I have two solutions to the problem and I can't figure out which is correct. One of the solutions is on this wiki page, and another is offered by my professors as follows:
${13 \choose 1}{4 \choose 3}{12 \choose 1}{4 \choose 2}({{46 \choose 2}-1})$
The reasoning for the last term is as follows: a full house with 7 cards means your hand contains 3 of a kind, 2 of another, and contains no better hand, specifically no four of a kind. If the first five cards you pick are K,K,K,Q,Q, the 6th and 7th cards must be different (ruling out 5 cards)- neither can be a K, and they both can't be a Q (but it's OK if only one is).