# 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?

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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