13 Oct 2007, 08:38
[quote="KillerSquirrel"]10C3 = 120 ---> choosing any three people from ten.

5C1= 5 ---> choosing a married couple out of five

8C1 = 8 ---> choosing the third person on the committee.

120 - 5*8 = 80

read more ---> http://www.themathpage.com/aPreCalc/per ... ations.htm

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Director
Joined: 10 Feb 2006
Posts: 660

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19 Oct 2007, 23:13
8C1 = 8 ---> choosing the third person on the committee.

How did you get that 8 from?
_________________

GMAT the final frontie!!!.

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VP
Joined: 08 Jun 2005
Posts: 1144

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19 Oct 2007, 23:52
8C1 = 8 ---> choosing the third person on the committee.

How did you get that 8 from?

Since started with ten people, and already choose a married couple (i.e. two people) you are left with 8 people to choose from (i.e. 10-2 = 8) and you need to fill one more place on the committee.

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VP
Joined: 28 Mar 2006
Posts: 1367

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21 Oct 2007, 07:33
10*8*6/3! = 80

Divide by 3! to remove the duplication of count

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Intern
Joined: 10 Aug 2007
Posts: 22

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21 Oct 2007, 09:35
[quote="trivikram"]10*8*6/3! = 80

Divide by 3! to remove the duplication of count[/quote]

Could you plz explain your fomular a little bit more? I dont understand the way to put these numbers together. thanks

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21 Oct 2007, 09:35
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