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

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Director
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Two members of a club are to be selected to represent the [#permalink] New post 17 Sep 2006, 16:22
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
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 [#permalink] New post 17 Sep 2006, 20:00
.
.
.
.
.

answer is A.


you get the equation x^2 -x - 380 = 0.

try the answers, and u get A.
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 [#permalink] New post 17 Sep 2006, 22:15
Yeah - here you have nC2 = 190.

Figure out n by plugging in numbers.
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 [#permalink] New post 17 Sep 2006, 22:17
This is a combinations thing (ie the order in which the members are picked does not matter). The number of ways in which 2 members can be picked is 190, so:

(x!)
190 = ------------
(2!)*(x-2)!

which can be worked to:

380 = x*(x-1)

Then plug away! Since the answers are given in increasing value start with the middle one (C). You get 40*39 which is >1000 so then move "up" to a smaller answer. Anwers B ia also too big. Then finally to A. 20*19 = 380 so the answer is A. By starting in the middle you know which way to move ("up" or "down") and that way you only need to plug in 3 out of the 5 answers, saving a little time.
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 [#permalink] New post 17 Sep 2006, 22:20
crap sorry about the formatting of the first equation:

190 = (x!)/[2!*(x-2)!]
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Re: Club [#permalink] New post 17 Sep 2006, 22:35
alimad 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


straight A. 20C2 = 190
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 [#permalink] New post 17 Sep 2006, 23:13
A.

n=20
r=2

Combination, order doesnt matter...

n!/(n-r)!(r!)
= (20!)/(18!)(2!)
= 20x19/2
= 380/2
= 190
  [#permalink] 17 Sep 2006, 23:13
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