This question has probably been asked before, but just to be clear here, I am NOT asking for the answer, I know the answer. What i want to know is why my solution is not equivalent to the actual solution.

Three of a kind in a deck means you draw 3 cards of the same rank, and 2 other cards each of different ranks.

So. The solution in my book is the following: (13 C 1)(4 C 3) * (12 C 2)(4 C 1)(4 C 1)
The logic is the following:
1) Choose 1 out of the 13 ranks, then for the 4 cards in the chosen rank, pick 3.
2) Choose a combination of 2 ranks from the remaining 12, then for the 4 cards in each rank, choose 1.
Total combinations: 54912

My solution is the following:
((13 C 1)(4 C 3)) * ((12 C 1)(4 C 1)) * ((11 C 1)(4 C 1))
1) My logic for the 3 of a kind is the same as the solution.
2) Choose 1 of the remaining 12 ranks, and 1 card from the 4 cards in the rank.
3) Choose 1 of the remaining 11 ranks, and 1 card from the 4 cards in the rank.
Total combinations: 109824

My solution says that the combinations are exactly double of the book's solution. I'm just very confused as to why this is. If anyone could explain this to me I would be very appreciative, thank you.

  • 1
    $\begingroup$ suppose that you have $3$ aces. Then you choose a $9$ and after that an $8$. If you choose first an $8$ and after that a $9$ then you get the same result. You are actually double-counting this result. To repair that your result $109824$ must be divided by $2$. $\endgroup$ – drhab Dec 14 '14 at 19:41

In your method, for your 2 other cards, you are counting, say, "10 of spades and 9 of hearts" and "9 of hearts and 10 of spades" as two different occurrences I think. When, in fact, it's the same hand.

  • $\begingroup$ Indeed, the reason why the OP is getting twice the result is because he is double-counting. $\endgroup$ – Newb Dec 14 '14 at 19:42

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