# Thread: How many 7-card hands can be chosen

1. ## How many 7-card hands can be chosen

How many 7-card hands can be chosen from a standard deck of 52 cards such that no two cards are of the same rank?

Is it as simple as $\displaystyle {13\choose 7}$ or am I missing something? I tried starting with all $\displaystyle {52 \choose 7}$ possibilities and then subtracting the set of hands where there was exactly one pair, then exactly 3 the same etc. but it became quite convoluted.

It seems like I am missing a lot of potential hands with $\displaystyle {13\choose 7}$ as there are four valid choices from each of the 13 ranks each time you choose - not just one.

Would it be:

$\displaystyle {13\choose 1}\times 4 \times {12\choose 1}\times 4 ... \times {7\choose 1}\times 4 = \frac{13!}{6!}\times 4^7$ ? but then what would I divide by to remove equivalent permutations?

2. ## Re: How many 7-card hands can be chosen Originally Posted by terrorsquid Would it be:

$\displaystyle {13\choose 1}\times 4 \times {12\choose 1}\times 4 ... \times {7\choose 1}\times 4 = \frac{13!}{6!}\times 4^7$ ? but then what would I divide by to remove equivalent permutations?
I think you divide by 7! because the order of cards does not matter. So, this is $\displaystyle \binom{13}{7}\cdot4^7$: first you select the rank and then for each rank you select the suit.

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