Basic holdem combination question.

[BruceZ]

I get that there are 13 pairs, and 78 suited cards (2 card combinations out of 13 cards - suit is irrelevant, so no need to multipy by 4). What I don't really understand is how the 78 unsuited combinations are calculated.

Exactly the same way as the suited cards. For every suited hand there is a corresponding offsuit hand, and visa versa. Of course there are 3 times more offsuit hands (12 vs. 4) but we are counting all 12 of them as the same hand.