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.


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


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 $4\cdot5!$.

  • $\begingroup$ 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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