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Some fourth graders are choosing foursquare teams at

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Some fourth graders are choosing foursquare teams at

by VJesus12 » Thu May 03, 2018 10:32 am

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Some fourth graders are choosing foursquare teams at recess. What is the total possible number of combinations of four-person teams that could be chosen from a group of six kids?

(A) 6
(B) 15
(C) 120
(D) 360
(E) 98,280

The OA is the option B.

What is the formula that I should use here? May someone helps me? Please.
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by Brent@GMATPrepNow » Thu May 03, 2018 2:39 pm
VJesus12 wrote:Some fourth graders are choosing foursquare teams at recess. What is the total possible number of combinations of four-person teams that could be chosen from a group of six kids?

(A) 6
(B) 15
(C) 120
(D) 360
(E) 98,280
This is a somewhat ambiguous question since it's hard to tell whether each team member's position matters.
For example, if the question were rephrased to ask how many 9-person baseball teams can be created from 9 people, we might say the correct answer is 1.
However, if we consider the different positions that each player can have, we could also say the correct answer is 9!

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by [email protected] » Thu May 03, 2018 4:32 pm
Hi VJesus12,

We're asked for the total possible number of combinations of four-person foursquare teams that could be chosen from a group of six kids. I agree with Brent's assessment (it's not perfectly clear whether we're supposed to be treating this question as a Permutation or a Combination). That having been said, the 'intent' of this question is that it's meant to be a Combination Formula question:

N!/(K!)(N-K)! where N is the total number of people and K is the size of the 'subgroup'

With 6 children and a group of 4, we have.... 6!/(4!)(2!) = (6)(5)/(2)(1) = 15 possible groups of 4.

Final Answer: B

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by Jeff@TargetTestPrep » Mon May 07, 2018 10:15 am
VJesus12 wrote:Some fourth graders are choosing foursquare teams at recess. What is the total possible number of combinations of four-person teams that could be chosen from a group of six kids?

(A) 6
(B) 15
(C) 120
(D) 360
(E) 98,280
The number of ways 4 people can be chosen from 6 people is 6C4 = 6!/(4! x 2!) = (6 x 5)/( 2) = 30/2 = 15.

Answer: B
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