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

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Senior Manager
Joined: 02 Jan 2017
Posts: 290
At a high school track tryout, there are eight women and five men comp  [#permalink]

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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
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4476
Re: At a high school track tryout, there are eight women and five men comp  [#permalink]

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24 Feb 2017, 16:58
1
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

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)
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Joined: 31 Aug 2017
Posts: 3
Re: At a high school track tryout, there are eight women and five men comp  [#permalink]

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02 Sep 2017, 14:15
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
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Joined: 02 Sep 2009
Posts: 60627
Re: At a high school track tryout, there are eight women and five men comp  [#permalink]

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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]

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03 Dec 2019, 11:43
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Re: At a high school track tryout, there are eight women and five men comp   [#permalink] 03 Dec 2019, 11:43
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# At a high school track tryout, there are eight women and five men comp

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