There are 10 children in a company's day-care center, and a pair of

First way

1. first child out of 10 - 10 possibilities
2. second child out of 9 - 9 possibilities. the total number of pairs = 9*10=90
3. exclude xy yx cases: 90/2=45

Second way

N=10C2=10!/(8!2!)=10*9/2=45

Hi All,

This is an example of a fairly-straightforward Combinations question. It asks how many different pairs of children (meaning "sets of 2") are there that can be selected from a group of 10 children. In these situations, the ORDER of the two children does not matter, so if 'child A' plays 'child B' then that pair is the SAME as when 'child B' plays 'child A.' As such, you are NOT supposed to count that pairing twice. The Combinations Formula removes all of those duplicate pairs.

Combinations = N!/[K!(N-K)!] Where N is the total number of children and K is the number that you are going to pick.

Here, we have N = 10 and K = 2

10!/[2!8!] = (10)(9)/(2)(1) = 45 different pairs of 2 children.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Hello all,

I also did it like this:

10 children, one pair (2 children) selected and 8 children not:

10!/2!*8! = 9 * 10 / 1*2 = 90 / 2 = 45.

However, if we want to use pairs instead of the number of children, could we also say that out of the 5 pairs we are selecting 1 pair? And how would this go using combinations?

hi pacifist,
the meaning of question will change if you write out of the 5 pairs we are selecting 1 pair?..
it means each pair is a separate identity and you have to choose one out of it ..
the ans will be 5c1=5...
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Yes, I also tested it and ended up with 5. But I wnted to see if there would be another approach to correct for it. For example, 5*9=45. But is this logical?

no pacifist, it will not be correct.
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Solving tricky questions like this is really difficult until you do not have taken good basic educations and that is why I have decided to send my daughter to Phoenix kindergarten for her better future.

Out of 10 children at list a pair means 2

Therefore 10C2 = 10×9/2×1 = 5×9 =45

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Solution:

Since 2 children are selected from 10 and the order of selection doesn’t matter, there are 10C2 = (10 x 9)/2 = 45 different pairs.

Answer: D
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