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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, 08:21
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Difficulty:

5% (low)

Question Stats:

86% (01:39) correct 14% (01:59) wrong based on 99 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, 08:15.
Last edited by Bunuel on 10 May 2012, 08:21, edited 1 time in total.
Edited the question
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Joined: 02 Sep 2009
Posts: 55228
Re: Two members of a club are to be selected to represent the  [#permalink]

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10 May 2012, 08:28
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2
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.
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Joined: 26 Nov 2011
Posts: 13
Re: Two members of a club are to be selected to represent the  [#permalink]

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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:
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, 12:10
1
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  [#permalink]

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13 Feb 2019, 15:18
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nC2, or n(n-1)/2 = 190
n(n-1) = 380
We can use prime factorization: 380 = 2*2*5*19
Becomes pretty obvious that 19 is one of the numbers (n-1) and the other is n=20.
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Re: Two members of a club are to be selected to represent the  [#permalink]

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15 Feb 2019, 18:32
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

The order of selecting the two representatives is of no importance; thus, we use combinations. Letting n = the total number of members of the club, we can create the equation:

nC2 = 190

nC2 = n!/[(n - 2)! x 2!] =[ n x (n-1) x (n-2) x (n-1) x … x 1] / {[(n-2) x (n-1) x … x 1] x 2!}

We see that all the factors in the numerator cancel with those in the denominator, except n x (n-1). Thus, we have:

(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 n can’t be negative, then n must be 20.

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

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Joined: 24 Jun 2008
Posts: 1520
Re: Two members of a club are to be selected to represent the  [#permalink]

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16 Feb 2019, 16:35
(n)(n-1)/2! = 190
(n)(n-1) = 380
n^2 - n = 380

Since n is a fairly large positive integer, n^2 is much larger than n, so we want a value of n^2 that is very close to 380. Glancing at the answer choices, n = 20 is the only realistic candidate, and it's easy to plug it back in to confirm it works.
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Re: Two members of a club are to be selected to represent the   [#permalink] 16 Feb 2019, 16:35
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