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From a group of 10 students, 7 girls and 3 boys, a teacher must choose 2

Expert replies
by BTGModeratorVI » Sun Jul 26, 2020 6:39 am
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From a group of 10 students, 7 girls and 3 boys, a teacher must choose 2 girls and 2 boys to present book reports. How many different arrangements of students, in order, are possible?

A. 252
B. 504
C. 1,008
D. 1,512
E. 5,040

Answer: D
Source: Princeton Review
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Source: — Problem Solving |

BTGModeratorVI wrote: ↑
Sun Jul 26, 2020 6:39 am
From a group of 10 students, 7 girls and 3 boys, a teacher must choose 2 girls and 2 boys to present book reports. How many different arrangements of students, in order, are possible?

A. 252
B. 504
C. 1,008
D. 1,512
E. 5,040

Answer: D
Source: Princeton Review
The required no. of ways to selected 4 students = 7C2*3C2 = (7*6/1*2)*(3*2/1*2) = 21*3 = 63

Thus, no. of ways to arrange 4 selected students = 4!*63 = 1,512

Correct answer: D

Hope this helps!

-Jay
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Manhattan Review GMAT Prep

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BTGModeratorVI wrote: ↑
Sun Jul 26, 2020 6:39 am
From a group of 10 students, 7 girls and 3 boys, a teacher must choose 2 girls and 2 boys to present book reports. How many different arrangements of students, in order, are possible?

A. 252
B. 504
C. 1,008
D. 1,512
E. 5,040

Answer: D
Source: Princeton Review
Take the task of arranging students and break it into stages.

Stage 1: Select two girls
Since the order in which we select the women does not matter, we can use combinations.
We can select 2 girls from 7 girls in 11C2 ways (21 ways)
So, we can complete stage 1 in 21 ways

If anyone is interested, here's a video on calculating combinations (like 7C2) in your head: https://www.gmatprepnow.com/module/gmat- ... /video/789

Stage 2: Select two boys
We can select 2 boys from 3 boys in 3C2 ways (3 ways)
So, we can complete stage 2 in 3 ways

Stage 3: Arrange the 4 children in a row
We can arrange n unique objects in n! ways
So, we can arrange 4 unique children in 4! ways (=24 ways)
We can complete this stage in 24 ways.

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus select and arrange 4 children) in (21)(3)(24) ways (= 1512ways)

Answer: D
Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch this 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
Brent Hanneson - Creator of GMATPrepNow.com
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