# Card Counting Problem

I tried these homework problems but think I ended up double counting on a few. I will post the question followed by my answer. I need help checking that they are correct or what the answer really is.

Consider a game where 4 cards are dealt from a standard 52 card deck. Determine the number of hands that contain:

(a) 4 different suits

$13C1^4 \cdot 4C1 \cdot 3C1 \cdot 2C1 \cdot 1C1 = 685,464$

(b) 4 different suits and 4 different weights

$13C1 \cdot 4C1 \cdot 12C1 \cdot 3C1 \cdot 11C1 \cdot 2C1 \cdot 10C1 \cdot 1C1 = 411,840$

(c) 4 consecutive ranks

$11C1 \cdot 4C1^4 = 2816$

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+1 For showing what you tried. –  JavaMan Dec 6 '12 at 18:35

If we consieder hands as unordered sets of cards (i.e. there are ${52\choose 4}$ hands), then there are $13^4$ hands with four different suits ($13$ choices for each suit), $4!{13\choose 4}=\frac{13!}{9!}$ hands with four different suits and four different weights (choose four weigts and permute them to the suits), $(13-3)\cdot 4^4$ hands with consecutive ranks (choose wone of $10$ "starting points" and assign suits).

If you consider hands as ordered sets instead (so that there are $\frac{52!}{48!}$ hands in total), each of the above results must be multiplied with $4!$.

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(a) We need to choose a card from each suit. The $\spadesuit$ can be chosen in $\dbinom{13}{1}$ ways. For each such choice, the $\heartsuit$ can be chosen in $\dbinom{13}{1}$ ways, and so on. That gives a total of $\dbinom{13}{1}^4$.

You had this as a component of your answer. The rest is not right. You multiplied the correct number by $4!$. The product counts the number of ways of choosing such a hand and laying it out in a row.

(b) The $\spadesuit$ can be chosen in $13$ ways. For each such choice, the $\heartsuit$ can be chosen in $12$ ways, and so on, for a total of $13\cdot 12\cdot 11\cdot 10$. Your answer again has an extra factor of $4!$.

(c) Let's decide Ace is at the bottom, it doesn't matter. Then the "top" rank can be any of King down to $4$, a total of $10$ choices. For each rank, the actual card can be chosen in $4$ ways, for a total of $10\cdot 4^4$. That is almost your answer. The procedure was right, slight miscount.

So everything was going fine, except you need to remember what a hand means: order of cards is irrelevant to the count. There is only one $5$-card poker hand with all four Aces and the King of $\clubsuit$.

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Thank you for explaining it. I see that I considered hands as ordered sets. –  Joseph Schmidt Dec 6 '12 at 18:46