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

Question

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

amitdgr · Answer

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)/2

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

so n=9

is the approach right ?

scthakur · Answer

I used exactly the same approach.

AdmitJA · Answer

I used the same approach too. Don't see another way. Do you have an OA/OE?

Bunuel · Answer

Solution above is correct. Similar question to practice: two-members-of-a-club-are-to-be-selected-to-represent-the-132340.html

kjax · Answer

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.

GMATinsight · Answer

Method-1

Total No. of Selection of r out of n objects are defined by n C r = n! /

i.e. If total member = n
then n C 2 = n! /   = 36

i.e. n*(n-1)* (n-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

Answer: Option E

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

Answer: Option E

BrentGMATPrepNow · Answer

Let n = TOTAL number of members.

Since the order in which we select the two members does not matter, this is a COMBINATION question. 
We can write: nC2 = 36

This means (n)(n-1)/(2)(1) = 36
In other words, (n)(n-1) = 72
We COULD try to solve this equation, but it might be faster to check the answer choices.

A) 5
If n = 5, then we get: (n)(n-1) = (5)(5-1) = 72 NOT TRUE
.
.
.
E) 9
If n = 9, then we get: (n)(n-1) = (9)(9-1) = 72 WORKS!!

Answer: E

Cheers,
Brent

RELATED VIDEO
 https://www.youtube.com/watch?v=05a0A7vjG8I

Archit3110 · Answer

use combinatorics to solve ; ncr ; nc2 is given relation 
out of given option 9c2 = 36 so members = 9
IMO E

EphraimMat · Answer

I used the slots method and used answers choices in them
for example 2 member means -> ___ ___

and remember to think of overcounting, so any answer would be divided by 2! = 2*1 = 2, since they are counted these many times again.

1) 5 ->  5  *  4  = 20/2 = 10= not equal to 36
similarly for other options will give you 35/2 , 42/2 , 56/2 and finally 72/2 which is equal to 36

therefore our answer is 9.

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

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