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# Two members of a certain club are selected to speak at the

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VP
Joined: 30 Jun 2008
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Two members of a certain club are selected to speak at the [#permalink]  20 Sep 2008, 22:13
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Two members of a certain club are selected to speak at the next club meeting. If there are 36 different possible selections of the 2 club members, how many members does the club have?
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VP
Joined: 30 Jun 2008
Posts: 1048
Followers: 11

Kudos [?]: 285 [0], given: 1

Re: PS: How many members ? [#permalink]  20 Sep 2008, 23:46
I did it this way.

we know total number of ways of selecting 2 People out of n people is nC2 = 36(given in the question)

ie n!/(2!*(n-2)!) = [n(n-1)*(n-2)!] / [(2!*(n-2)!)] = n(n-1)/2

so n(n-1)/2=36
n(n-1)=72
we know 72 = 9*8

so n=9

is the approach right ?
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SVP
Joined: 17 Jun 2008
Posts: 1579
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Kudos [?]: 185 [0], given: 0

Re: PS: How many members ? [#permalink]  22 Sep 2008, 00:52
amitdgr wrote:
I did it this way.

we know total number of ways of selecting 2 People out of n people is nC2 = 36(given in the question)

ie n!/(2!*(n-2)!) = [n(n-1)*(n-2)!] / [(2!*(n-2)!)] = n(n-1)/2

so n(n-1)/2=36
n(n-1)=72
we know 72 = 9*8

so n=9

is the approach right ?

I used exactly the same approach.
Re: PS: How many members ?   [#permalink] 22 Sep 2008, 00:52
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