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

Expert replies
by Gmat_mission » Mon Jun 10, 2019 4:53 am

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty

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

[spoiler]OA=D[/spoiler]

Source: Princeton Review
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Source: — Problem Solving |

by Brent@GMATPrepNow » Mon Jun 10, 2019 5:00 am
Gmat_mission wrote: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
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 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
Brent Hanneson - Creator of GMATPrepNow.com
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by Scott@TargetTestPrep » Tue Jun 11, 2019 7:09 pm
Gmat_mission wrote: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

[spoiler]OA=D[/spoiler]

Source: Princeton Review
We are selecting 2 of 3 boys. The number of boys can be selected in 3C2 = 3 ways.

We are selecting 2 of 7 girls. The number of girls can be selected in 7C2 = (7 x 6)/2! = 21 ways,

The total number of ways to select the group is 3 x 21 = 63.

The group of 4 children can then be ordered in 4! = 24 ways.

So, the total number of ways to select and then order the group is 63 x 24 = 1,512.

Alternate Solution:

First, let's find the number of ways to have the book reports presented by two girls, followed by two boys.

2 girls from a group of 7 girls can be selected and ordered in 7P2 = 7!/(7 - 2)! = 7 x 6 = 42 ways.

2 boys from a group of 3 boys can be selected and ordered in 3P2 = 3!/(3 - 2)! = 3 x 2 = 6 ways.

Thus, there are 42 x 6 = 252 ways to have two girls present their book report, followed by two boys.

Now, let's consider the different arrangements of genders to present the book report. We are looking for the arrangements of GGBB and by the indistinguishable permutations formula, there are 4!/(2!*2!) = (4 x 3)/2 = 6 of such arrangements.

Thus, the total number of different possible orderings for the presentation is 252 x 6 = 1,512.

Answer: D

Scott Woodbury-Stewart
Founder and CEO
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