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From a group of three boys and four girls . .

This topic has 2 expert replies and 0 member replies

Top Member

From a group of three boys and four girls . .

Post Mon Nov 06, 2017 6:52 am
From a group of three boys and four girls, a school hallway monitor group is to be selected. If the group is to be formed of two boys and three girls, how many different such groups can be formed?

A. 3
B. 4
C. 12
D. 72
E. 144

The OA is C.

How could I determine the possibilities? With permutations?

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GMAT/MBA Expert

Post Fri Nov 10, 2017 9:36 am
M7MBA wrote:
From a group of three boys and four girls, a school hallway monitor group is to be selected. If the group is to be formed of two boys and three girls, how many different such groups can be formed?

A. 3
B. 4
C. 12
D. 72
E. 144
Take the task of creating the hallway monitor group and break it into stages.

Stage 1: Select two boys for the group
Since the order in which we select the boys does not matter, we can use combinations.
We can select 2 boys from 3 boys in 3C2 ways (3 ways)
So, we can complete stage 1 in 3 ways

If anyone is interested, we have a free video on calculating combinations (like 3C2 or 11C2) in your head: http://www.gmatprepnow.com/module/gmat-counting/video/789

Stage 2: Select two girls for the group
We can select 3 girls from 4 girls in 4C3 ways (4 ways)
So, we can complete stage 2 in 4 ways

By the Fundamental Counting Principle (FCP), we can complete the two stages (and thus create the hallway monitor group) in (3)(4) ways (= 12 ways)

Answer: C
--------------------------

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: http://www.gmatprepnow.com/module/gmat-counting/video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat-counting/video/776

Then you can try solving the following questions:

EASY
- http://www.beatthegmat.com/what-should-be-the-answer-t267256.html
- http://www.beatthegmat.com/counting-problem-company-recruitment-t244302.html
- http://www.beatthegmat.com/picking-a-5-digit-code-with-an-odd-middle-digit-t273110.html
- http://www.beatthegmat.com/permutation-combination-simple-one-t257412.html
- http://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- http://www.beatthegmat.com/combinatorics-solution-explanation-t273194.html
- http://www.beatthegmat.com/arabian-horses-good-one-t150703.html
- http://www.beatthegmat.com/sub-sets-probability-t273337.html
- http://www.beatthegmat.com/combinatorics-problem-t273180.html
- http://www.beatthegmat.com/digits-numbers-t270127.html
- http://www.beatthegmat.com/doubt-on-separator-method-t271047.html
- http://www.beatthegmat.com/combinatorics-problem-t267079.html


DIFFICULT
- http://www.beatthegmat.com/wonderful-p-c-ques-t271001.html
- http://www.beatthegmat.com/permutation-and-combination-t273915.html
- http://www.beatthegmat.com/permutation-t122873.html
- http://www.beatthegmat.com/no-two-ladies-sit-together-t275661.html
- http://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent

_________________
Brent Hanneson – Founder of GMATPrepNow.com
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GMAT/MBA Expert

ErikaPrepScholar Master | Next Rank: 500 Posts
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Post Mon Nov 06, 2017 9:15 am
Let's find the number of possibilities for just boys and just girls first.

Boys: 3, choose 2
$$\frac{3!}{2!}=3$$

Girls: 4, choose 3
$$\frac{4!}{3!}=4$$

So there are 3 possible groups of boys and 4 possible groups of girls. Now, any of these groups of boys can be paired with any of these groups of girls.So the total number of possibilities is 3 * 4 = 12.

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