phoenixhazard wrote:Sounds like such an easy question but I can't find how to do it fast and simple.

"There are 10 children in a day care, and a pair of children is to be selected to play a game. At most, how many different pairs are possible?"

This time I am correct dude (think so)
It is a Combinatorics problem.
Out of 10 Children 2 can be selected in C(10,2) ways.

C(10,2) = 10!/(2!*8!) = 45

awesome thanks

so its (Total Number of students)/(Number per group)*(other options)?

phoenixhazard wrote:awesome thanks

so its (Total Number of students)/(Number per group)*(other options)?

This is a formula in General:

Let n be the total number of choices
r be the choices to be selected from n

If the arrangement of r choices does not matter then use Combinatorics:
C(n,r) = n!/[r!*(n-r)!]

If the arrangement of r choices matters then use Permutations:
P(n,r) = n!/(n-r)!

One Q for you

Q: "There are 10 children in a day care, and 8 children are to be selected to play a game. At most, how many different groups are possible?"

Nice one, Shovan.