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From 7 boys and 4 girls, how many different committees can be selected consisting of 3 boys and 2 girls?

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by BTGModeratorVI » Sun Jul 19, 2020 1:43 pm
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From 7 boys and 4 girls, how many different committees can be selected consisting of 3 boys and 2 girls?

A. 108
B. 168
C. 210
D. 330
E. 462

Answer: C
Source: Kaplan
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Re: From 7 boys and 4 girls, how many different committees can be selected consisting of 3 boys and 2 girls?

by Jay@ManhattanReview » Wed Jul 22, 2020 12:38 am
BTGModeratorVI wrote: ↑
Sun Jul 19, 2020 1:43 pm
 From 7 boys and 4 girls, how many different committees can be selected consisting of 3 boys and 2 girls?

A. 108
B. 168
C. 210
D. 330
E. 462

Answer: C
Source: Kaplan
# of committees = 7C3*4C2 = [7.6.5/1.2.3]*[4.3/1.2] = 210

Correct answer: C

Hope this helps!

-Jay
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Re: From 7 boys and 4 girls, how many different committees can be selected consisting of 3 boys and 2 girls?

by Brent@GMATPrepNow » Wed Jul 22, 2020 6:29 am
BTGModeratorVI wrote: ↑
Sun Jul 19, 2020 1:43 pm
 From 7 boys and 4 girls, how many different committees can be selected consisting of 3 boys and 2 girls?

A. 108
B. 168
C. 210
D. 330
E. 462

Answer: C
Source: Kaplan
Take the task of creating the committee and break it into stages.

Stage 1: Select 3 boys to be on the committee
Since the order in which we select the boys does not matter, we can use combinations.
We can select 3 boys from 7 boys in 7C3 ways (35 ways)
So, we can complete stage 1 in 35 ways

Stage 2: Select 2 girls to be on the committee
Since the order in which we select the girls does not matter, we can use combinations.
We can select 2 girls from 4 girls in 4C2 ways (6 ways)
So, we can complete stage 2 in 6 ways

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus create a committee) in (35)(6) ways (= 210 ways)

Answer: C
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