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Combinatorics - How many combinations out of two groups?

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Combinatorics - How many combinations out of two groups?

by Bens4vcobra » Thu Jun 16, 2011 6:04 am
A 4 person task force is to be formed from 4 men and 3 women who work in Company G's HR dept. If there are to be 2 men and 2 women, how many different task forcers can be formed?
A)14
B)18 (correct answer)
C)35
D)56
E)144

I am trying to use the "anagram" method prefered by MGMAT but I must be doing something wrong because I cannot come up with the correct answer.
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by Brent@GMATPrepNow » Thu Jun 16, 2011 6:12 am
Bens4vcobra wrote:A 4 person task force is to be formed from 4 men and 3 women who work in Company G's HR dept. If there are to be 2 men and 2 women, how many different task forcers can be formed?
A)14
B)18
C)35
D)56
E)144
We can take the task of "building" a task force and break it into stages.

Stage 1: select the 2 women for the team
Stage 2: select the 2 men for the team

Stage 1: since the order of the selected women does not matter, this is a combination question.
There are 3 women, and we want to choose 2 of them.
This can be accomplished in 3C2 ways, which equals 3

Stage 2: since the order of the selected men does not matter, this is a combination question.
There are 4 men, and we want to choose 2 of them.
This can be accomplished in 4C2 ways, which equals 6

At this point, we can apply the Fundamental Counting Principle and find the product of the two stages to get (3)(6) = 18

So the answer is B

Cheers,
Brent

PS: To learn a shortcut for calculating combinations in your head, see https://youtu.be/-bbu2h-07iA
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by Frankenstein » Thu Jun 16, 2011 6:13 am
Hi,
From 4 men, 2 men can be selected in 4C2 = 6ways
From 3 women, 2 women can be selected in 3C2 = 3ways
So, number of different task force teams that can e formed is 6*3 = 18 ways.
Cheers!

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by Anurag@Gurome » Thu Jun 16, 2011 6:15 am
Bens4vcobra wrote:A 4 person task force is to be formed from 4 men and 3 women who work in Company G's HR dept. If there are to be 2 men and 2 women, how many different task forcers can be formed?
A)14
B)18 (correct answer)
C)35
D)56
E)144
Number of different task forces = (Number of ways to select 2 men out of 4)*(Number of ways to select 2 women out of 3) = (4C2)*(3C2) = 6*3 = 18

The correct answer is B.
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by Bens4vcobra » Thu Jun 16, 2011 7:09 am
Thanks all.

What does the "C2" mean? I've never seen this before...
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by thephoenix » Thu Jun 16, 2011 7:30 am
Bens4vcobra wrote:Thanks all.

What does the "C2" mean? I've never seen this before...
C stands for combination
P stands for permutation

nCr means n!/(r!*(n-r)!)
4C2=4!(2!*(4-2)!)=4!/(2!*2!)

similarly
npr n!/(n-r)!
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by Bens4vcobra » Thu Jun 16, 2011 8:03 am
thephoenix wrote:
Bens4vcobra wrote:Thanks all.

What does the "C2" mean? I've never seen this before...
C stands for combination
P stands for permutation

nCr means n!/(r!*(n-r)!)
4C2=4!(2!*(4-2)!)=4!/(2!*2!)

similarly
npr n!/(n-r)!
Ah so its just an abreviation for using the combinatorics formula. Thanks.
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