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Ten telegenic contestants with a variety of personality disorders are to be divided into

Expert replies
by BTGModeratorVI » Sun Jul 19, 2020 1:34 pm

Timer

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Answers

A

B

C

D

E

Stats

Difficulty—

Ten telegenic contestants with a variety of personality disorders are to be divided into two “tribes” of five members each, tribe A and tribe B, for a competition. How many distinct groupings of two tribes are possible?

A. 120
B. 126
C. 252
D. 1200
E. 1260

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

BTGModeratorVI wrote: ↑
Sun Jul 19, 2020 1:34 pm
Ten telegenic contestants with a variety of personality disorders are to be divided into two “tribes” of five members each, tribe A and tribe B, for a competition. How many distinct groupings of two tribes are possible?

A. 120
B. 126
C. 252
D. 1200
E. 1260

Answer: C
Source: Princeton Review
# of ways 5 contestants can be chosen for tribe A = 10C5 = (10.9.8.7.6)/(1.2.3.4.5) = 252.

Since upon the selection of 5 contestants, the remaining 5 contestants would automatically be selected for tribe B, we need not find the no. of ways for such selection.

Correct answer: C

Hope this helps!

-Jay
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Manhattan Review GMAT Prep

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BTGModeratorVI wrote: ↑
Sun Jul 19, 2020 1:34 pm
Ten telegenic contestants with a variety of personality disorders are to be divided into two “tribes” of five members each, tribe A and tribe B, for a competition. How many distinct groupings of two tribes are possible?

A. 120
B. 126
C. 252
D. 1200
E. 1260

Answer: C
Source: Princeton Review
Let's take the task of creating the teams and break it into stages.

Stage 1: Select two 5 contestants to be in tribe A
Since the order in which we select the contestants does not matter, we can use combinations.
We can select 5 contestants from 10 contestants in 10C5 ways
10C5 = (10)(9)(8)(7)(6)/(5)(4)(3)(2)(1) = 252
So, we can complete stage 1 in 252 ways

Stage 2: Place the remaining 5 people in tribe B
There's only 1 way to place all 5 remaining people in tribe B
So we can complete this stage in 1 way.

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus create 2 tribes of 5 contestants each) in (252)(1) ways (= 252 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 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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