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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).

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It may be fun to use Brian's comment to see if you can generate the same answer as wiki. (using what you have as a start) –  picakhu Nov 14 '12 at 21:49
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up vote 2 down vote accepted

The derivation on the Wikipedia page is correct; the expression that you’ve given counts each hand consisting of a three-of-a-kind and two pairs twice, once for each pair. It also overcounts the hands with two threes-of-a-kind: $KKKQQQJ$ gets counted three times as $KKKQQxx$, once for each of the $3$ ways to pick two of the three queens, and three more times as $QQQKKxx$.

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So this counts KKKQQQJ and KKKQQJQ as different hands? –  user1038665 Nov 14 '12 at 21:54
    
@user1038665: It counts $K_1K_2K_3Q_1Q_2Q_3J$ as $K_1K_2K_3Q_1Q_2$ with $Q_3J$ extra, as $K_1K_2K_3Q_1Q_3$ with $Q_2J$ extra, as $K_1K_2K_3Q_2Q_3$ with $Q_1J$ extra, and as three more with the kings and queens interchanged. –  Brian M. Scott Nov 14 '12 at 21:56
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