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

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