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How many five card hands have at least one card from each suit?

I'm perplexed. I want to say that the answer is:

$\binom{4}{1} \cdot \binom{13}{1} \cdot \binom{3}{1} \cdot \binom{13}{1} \cdot\binom{2}{1} \cdot \binom{13}{1} \cdot \binom{1}{1} \cdot \binom{13}{2}$,

since we have to choose a suit, then a card in that suit, then three suits to choose from and then a card in this next suit, etc.

Is this correct? What is difficult for me is that I can't tell if my choices are independent or not. This has always been unclear for me using the Product rule for counting.

Another possible answer I can think of is $\binom{13}{1} \binom{13}{1} \binom{13}{1} \binom{13}{2} \div 5!$, since there is $5!$ ways to be dealt the same hand. Why is this not correct?

Thank you for you help.

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Your first calculation overcounts, and your second undercounts.

A five-card hand with at least one card in each suit must have two cards in one suit and one card in each of the other three suits. There are $4$ ways to pick the suit with two cards. Once it’s been picked, there are $\binom{13}2$ ways to pick two cards from it and $13^3$ ways to pick one card from each of the other three suits. Thus, there are

$$4\cdot\binom{13}2\cdot13^3=685,464$$

such hands.

Your first calculation errs in two ways. First, you’re distinguishing the order of the suits with your factors of $4,3,2$, and $1$. Secondly, you’re not distinguishing the suit with $2$ cards from the others. The first error introduces a spurious factor of $4!=24$; the second omits the important factor of $4$ for picking out the suit with $2$ cards. The net effect is that your answer is $\frac{24}4=6$ times the correct answer.

Your second calculation omits that same factor of $4$ for picking out the suit with two cards and then gets rid of an overcounting that you did not actual perform; the net effect is that it gives a result that is too small by a factor of $4\cdot5!$.

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  • $\begingroup$ I see, so choose the 2 cards in the same suit first. Thanks. $\endgroup$ – user43666 Dec 7 '13 at 0:29
  • $\begingroup$ @user43666: I think that that’s the easiest way to analyze it correctly, yes. $\endgroup$ – Brian M. Scott Dec 7 '13 at 0:30

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