Two members of a club are to be selected to represent the

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Two members of a club are to be selected to represent the [#permalink]

10 May 2012, 08:15

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Two members of a club are to be selected to represent the club at a national meeting. if there are 190 different possible selections of the 2 members, how many members does the club have?

A. 20
B. 27
C. 40
D. 57
E. 95

I know I can just plug in answer choices in to the formula n!/(n-r)!*r!=190

n=20

but I saw a explanation on this problem where they showed this step

n!/(n-2)! = n(n-1)

How is this step possible ?

[Reveal] Spoiler: OA

Last edited by Bunuel on 10 May 2012, 08:21, edited 1 time in total.

Edited the question

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Re: Two members of a club are to be selected to represent the [#permalink]

10 May 2012, 08:28

arnijon90 wrote:

Welcome to GMAT Club. Below is an answer to your question.

C^2_n=190 --> \frac{n!}{2!*(n-2)!}=190. Now, notice that n!=(n-2)!*(n-1)*n, so \frac{(n-2)!*(n-1)*n}{2!*(n-2)!}=190 --> \frac{(n-1)*n}{2}=190 --> (n-1)n=380 --> n=20.

Answer: A.

P.S. Please always post answer choices for PS questions. On the PS section one should always look at the answer choices before starts to solve a problem. They might often give a clue on how to approach the question.
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Re: Two members of a club are to be selected to represent the [#permalink]

21 Jun 2012, 01:30

I have done this problem by substituting numbers

We have n!/2!(n-2)! = 190

or n(n-1)/2 = 190

n=20 satisfies the equation

arnijon90 wrote:

Re: Two members of a club are to be selected to represent the [#permalink] 21 Jun 2012, 01:30

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Two members of a club are to be selected to represent the

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Two members of a club are to be selected to represent the

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