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Probability..help

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by jkwan » Thu Feb 05, 2009 10:08 pm
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: Probability..help

by betamax » Thu Feb 05, 2009 10:33 pm
jkwan 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
There are a total of 11!/2!(11-2)! * 9!/2!(9-2)! = 1980

11!/2!9! = 11*10*9!/2!9!=11*10/2
9!/2!7!=9*8*7!/2!7!=9*8/2
55*36=1980

I think this is right~
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by gaggleofgirls » Thu Feb 05, 2009 10:36 pm
The number of possible committees = number of ways to chose 2 from the 11 women * number of possible ways to chose 2 from the 9 men.

11c2 * 9c2

(11! / 2!*9!) * (9! / 2!*7!)

(11*10 / 2) * (9*8/2)

(11*5) * (9*4)

55 *36

1980

Answer = E

-Carrie
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by Scott@TargetTestPrep » Tue Dec 05, 2017 5:11 pm
jkwan 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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