Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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).

share|cite|improve this question
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
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 $QQQKKxx$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.