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

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M7MBA: Moderator; Posts: 2058; Joined: Sun Oct 29, 2017 4:24 am; Thanked: 1 times; Followed by:5 members

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

by M7MBA » Sun Jan 07, 2018 11:28 am

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

The OA is E.

Why is 9 the correct option? I couldn't solve this PS question. I didn't know how to afford it. Experts, may you give me some help? Thanks in advanced.

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Source: Beat The GMAT — Problem Solving | Sun Jan 07, 2018 11:28 am

GMAT/MBA Expert

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by [email protected] » Sun Jan 07, 2018 4:30 pm

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

Contact Rich at [email protected]

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regor60: Master | Next Rank: 500 Posts; Posts: 416; Joined: Thu Oct 15, 2009 11:52 am; Thanked: 27 times

by regor60 » Mon Jan 08, 2018 6:42 am

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

The OA is E.

Why is 9 the correct option? I couldn't solve this PS question. I didn't know how to afford it. Experts, may you give me some help? Thanks in advanced.

Another way is to use the combination formula, in this case X!/(2!*(X-2)!) = 72, where X denotes the number of members and 2 represents the selection of 2 club members.

X!/(X-2)! is just X*(X-1), so the above reduces to X*(X-1) = 72.

Rearranging: X^2-X-72=0. You could use the quadratic formula, but if you recognize that 9*8 = 72 conveniently yields the difference of 1 in the quadratic equation, you know it's going to be 9 or 8. Using a brief trial and error for plus or minus yields [spoiler]9{/spoiler]

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8085; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Mon Aug 12, 2019 10:43 am

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

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Founder and CEO
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