Counting the amount of hands of cards make use of all 4 matches

From a typical $52$-card deck, the amount of means exist to select a hand of $k$ cards that consists of one card from all 4 matches?

I recognize that for any kind of details $k$, it's feasible to damage it up right into instances based upon the dividings of $k$ right into $4$ components. As an example, if I intend to pick a hand of 6 cards, I can damage it up right into 2 instances based upon whether there are $(1)$ 3 cards from one match and also one card from each of the various other 3 or $(2)$ 2 cards from each of 2 matches and also one card from each of the various other 2.

Exists a less complex, extra basic remedy that does not call for splitting the trouble right into several instances?

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2019-05-07 01:59:58
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Count the variety of hands that do not have at the very least one card from every match and also subtract from the complete variety of k - card hands. To count the variety of hands that do not have at the very least one card from every match, usage incorporation - exemption considering what matches are not in an offered hand. That is, allowing $N(\dots)$ suggest the variety of hands fulfilling the offered standards, $$\begin{align} &N(\mathrm{no\ }\heartsuit)+N(\mathrm{no\ }\spadesuit)+N(\mathrm{no\ }\clubsuit)+N(\mathrm{no\ }\diamondsuit) \\ &\quad\quad-N(\mathrm{no\ }\heartsuit\spadesuit)-N(\mathrm{no\ }\heartsuit\clubsuit)-N(\mathrm{no\ }\heartsuit\diamondsuit)-N(\mathrm{no\ }\spadesuit\clubsuit)-N(\mathrm{no\ }\spadesuit\diamondsuit)-N(\mathrm{no\ }\clubsuit\diamondsuit) \\ &\quad\quad+N(\mathrm{no\ }\heartsuit\spadesuit\clubsuit)+N(\mathrm{no\ }\heartsuit\spadesuit\diamondsuit)+N(\mathrm{no\ }\heartsuit\clubsuit\diamondsuit)+N(\mathrm{no\ }\spadesuit\clubsuit\diamondsuit) \\ &\quad\quad-N(\mathrm{no\ }\heartsuit\spadesuit\clubsuit\diamondsuit) \\ &=4{39 \choose k}-6{26 \choose k}+4{13 \choose k}-{0 \choose k}. \end{align}$$

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

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2019-05-09 00:26:31
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