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# There are 11 women and 9 men in a certain club. If the club

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There are 11 women and 9 men in a certain club. If the club [#permalink]  07 Sep 2008, 10:08
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There are 11 women and 9 men in a certain club. If the club is to select a committee of 2 women and 2 men, how many different such committees are possible?

A. 120
B. 720
C. 1,060
D. 1,520
E. 1,980
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Re: men and women [#permalink]  08 Sep 2008, 13:10
E

(11!/2!9!)*(9!/2!7!) = 1980
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Re: men and women [#permalink]  06 Nov 2008, 00:40
guys
can some one tell my why in this case we multiply the 2 parts and do not add them up?
is there a general rule for either multiplying od adding up in permutations?

cheers
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Re: men and women [#permalink]  06 Nov 2008, 00:46
Thats because its an "and" condition..and implies multiply...
if its an "OR" U have to add..

its simple to understand if ur familiar with Binary logic...
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Re: men and women [#permalink]  11 Sep 2011, 03:35

I'm guessing that would've been the number of 2-member committees which had men and women separately. Is that right?
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petrifiedbutstanding

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Re: men and women [#permalink]  11 Sep 2011, 09:56
these are two independent events. meaning picking men from a group of men has no effect on picking women from group of women. so you have to multiply here.

11c2*9c2 =1980

If you are trying to find possible number of 2 member committee then the answer would be
11c2+9c2+11c1*9c1 (you can pick 2M or 2W or 1M,1W)

petrifiedbutstanding wrote:

I'm guessing that would've been the number of 2-member committees which had men and women separately. Is that right?
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Re: men and women [#permalink]  12 Sep 2011, 05:10
Independent events must always be multiplied to get the total combinations..

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Re: men and women   [#permalink] 12 Sep 2011, 05:10
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