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From a standard $52$-card deck, how many ways are there to pick a hand of $k$ cards that includes one card from all four suits?

I know that for any specific $k$, it's possible to break it up into cases based on the partitions of $k$ into $4$ parts. For example, if I want to choose a hand of six cards, I can break it up into two cases based on whether there are $(1)$ three cards from one suit and one card from each of the other three or $(2)$ two cards from each of two suits and one card from each of the other two.

Is there a simpler, more general solution that doesn't require splitting the problem into many different cases?

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Count the number of hands that do not contain at least one card from every suit and subtract from the total number of k-card hands. To count the number of hands that do not contain at least one card from every suit, use inclusion-exclusion considering what suits are not in a given hand. That is, letting $N(\dots)$ mean the number of hands meeting the given criteria, $$\begin{align} &N(\mathrm{no\ }\heartsuit)+N(\mathrm{no\ }\spadesuit)+N(\mathrm{no\ }\clubsuit)+N(\mathrm{no\ }\diamondsuit) \\ &\quad\quad-N(\mathrm{no\ }\heartsuit\spadesuit)-N(\mathrm{no\ }\heartsuit\clubsuit)-N(\mathrm{no\ }\heartsuit\diamondsuit)-N(\mathrm{no\ }\spadesuit\clubsuit)-N(\mathrm{no\ }\spadesuit\diamondsuit)-N(\mathrm{no\ }\clubsuit\diamondsuit) \\ &\quad\quad+N(\mathrm{no\ }\heartsuit\spadesuit\clubsuit)+N(\mathrm{no\ }\heartsuit\spadesuit\diamondsuit)+N(\mathrm{no\ }\heartsuit\clubsuit\diamondsuit)+N(\mathrm{no\ }\spadesuit\clubsuit\diamondsuit) \\ &\quad\quad-N(\mathrm{no\ }\heartsuit\spadesuit\clubsuit\diamondsuit) \\ &=4{39 \choose k}-6{26 \choose k}+4{13 \choose k}-{0 \choose k}. \end{align}$$

So, the number of hands of k cards that include at least one card from every suit is $${52 \choose k}-4{39 \choose k}+6{26 \choose k}-4{13 \choose k}+{0 \choose k}.$$ [Drop terms as appropriate for larger values of k.]

  • 2
    This is known as the [inclusion-exclusion principle](http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle)2010-07-23
  • 0
    This is a fantastic solution that I hadn't thought of. Thanks!2010-07-30