I have twenty balls numbered 1 through 20 in a bin.  In how many ways can I select 2 balls, if the order in which I draw them doesn't matter?
I want to count the number of combinations of 2 balls out of 20, where the order doesn't matter. This is a common problem in combinatorics, and there is a formula for it: n choose r, or nCr, which is n!/(r!(n-r)!), where n is the total number of balls, r is the number of balls I select, and ! means factorial. In this case, n = 20 and r = 2, so nCr = 20!/(2!(20-2)!) = 20!/(2!18!). I can simplify this by canceling out some common factors: 20! = 20*19*18!, and 2! = 2*1, so 20!/(2!18!) = 20*19/2 = 190. 