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

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

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

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

by Jay@ManhattanReview » Wed Jul 22, 2020 12:53 am
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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Re: Ten telegenic contestants with a variety of personality disorders are to be divided into

by Brent@GMATPrepNow » Wed Jul 22, 2020 6:27 am
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:

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MEDIUM
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Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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