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How many committees can be formed comprising 2 male members selected f

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How many committees can be formed comprising 2 male members selected f  [#permalink]

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New post 04 Mar 2019, 02:02
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Question Stats:

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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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Re: How many committees can be formed comprising 2 male members selected f  [#permalink]

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New post 04 Mar 2019, 02:30
imo B .

2 men of 4 , 3 women out of 5 and 3 juniors out of 6


4C2*5C3*6C3 = 1200

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Re: How many committees can be formed comprising 2 male members selected f  [#permalink]

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New post 04 Mar 2019, 04:05
We could almost rephrase this as three different questions:

How many ways to choose 2 from 4?
How many ways to choose 3 from 5?
How many ways to choose 3 from 6?

For this question, it's easier if you know the Combinations formula, but you could technically count.

Combinations formula = n! / k! (n-k)!

n = the bigger number (what we're choosing from)
k = the smaller number (how many we're choosing)

For example: how many ways to choose 2 from 4?

4! / 2! (4-2)!
4! / 2!2!
(4 x 3 x 2 x 1) / (2 x 1)(2 x 1)
24/4 = 6

But if we counted this and said the four males were ABCD, we could just list AB, AC, AD, BC, BD, and CD, and you can see it also equals 6.

Anyway, finding the other two:

How many ways to choose 3 from 5?

5! / 3! (5-3)!
5! / 3! 2!
We can cancel out 3! from both numerator and denominator.
5 x 4 / 2
20/2 = 10

How many ways to choose 3 from 6?

6! / 3! (6-3)!
6! / 3! 3!
We can cancel out one 3! from both numerator and denominator.
6 x 5 x 4 / 3 x 2
120 / 6 = 20

Back to our original questions:

How many ways to choose 2 from 4? 6
How many ways to choose 3 from 5? 10
How many ways to choose 3 from 6? 20

Now, multiply all those numbers together! :)

6 x 10 x 20 = 1200
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Re: How many committees can be formed comprising 2 male members selected f  [#permalink]

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New post 06 Mar 2019, 02:02
=>

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 = \({\frac{(4*3)}{(2*1)}}{\frac{(5*4*3)}{(3*2*1)}}{\frac{(6*5*4)}{(3*2*1)}} = 6*10*20 = 1200.\)

Therefore, B is the answer.
Answer: B
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Re: How many committees can be formed comprising 2 male members selected f  [#permalink]

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New post 06 Mar 2019, 02:11
MathRevolution 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



4c2 * 5c3*6c3 = 1200
IMO B
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Re: How many committees can be formed comprising 2 male members selected f  [#permalink]

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New post 09 Mar 2019, 04:30
1
MathRevolution 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



4c2 * 5c3*6c3 = 1200
Answer B

P.S. Question should be tagged Combinations rather than probability.
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Re: How many committees can be formed comprising 2 male members selected f   [#permalink] 09 Mar 2019, 04:30
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