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Two members of a club are to be selected to represent...

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by swerve » Fri Dec 15, 2017 1:19 pm
two members of a club are to be selected to represent the club at a national meeting. If there are 190 different possible selections of the two members, how many members does the club have?

A. 20
B. 27
C. 40
D. 57
E. 95

The OA is A.

I know I can just plug in answer choices in to the formula n!/(n-r)!*r!=190, then

n=20

but I saw a explanation on this problem where they showed this step,

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

How is this step possible? Experts, can you assist me with this PS question? Thanks!
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by [email protected] » Fri Dec 15, 2017 3:50 pm
Hi swerve,

The 'math step' that you're referring to is a way to 'rewrite' part of the Combination Formula. Here's why it could be used here: if we have 20 total people and we're choosing 'groups of 2', we can use the Combination Formula and create the following calculation:

20!/(2!)(18!) = (20)(19)(18)(17).....(3)(2)(1) / (2)(1)(18)(17)....(2)(1)

Each of the values from (18)(17)...(3)(2)(1) 'cancel out' in both the numerator and denominator. Algebraically, this whole thing can be rewritten as...

(N)(N-1)(N-2)! / (2)(1)(N-2)!

"N" represents 20, "N-1" represents 19, "N-2"! represents 18! Notice that (N-2)! appears in both the numerator and the denominator, so we can 'cancel them out.' This leaves us with....

(N)(N-1)/(2)(1)

You still have to divide (N)(N-1) by 2 to get the correct answer though.

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by GMATWisdom » Sat Dec 16, 2017 7:38 am
swerve wrote:two members of a club are to be selected to represent the club at a national meeting. If there are 190 different possible selections of the two members, how many members does the club have?

A. 20
B. 27
C. 40
D. 57
E. 95

The OA is A.

I know I can just plug in answer choices in to the formula n!/(n-r)!*r!=190, then

n=20

but I saw a explanation on this problem where they showed this step,

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

How is this step possible? Experts, can you assist me with this PS question? Thanks!
N(N-1)(N-2)!/(N-2)!=N(N-1)
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by Scott@TargetTestPrep » Mon Sep 23, 2019 4:34 pm
swerve wrote:two members of a club are to be selected to represent the club at a national meeting. If there are 190 different possible selections of the two members, how many members does the club have?

A. 20
B. 27
C. 40
D. 57
E. 95
The order of selecting the two representatives is of no importance; thus, we use combinations. Letting n = the total number of members of the club, we can create the equation:

nC2 = 190

nC2 = n! / [(n - 2)! x 2!] = [ n x (n-1) x (n-2) x (n-3) x ... x 1] / {[(n-2) x (n-1) x ... x 1] x 2!}

We see that all the factors in the numerator cancel with those in the denominator, except n x (n - 1). Thus, we have:

(n)(n - 1)/2 = 190

n^2 - n = 380

n^2 - n - 380 = 0

(n - 20)(n + 19) = 0

n = 20 or n = -19

Since n can't be negative, then n must be 20.

Answer: A

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