Nevermind! yay for the internet.

I guess this is a combination problem - which i am not sure was covered in OG11....

10!
-----
2! x (10-2)!

shortens down to

10 x 9
-----
2

45.

yay

Here is another way to look at this which of course is consistent with Dime0000's answer (which is the shortcut). If the 10 students are a1, a2, a3, a4, a5, a6, a7, a8, a9, and a10
the 1st kid (a1) has 9 other kids to play with a2, a3, a4, ... a10
the 2nd kid (a2) has 8 more kids to play with a3, a4, a5, ...a10
the 3rd kid (a3) has 7 more kids to play with a4, a5, a6, a7, a8, a9, a10
and so on

Eventaually, you have 9+8+7+6+5+4+3+2+1=45 groups

dime0000 wrote:I don't believe I saw this question discussed.

There are 10 children in a company's day-care center, and a pair of children will be selected to play a game. At most, how many different pairs are possible?

100
90
50
45
25

I guess the answer is 45.. I dont know how they got to that.

Can anyone explain?