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

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There are 11 women and 9 men in a certain club. If the club [#permalink]  12 Jan 2008, 02:07
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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
[Reveal] Spoiler: OA
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Re: easy comb question [#permalink]  07 Sep 2009, 10:24
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or another way.

choose 2 from 11: 11*10
choose 2 from 9: 9*8

multiply results and divide by 4, since there can be 4 different orders in 1 pair with same people.

E. 1980
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Re: There are 11 women and 9 men in a certain club. If the club is to sele [#permalink]  18 Dec 2010, 13:01
Expert's post
ajit257 wrote:
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

$$C^2_{11}*C^2_{9}=1,980$$: $$C^2_{11}$$ - # of ways to choose 2 women out of 11 and $$C^2_9$$ - # of ways to choose 2 men out of 9. We multiply these two because if one event can occur in $$m$$ ways and a second can occur independently of the first in $$n$$ ways, then the two events can occur in $$mn$$ ways (this is called Principle of Multiplication).

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Re: There are 11 women and 9 men in a certain club. If the club is to sele [#permalink]  18 Dec 2010, 13:02
Bunuel,
Thanks for confirming the ans
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Re: easy comb question [#permalink]  31 Mar 2011, 00:40
11C2 * 9C2 = 11*10/2 * 9*8/2

= 11 * 10 * 9 * 2

= 990 * 2

= 1980

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Re: There are 11 women and 9 men in a certain club. If the club [#permalink]  19 Feb 2015, 14:34
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Re: There are 11 women and 9 men in a certain club. If the club [#permalink]  19 Feb 2015, 19:26
Expert's post
Hi All,

Both barakhaiev and subhashghosh have correctly answer this question, albeit doing the "math" in slightly different ways.

This type of question is relatively rare on the GMAT, but it's more likely to show up when you're scoring at a high level in the Quant section, so the basic ideas behind it are worth memorizing. You have to figure out each individual combination, then multiply the results.

Sometimes the answer choices are "spread out" enough that you can take advantage of that pattern and avoid a little bit of work (and save a little bit of time).

Here, choosing 2 women from a group of 11 is 11c2 = 11!/(2!9!) = (11)(10)/(2)(1) = 55

Choosing 2 men from a group of 9 men is 9c2 = 9!/(2!7!) = (9)(8)/(2)(1) = 36

Now we just have to multiply these two values together: (55)(36).

Since (50)(30) = 1500, and (55)(36) is going to be MUCH LARGER than 1500, there's really only one answer that could possibly be correct.

[Reveal] Spoiler:
E

Keep the answer choices in mind when you're doing your work; they often provide 'clues' as how best to proceed and potential patterns or shortcuts in how the "math" will work.

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Re: There are 11 women and 9 men in a certain club. If the club   [#permalink] 19 Feb 2015, 19:26
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