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Ten telegenic contestants with a variety of personality diso

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Ten telegenic contestants with a variety of personality diso

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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

OA C

Source: Princeton Review

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GMAT/MBA Expert

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BTGmoderatorDC wrote:
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
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 our free video: http://www.gmatprepnow.com/module/gmat-counting/video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat-counting/video/776

Then you can try solving the following questions:

EASY
- http://www.beatthegmat.com/what-should-be-the-answer-t267256.html
- http://www.beatthegmat.com/counting-problem-company-recruitment-t244302.html
- http://www.beatthegmat.com/picking-a-5-digit-code-with-an-odd-middle-digit-t273110.html
- http://www.beatthegmat.com/permutation-combination-simple-one-t257412.html
- http://www.beatthegmat.com/simple-one-t270061.html


MEDIUM
- http://www.beatthegmat.com/combinatorics-solution-explanation-t273194.html
- http://www.beatthegmat.com/arabian-horses-good-one-t150703.html
- http://www.beatthegmat.com/sub-sets-probability-t273337.html
- http://www.beatthegmat.com/combinatorics-problem-t273180.html
- http://www.beatthegmat.com/digits-numbers-t270127.html
- http://www.beatthegmat.com/doubt-on-separator-method-t271047.html
- http://www.beatthegmat.com/combinatorics-problem-t267079.html


DIFFICULT
- http://www.beatthegmat.com/wonderful-p-c-ques-t271001.html
- http://www.beatthegmat.com/permutation-and-combination-t273915.html
- http://www.beatthegmat.com/permutation-t122873.html
- http://www.beatthegmat.com/no-two-ladies-sit-together-t275661.html
- http://www.beatthegmat.com/combinations-t123249.html


Cheers,
Brent

_________________
Brent Hanneson – Creator of GMATPrepNow.com
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GMAT/MBA Expert

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Hi All,

For Test Takers who know the Combination Formula, this question is a fairly straight-forward prompt. If you DON'T know the Combination Formula though, then here's what it is and how to use it. Any time a prompt asks for "groups" or "combinations" of things, then the order of the things DOES NOT MATTER.

For example, if a 2-person team consists of A and B, then A,B is the same as B,A --> thus, order does NOT matter. Mathematically though, you're allowed to count this team TWICE - A,B and B,A are the same team, so it should only be counted ONCE. The Combination Formula removes all of the "duplicates", leaving you with the unique combinations for whatever situation you're working with.

The Combination Formula itself is:

N!/[K!(N-K)!]

N = the total number of items/people
K = the size of the subgroup

Here, we have 10 people and we're asked to form 2 groups of 5.

For the first group, N = 10 and K = 5....

10!/[5!5!] = (10)(9)(8)(7)(6)/(5)(4)(3)(2)(1) = 256 unique groups of 5 people

Once you have formed that group of 5, then the remaining 5 form the other group (so there's no more math to do)

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

_________________
Contact Rich at Rich.C@empowergmat.com

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BTGmoderatorDC wrote:
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

OA C

Source: Princeton Review
The number of ways to select the first tribe is 10C5:

10C5 = (10!)/(10 - 5)! x 5!

= (10 x 9 x 8 x 7 x 6)/(5 x 4 x 3 x 2 x 1)

= 3 x 2 x 7 x 6 = 252

The next tribe can be selected in 5C5 = 1 way.

So there are 252 x 1 = 252 ways to select the two tribes.

Answer: C

_________________
Scott Woodbury-Stewart Founder and CEO

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