# 7 card stud counting problem

Consider a standard deck of 52 cards. A game called 7 card stud is played where each player is dealt 7 cards. How many possible:

1. 7 card hands are there? --my answer:133,784,560
2. hands from (1) contain a 4-of-a-kind?
3. hands from (1) contain at least 5 cards of the same suit? ie- a flush, or straight flush. do not worry about other hands that can be made.
4. hands from (1) contain exactly 2 3-of-a-kinds? -- my answer:109,824

I'm stuck on questions 2 and 3!

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If you explain why it is that you were able to do 4 but not 2, more useful answers might be given. –  joriki Nov 13 '12 at 7:06
1. You need to choose 7 card from 52, so ${52 \choose 7} = 133784560$ (as the OP found).
2. You need to choose the rank of the poker, and another 3 card from 48, so $13*{48 \choose 3} = 224848$
3. You need to fix the suit, and then choose 7 card from 13, or 6 from 13 and 1 from 39, or 5 from 13 and 2 from 39, so $4* \left( {13 \choose 7} + {13 \choose 6}*{39 \choose 1} + {13 \choose 5}*{39 \choose 2} \right) = 4089228$
4. You need to choose the 2 rank of the tris, witch of the suit exclude from the tris, and another card from 44, so ${13 \choose 2}*4*4*44 = 54912$ (note that you forgot a factor 1/2)