How do you get 32% for the odds of hitting the flop with unpaired hole cards? You have 2 cards in your hand, leaving 50 in the deck. 6 of those cards left in the deck will pair your hand (or better), which means 44 won't. So from the (50 choose 3) total possible flops (44 choose 3) of them don't pair your hand. That means the odds of NOT hitting the flop are 44x43x42/50x49x48=67.6%. Therefore the odds of hitting the flop are 1 67.6%=32.4%. This counts all hands that include a pair which hits your hand, i.e. one pair, two pair, trips, full house, and quads, but not any straights or flushes. It also does not count the chance of the board making a pair or trips by itself. But you probably don't want to count those anyway because then everyone has that hand and you didn't really improve.
If you want the odds of flopping one pair exactly, and nothing better, look at the combinations that give that result. There are 3 cards that will pair your first card and 3 that pair your second. For each of those you have 44 cards left that will not pair your hole cards. But you can't select just any two of those because you don't want another pair on the board, giving you two pair, which is the way I have seen it posted several times. Instead, you select 2 of the 11 remaining ranks, then give each one of the four suits. So the calculation looks like this: 3x2x(11 choose 2)x4x4 = 5280. Again the total number of flops is (50 choose 3). So the odds of flopping exactly one pair, using one of the unpaired cards in your hand, is 5280/19600 = 26.94%. (The mistaken calculation, which includes two pair, gives 28.96%)
