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by MBA.Aspirant » Fri Jun 24, 2011 7:19 am
There are x people at a dance where x is an integer and greater than 1. The number of different pairs that can be partnered together is z. Which of the following could be the value of z?

2
6
9
14
20
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by Frankenstein » Fri Jun 24, 2011 7:35 am
Hi,
Number of pairs that can be formed by x people is xC2
Go from the options 6 = 4C2.

Hence, B
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by MBA.Aspirant » Fri Jun 24, 2011 8:08 am
Thanks Frankenstein. But why go for 6 directly?

2 is out cause x>1

why not the others? Thanks
Last edited by MBA.Aspirant on Fri Jun 24, 2011 8:46 am, edited 1 time in total.
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by Frankenstein » Fri Jun 24, 2011 8:22 am
MBA.Aspirant wrote:Thanks Frankenstein. But why go for 6 directly?

2 is out cause x>1
14 is out cause the no. of pairs should be even.

why not the others? Thanks
Hi,
I meant to say, check all options. Just by going in order, when 6 fits, we don't need to check others.
Btw, the number of pairs need not be even. If there are 3 people, the number of pairs is 3C2 = 3(odd)
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by Ozlemg » Fri Jun 24, 2011 8:37 am
could you elaborate little bit? I did not understand :(
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by Brent@GMATPrepNow » Fri Jun 24, 2011 8:47 am
In this question, we are selecting 2 people to create a dance pairing, and the order of the 2 selected people does not matter (for example, selecting A and B to be a pair is the same as selecting B and A).

Since the order does not matter here, this is a combination question.

We can select/choose 2 people from x people in xC2 different ways (this notation, xC2, can be read as "x choose 2") where xC2 = x!/(x-2)!(2!)

From here, we should plug different values of x (2, 3, 4, 5, etc) until we find an output that matches one of the answer choices.

To learn a quick way to calculate combinations in your head, you can watch this video: https://youtu.be/-bbu2h-07iA

Cheers,
Brent
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