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by yvonne12 » Wed Mar 07, 2007 3:27 pm

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2 members of a club are to be selected to represent the club at a national meeting. if there are 190 different possible selections of 2 members, how many members does the club have?

please explain
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by Argen » Wed Mar 07, 2007 4:15 pm
nC2=n!/((n-2)!2!)=190
n(n-1)=380
n=20
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by yvonne12 » Wed Mar 07, 2007 9:23 pm
thank you
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by Cybermusings » Tue Mar 27, 2007 1:58 am
190 selections of "n" members taken 2 at a time...

Which means nC2 = 190

n! /(n-2)! * 2! = 190

n*n-1*n-2!/n-2!*2=190

n*n-1=380
Therefore n=20
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by Scott@TargetTestPrep » Mon Jun 25, 2018 11:12 am
yvonne12 wrote:2 members of a club are to be selected to represent the club at a national meeting. if there are 190 different possible selections of 2 members, how many members does the club have?
Letting n = the number of members in the club, we have:

nC2 = 190

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 the number of members can't be negative, n = 20.

Answer: 20

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