So, how many ways can we have three aces? There are four aces, and the number of unordered collections of 3 objects from a set of 4 is:

4 C 3 = 4

Now, I assume that we don’t want the other ace as part of our non-pair. So, we have 48 other cards to choose from. How many ways can we take 2 cards from 48?

48 C 2 = 1128

But, we’ve counted too many. We’ve included all the pairs in our counting. So, if we count the number of pairs we can make from two cards, we can subtract that number from 1128 to obtain the number of 2-card hands that aren’t pair. There are 12 ranks left (since we don’t have the aces), and there are 4 cards per rank. Given any rank, how many pairs can we form? Well, we want a choice of 2 cards from 4, so:

4 C 2 = 6

Thus, over all the ranks, there are:

12 x 6 = 36

possible pairs we can make with 2 cards, excluding the aces. So, we therefore have:

1128 - 36 = 1092

possible two card hands that aren’t pairs. Finally, since the choices of aces and the other two cards are completely independent, the number of ace triples that aren’t full houses will be the product of the number of ace triples (i.e. 4) and the number of two card combinations that aren’t pairs (i.e. 1092). Thus the final answer is:

4 x 1092 = 4368