Two members of a certain club are selected to speak . . . .

Hi M7MBA,

We're told that 2 members of a certain club are selected to speak at the next club meeting and that there are 36 different possible selections of the 2 club members. We're asked for the total number of members in the club. While the question does not state it, we are meant to assume that the 'order' of the two speakers does NOT matter - so we should treat this as a 'Combination' question. This question can be solved by TESTing THE ANSWERS and using the Combination Formula.

Let's TEST Answer B: 6 people
IF.... the club has 6 members, then there are 6!/2!4! = (6)(5)/(2)(1) = 30/2 = 15 possible pairs of speakers.
This is TOO SMALL (there's supposed to be 36 different pairs), so there must be MORE members...
Eliminate Answers A and B

Let's TEST Answer D: 8 people
IF.... the club has 8 members, then there are 8!/2!6! = (8)(7)/(2)(1) = 56/2 = 28 possible pairs of speakers.
This is TOO SMALL (there's supposed to be 36 different pairs), so there must be MORE members...
Eliminate Answers C and D

There's only one answer remaining...

Final Answer: E

GMAT assassins aren't born, they're made,
Rich

M7MBA 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

We can create the equation:

nC2 = 36

(n)(n - 1)/2! = 36

n^2 - n = 72

n^2 - n - 72 = 0

(n - 9)(n + 8) = 0

n = 9 or n = -8

Since n can’t be negative, n = 9.

Answer: E