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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 four 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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1 Answer 1

up vote 7 down vote accepted

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.]

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2  
This is known as the inclusion-exclusion principle –  Casebash Jul 23 '10 at 23:40
    
This is a fantastic solution that I hadn't thought of. Thanks! –  Ben Alpert Jul 30 '10 at 3:49

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