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How many committees can be formed comprising 2 male members

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by Max@Math Revolution » Mon Mar 04, 2019 12:45 am

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[GMAT math practice question]

How many committees can be formed comprising 2 male members selected from 4 men, 3 female members selected from 5 women, and 3 junior members selected from 6 juniors?

A. 900
B. 1200
C. 1500
D. 1800
E. 2400
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by Max@Math Revolution » Wed Mar 06, 2019 12:43 am
=>

There are 4C2 ways to select 2 men from 4 men, 5C3 ways to select 3 women from 5 women and 6C3 ways to select 3 juniors from 6 juniors. Therefore, the total number of possible committees is
4C2*5C3*6C3 = {(4*3)/(2*1)}{(5*4*3)/(3*2*1)}{(6*5*4)/(3*2*1)} = 6*10*20 = 1200.

Therefore, B is the answer.
Answer: B
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by Scott@TargetTestPrep » Wed Mar 06, 2019 6:52 pm
Max@Math Revolution wrote:[GMAT math practice question]

How many committees can be formed comprising 2 male members selected from 4 men, 3 female members selected from 5 women, and 3 junior members selected from 6 juniors?

A. 900
B. 1200
C. 1500
D. 1800
E. 2400
The number of committees that can be formed is:

4C2 x 5C3 x 6C3 = (4 x 3)/2 x (5 x 4 x 3)/(3 x 2) x (6 x 5 x 4)/(3 x 2) = 6 x 10 x 20 = 1200

Answer: B

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by Scott@TargetTestPrep » Fri Mar 08, 2019 6:51 am
Max@Math Revolution wrote:[GMAT math practice question]

How many committees can be formed comprising 2 male members selected from 4 men, 3 female members selected from 5 women, and 3 junior members selected from 6 juniors?

A. 900
B. 1200
C. 1500
D. 1800
E. 2400
The number of ways to select the male members is 4C2:

(4 x 3)/2! = 6 ways

The number of ways to select the female members is 5C3:

(5 x 4 x 3)/3! = 10 ways

The number of ways to select the 3 juniors is 6C3:

(6 x 5 x 4)/3! = 20 ways

So the total number of ways to select the committee is 6 x 10 x 20 = 1200 ways.

Answer: B

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