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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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VP
Joined: 22 Nov 2007
Posts: 1044
There are 11 women and 9 men in a certain club. If the club  [#permalink]

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12 Jan 2008, 02:07
1
6
00:00

Difficulty:

5% (low)

Question Stats:

90% (01:20) correct 10% (01:42) wrong based on 292 sessions

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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
Manager
Joined: 10 Jul 2009
Posts: 107
Location: Ukraine, Kyiv

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07 Sep 2009, 10:24
2
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]

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18 Dec 2010, 13:01
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]

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18 Dec 2010, 13:02
1
Bunuel,
Thanks for confirming the ans
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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]

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19 Feb 2015, 19:26
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.

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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Joined: 28 Feb 2017
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28 Feb 2017, 19:21
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) 1060
d) 1520
e) 1980
Math Expert
Joined: 02 Aug 2009
Posts: 7099

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28 Feb 2017, 19:27
lyla_austin 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) 1060
d) 1520
e) 1980

Otherwise you have to select 2 out of 11 and 2 out of 9.. 11C2*9C2=$$\frac{11!}{9!2!}*\frac{9!}{7!/2!}=\frac{11*10*9*8*7!}{7!*4}=11*10*9*2=1980$$

E
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Re: There are 11 women and 9 men in a certain club. If the club is to sele  [#permalink]

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28 Feb 2017, 20:01
lyla_austin 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) 1060
d) 1520
e) 1980

Merging topics. Please refer to the discussion above.

Also, please follow the rules (https://gmatclub.com/forum/rules-for-po ... 33935.html) when posting a question.
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Re: There are 11 women and 9 men in a certain club. If the club is to sele  [#permalink]

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04 Jul 2018, 18:27
marcodonzelli 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

We need to determine how many different committees are possible with 2 women and 2 men from 11 women and 9 men. Let’s first determine the number of ways to select 2 women.

number of ways to select 2 women from 11 of them = 11C2 = (11 x 10)/2! = 55 ways

Next we determine the number of ways to select 2 men.

number of ways to select 2 men from 9 of them = 9C2 = (9 x 8)/2! = 36 ways

Thus, the number of ways to select 2 women and 2 men is 55 x 36 = 1,980 ways.

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Re: There are 11 women and 9 men in a certain club. If the club is to sele &nbs [#permalink] 04 Jul 2018, 18:27
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