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?
From a group of three boys and four girls . .
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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.
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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Take the task of creating the hallway monitor group and break it into stages.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
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: https://www.gmatprepnow.com/module/gmat- ... /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: https://www.gmatprepnow.com/module/gmat- ... /video/775
You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776
Then you can try solving the following questions:
EASY
- https://www.beatthegmat.com/what-should- ... 67256.html
- https://www.beatthegmat.com/counting-pro ... 44302.html
- https://www.beatthegmat.com/picking-a-5- ... 73110.html
- https://www.beatthegmat.com/permutation- ... 57412.html
- https://www.beatthegmat.com/simple-one-t270061.html
MEDIUM
- https://www.beatthegmat.com/combinatoric ... 73194.html
- https://www.beatthegmat.com/arabian-hors ... 50703.html
- https://www.beatthegmat.com/sub-sets-pro ... 73337.html
- https://www.beatthegmat.com/combinatoric ... 73180.html
- https://www.beatthegmat.com/digits-numbers-t270127.html
- https://www.beatthegmat.com/doubt-on-sep ... 71047.html
- https://www.beatthegmat.com/combinatoric ... 67079.html
DIFFICULT
- https://www.beatthegmat.com/wonderful-p- ... 71001.html
- https://www.beatthegmat.com/permutation- ... 73915.html
- https://www.beatthegmat.com/permutation-t122873.html
- https://www.beatthegmat.com/no-two-ladie ... 75661.html
- https://www.beatthegmat.com/combinations-t123249.html
Cheers,
Brent
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- Scott@TargetTestPrep
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The number of groups that can be formed is: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
The OA is C.
How could I determine the possibilities? With permutations?
3C2 x 4C3 = 3 x 4 = 12
Answer: C
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