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# GRE Math Challenge #4-From a group of 8 people

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GRE Math Challenge #4-From a group of 8 people [#permalink]  18 Aug 2014, 10:24
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100% (00:27) correct 0% (00:00) wrong based on 2 sessions
From a group of 8 people, it is possible to create 56 different k-person committees. Which of the following could be the value of k ?

Indicate all such values.

A)1
B)2
C)3
D)4
E)5
F)6
G)7
[Reveal] Spoiler: OA
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Re: GRE Math Challenge #4 [#permalink]  04 Sep 2014, 06:58
C,E
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Re: GRE Math Challenge #4-From a group of 8 people [#permalink]  03 Sep 2018, 18:00
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Re: GRE Math Challenge #4-From a group of 8 people [#permalink]  03 Sep 2018, 18:31
CE is correct

NixonDutt wrote:

Mathematically, 8 choose k = 56, therefore k=2 or 8. (Check out the formula for permutation)

You could also plug in each choice below and validate it. When you're picking 2 people out of 8, the first time you have 8 choices, second you have 7, so total 8*7=56,

6 also applies because picking 6 out of 8 people for the committee is the same as picking 2 out of 8 people NOT for the committee.

The other options don't make sense.
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Re: GRE Math Challenge #4-From a group of 8 people [#permalink]  14 Oct 2018, 17:22
To chose k people from 8 to form 56 k-person committees:

8!/x!(8-x)!= 56

I chose at random with 3, so 8!/3!5! = 8*7*6*5!/(3*2)5! = 8*7*6/6 = 56

So 3 and 5 are the answers because even when you switch 3! for 5! in the first half of the denominator, you still get the same value in the end.

I hope that makes sense!
Re: GRE Math Challenge #4-From a group of 8 people   [#permalink] 14 Oct 2018, 17:22
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# GRE Math Challenge #4-From a group of 8 people

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