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

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Intern
Joined: 18 Mar 2012
Posts: 13
Two members of a club are to be selected to represent the  [#permalink]

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Updated on: 10 May 2012, 07:21
1
1
00:00

Difficulty:

5% (low)

Question Stats:

92% (01:36) correct 8% (01:31) wrong based on 97 sessions

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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 ?

Originally posted by arnijon90 on 10 May 2012, 07:15.
Last edited by Bunuel on 10 May 2012, 07:21, edited 1 time in total.
Edited the question
Math Expert
Joined: 02 Sep 2009
Posts: 51072
Re: Two members of a club are to be selected to represent the  [#permalink]

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10 May 2012, 07:28
arnijon90 wrote:
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 ?

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

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.
_________________
Intern
Joined: 26 Nov 2011
Posts: 14
Re: Two members of a club are to be selected to represent the  [#permalink]

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21 Jun 2012, 00:30
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:
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 ?
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Re: Two members of a club are to be selected to represent the  [#permalink]

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25 Jun 2018, 11:10
arnijon90 wrote:
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

Letting n = the number of members in the club, we have:

nC2 = 190

n(n - 1)/2 = 190

n^2 - n = 380

n^2 - n - 380 = 0

(n - 20)(n + 19) = 0

n = 20 or n = -19

Since the number of members can’t be negative, n = 20.

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Re: Two members of a club are to be selected to represent the &nbs [#permalink] 25 Jun 2018, 11:10
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