Counting how many hands of cards use all four suits 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?
 A: 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.]
