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

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Two members of a certain club are selected to speak at the next club [#permalink]

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20 Sep 2008, 23:13
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5% (low)

Question Stats:

87% (01:03) correct 13% (01:45) wrong based on 134 sessions

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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?

A. 5
B. 6
C. 7
D. 8
E. 9
[Reveal] Spoiler: OA

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Last edited by Bunuel on 22 Nov 2014, 07:07, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.

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Re: Two members of a certain club are selected to speak at the next club [#permalink]

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21 Sep 2008, 00:46
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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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Re: Two members of a certain club are selected to speak at the next club [#permalink]

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22 Sep 2008, 01:52
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.

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Re: Two members of a certain club are selected to speak at the next club [#permalink]

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22 Nov 2014, 03:54
I used the same approach too. Don't see another way. Do you have an OA/OE?
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Re: Two members of a certain club are selected to speak at the next club [#permalink]

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22 Nov 2014, 07:10
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Expert's post
joseph0alexander wrote:
I used the same approach too. Don't see another way. Do you have an OA/OE?

Solution above is correct. Similar question to practice: two-members-of-a-club-are-to-be-selected-to-represent-the-132340.html
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Re: Two members of a certain club are selected to speak at the next club [#permalink]

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21 Jul 2015, 10:02
This one was easiest for me to solve by backsolving. I didn't bother with Algebra: Too much potential for mistakes.

Test answer B: 6C2 = 15. We need a number bigger than 6 to get us to 36. Eliminate A & B.
Test answer D: 8C2 = 28. Still not big enough. The answer must be E.

Only had to use 2 calculations. Proof: 9C2 = 36.

This is a great example of a problem that can be solved fastest by using the answers.

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Re: Two members of a certain club are selected to speak at the next club [#permalink]

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21 Jul 2015, 10:35
amitdgr wrote:
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?

A. 5
B. 6
C. 7
D. 8
E. 9

Method-1

Total No. of Selection of r out of n objects are defined by nCr = n! / [(r!)(n-r)!]

i.e. If total member = n
then nC2 = n! / [(2!)(n-2)!] = 36

i.e. n*(n-1)*(n-2)! / [(2!)(n-2)!] = 36
i.e. n*(n-1) = 72

but 9*8 = 72 (for Positive Values of n)
therefore, n*(n-1) = 9*8
i.e. n= 9

Method-2
Let, Total members = n
@n=2, no. of ways to select 2 out of 2(A,B) = (AB) = 1
@n=3, no. of ways to select 2 out of 3(A,B,C) = (AB, AC, BC) = 2+1 = 3
@n=4, no. of ways to select 2 out of 4(A,B,C,D) = (AB, AC, AD, BC, BD, CD) = 3+2+1 = 6

i.e. @n=5, Total Selection of 2 = 4+3+2+1 = 10
i.e. @n=6, Total Selection of 2 = 5+4+3+2+1 = 15
i.e. @n=7, Total Selection of 2 = 6+5+4+3+2+1 = 21
i.e. @n=8, Total Selection of 2 = 7+6+5+4+3+2+1 = 28
i.e. @n=9, Total Selection of 2 = 8+7+6+5+4+3+2+1 = 36

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Re: Two members of a certain club are selected to speak at the next club   [#permalink] 21 Jul 2015, 10:35
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