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.
Is there a strategic approach to this PS question?
I did it this way.
We know total number of ways of selecting 2 People out of n people is nC2 = 36(given in the question)
$$ie,\ \ n!/(2!*(n-2)!)=[n(n-1)*(n-2)!]/[(2!*(n-2)!)]=n(n-1)/2$$
so n(n-1)/2=36
n(n-1)=72
We know 72 = 9*8
so n=9
Is the approach right ?
A. 5
B. 6
C. 7
D. 8
E. 9
The OA is E.
Is there a strategic approach to this PS question?
I did it this way.
We know total number of ways of selecting 2 People out of n people is nC2 = 36(given in the question)
$$ie,\ \ n!/(2!*(n-2)!)=[n(n-1)*(n-2)!]/[(2!*(n-2)!)]=n(n-1)/2$$
so n(n-1)/2=36
n(n-1)=72
We know 72 = 9*8
so n=9
Is the approach right ?
