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88% (01:30) correct
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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 2 members, how many members does the club have?
A. 20
B. 27
C. 40
D. 57
E. 95
A. 20
B. 27
C. 40
D. 57
E. 95
I know I can just plug in answer choices in to the formula n!/(n-r)!*r!=190
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 ?
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 ?
10 May 2012, 08:28
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arnijon90
Welcome to GMAT Club. Below is an answer to your question.
\(C^2_n=190\) --> \(\frac{n!}{2!*(n-2)!}=190\). Now, notice that \(n!=(n-2)!*(n-1)*n\), so \(\frac{(n-2)!*(n-1)*n}{2!*(n-2)!}=190\) --> \(\frac{(n-1)*n}{2}=190\) --> \((n-1)n=380\) --> \(n=20\).
Answer: A.
P.S. Please always post answer choices for PS questions. On the PS section one should always look at the answer choices before starts to solve a problem. They might often give a clue on how to approach the question.
13 Feb 2019, 15:18
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nC2, or n(n-1)/2 = 190
n(n-1) = 380
We can use prime factorization: 380 = 2*2*5*19
Becomes pretty obvious that 19 is one of the numbers (n-1) and the other is n=20.
n(n-1) = 380
We can use prime factorization: 380 = 2*2*5*19
Becomes pretty obvious that 19 is one of the numbers (n-1) and the other is n=20.
