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At a high school track tryout, there are eight women and five men comp

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At a high school track tryout, there are eight women and five men comp [#permalink]

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New post 24 Feb 2017, 09:41
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At a high school track tryout, there are eight women and five men competing for the three male and three female spots on the decathlon team. How many different combinations of decathletes are possible on the final team for the six spots?

(A) 112,896
(B) 3,136
(C) 560
(D) 66
(E) 18
[Reveal] Spoiler: OA

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Re: At a high school track tryout, there are eight women and five men comp [#permalink]

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New post 24 Feb 2017, 16:58
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vikasp99 wrote:
At a high school track tryout, there are eight women and five men competing for the three male and three female spots on the decathlon team. How many different combinations of decathletes are possible on the final team for the six spots?

(A) 112,896
(B) 3,136
(C) 560
(D) 66
(E) 18

Dear vikasp99,

I'm happy to respond. :-) Like most Veritas questions, this is a great question. For computing combinations, see these two posts:
GMAT Permutations and Combinations
GMAT Math: Calculating Combinations

For the females, we have 8 women competing for three spots.
\(8C3 = \frac{8!}{(5!)(3!)} = \frac{8*7*6}{3*2*1} = 8*7 = 56\)

For the males, we have 5 men competing for three spots.
\(5C3 = \frac{5!}{(2!)(3!)} = \frac{5*4}{2} = 10\)

Any male set of three can be paired with any female set of three, so we multiply, because of the Fundamental Counting Principle.
total number = 56*10 = 560

OA = (C)

Does all this make sense?
Mike :-)
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Mike McGarry
Magoosh Test Prep

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Re: At a high school track tryout, there are eight women and five men comp [#permalink]

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New post 02 Sep 2017, 14:15
I dont get the last step 56*10. What in the question suggests they are paired?

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Re: At a high school track tryout, there are eight women and five men comp [#permalink]

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New post 03 Sep 2017, 04:31
churchntj wrote:
I dont get the last step 56*10. What in the question suggests they are paired?

Sent from my SM-G360H using GMAT Club Forum mobile app


We multiply because of FUNDAMENTAL PRINCIPLE OF COUNTING (Mike above mentions it and gives a link to the article explaining it), which states if an operation can be performed in ‘m’ ways and when it has been performed in any of these ways, a second operation that can be performed in ‘n’ ways then these two operations can be performed one after the other in ‘m*n’ ways.

We have 6 spots to fill (3 male and 3 female). We can choose 3 females from 8 in 56 ways and we can choose 3 males from 5 in 10 ways, thus according to the FPC the total number of ways to fill 6 spots is 56*10.

Check the links below:

21. Combinatorics/Counting Methods



For more:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread


Hope it helps.
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Re: At a high school track tryout, there are eight women and five men comp   [#permalink] 03 Sep 2017, 04:31
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