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A chess club with 8 members can send 3 people to an advanced strategy

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A chess club with 8 members can send 3 people to an advanced strategy  [#permalink]

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New post 16 Apr 2018, 05:26
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Question Stats:

94% (00:40) correct 6% (01:21) wrong based on 71 sessions

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A chess club with 8 members can send 3 people to an advanced strategy class. How many different groups of 3 members could the club possibly send?

A. 48
B. 56
C. 64
D. 72
E. 96

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A chess club with 8 members can send 3 people to an advanced strategy  [#permalink]

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New post 16 Apr 2018, 08:12
Bunuel wrote:
A chess club with 8 members can send 3 people to an advanced strategy class. How many different groups of 3 members could the club possibly send?

A. 48
B. 56
C. 64
D. 72
E. 96



3 people can be sent to advanced strategy class while 5 people will not go to the advanced class.

The solution will be 8!/(3!*5!) = (8*7*6)/3! = 8*7 = 56

Hence option = B =56 is the answer.
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Re: A chess club with 8 members can send 3 people to an advanced strategy  [#permalink]

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New post 21 Apr 2018, 09:23
8C3 = 56

choosing 3 members out of 8 hence 8C3
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Re: A chess club with 8 members can send 3 people to an advanced strategy  [#permalink]

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New post 21 Apr 2018, 09:51
Hi guys , how can we know if the question is part of combinations or permutations???


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Re: A chess club with 8 members can send 3 people to an advanced strategy  [#permalink]

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New post 21 Apr 2018, 10:01
Nirenjan wrote:
Hi guys , how can we know if the question is part of combinations or permutations???

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Hey Nirenjan ,

Whenever you select something, it is a combination.

In this question, you are selecting 3 people out of 8. Hence, we are using combinations formula.

Check COMBINATORICS for more details.
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Re: A chess club with 8 members can send 3 people to an advanced strategy  [#permalink]

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New post 21 Apr 2018, 10:03
Nirenjan wrote:
Hi guys , how can we know if the question is part of combinations or permutations???


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Attaching link to relevant topic, hope this is what u need

https://gmatclub.com/forum/permutations-and-combinations-simplified-150835.html
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Re: A chess club with 8 members can send 3 people to an advanced strategy  [#permalink]

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New post 21 Apr 2018, 10:50
A chess club with 8 members can send 3 people to an advanced strategy class. How many different groups of 3 members could the club possibly send?

8C3 = \(\frac{8!}{3! * 5!}\) = 56

Ans: B
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Re: A chess club with 8 members can send 3 people to an advanced strategy  [#permalink]

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New post 23 Apr 2018, 04:15
Bunuel wrote:
A chess club with 8 members can send 3 people to an advanced strategy class. How many different groups of 3 members could the club possibly send?

A. 48
B. 56
C. 64
D. 72
E. 96


Total member = 8

Needed member = 3

Selecting 3 out of 8 members = \(8C3\)

\(\frac{8!}{3!*5!}\)

56 (B)
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Re: A chess club with 8 members can send 3 people to an advanced strategy &nbs [#permalink] 23 Apr 2018, 04:15
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