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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]

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New post 12 Jan 2008, 03:07
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Question Stats:

88% (00:57) correct 12% (01:00) wrong based on 278 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
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Re: easy comb question  [#permalink]

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New post 07 Sep 2009, 11:24
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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New post 18 Dec 2010, 14:01
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).

Answer: E.
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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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New post 18 Dec 2010, 14:02
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Bunuel,
Thanks for confirming the ans
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Re: easy comb question  [#permalink]

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New post 31 Mar 2011, 01:40
11C2 * 9C2 = 11*10/2 * 9*8/2

= 11 * 10 * 9 * 2

= 990 * 2

= 1980

Answer - E
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Re: There are 11 women and 9 men in a certain club. If the club  [#permalink]

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New post 19 Feb 2015, 20:26
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.

Final Answer:

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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Probability Question  [#permalink]

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New post 28 Feb 2017, 20:21
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
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Re: Probability Question  [#permalink]

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New post 28 Feb 2017, 20:27
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


Please give topic name and OA for your posts..

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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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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New post 28 Feb 2017, 21:01
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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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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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New post 04 Jul 2018, 19:27
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.

Answer: E
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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, 19:27
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