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-1Total No. of Selection of r out of n objects are defined by nCr = n! / [(r!)(n-r)!]i.e. If total member = n

then n

C2 = 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

Answer: Option E

Method-2Let, 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 = 36Answer: Option E

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