This is how you can approach these problems in general on your own:
The first of the pair can be picked in N (=100) ways. You don't want to pick this item again, so the second of the pair can be picked in N-1 (=99) ways. In total you can pick 2 items out of N in N(N-1) (= 100*99=9900) different ways.
But hold on, this way you count also different orderings: AB and BA are both counted. Since every pair is counted twice you have to divide N(N-1) by two (the number of ways that you can order a list of two items). The number of subsets of two that you can make with a set of N is then N(N-1)/2 (= 9900/2 = 4950).