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

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

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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 ?

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Hello AAPL.

Your answer is very well.

Another way you could try it is: we have to choose 2 people (the order doesn't matter), we have two places: _____ _____.

For the first place we have n options and for the second place we have n-1 options. So, we have n*(n-1) options. But, we have to divide it by 2 (beacuse the order doesn't matter). That is to say, $$\frac{n\cdot\left(n-1\right)}{2}=36\ \leftrightarrow\ \ n\left(n-1\right)=72\ \leftrightarrow\ n=9.$$ I hope this explanation can help you.

Feel free to ask me again if you have a doubt.

Regards.

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