I have the following question:

I have a card game where the player can make a deck of a number of cards. That deck will consist of 25-40 cards (not relevant) and no card can be in the deck more than 3 times (very relevant).

So cards can be unique, can be there twice, or three times.

Now the player draws 10 cards and I want to know how many combinations there are. The order of drawing does not matter.

I know that for a deck of $n$ unique cards the answer is $\frac{n!}{10!(n-10)!}$. And I also came up with the following formula: $$\sum_{n=0}^{10}\Bigg(\sum_{m=0}^{n-10} \binom{u}{n}\binom{d}{m}\binom{t}{10-n-m}\Bigg)$$ $u$ is the amount of unique cards, $d$ the amount of double cards, $t$ the amount of triple cards. The sums go through all possible draw combinations (like 1 unique, 3 of the doubles and 6 triples, etc.).

What I am missing though is a way to exclude those combinations that I count multiple times, as I don't care which 2 cards I have out of one triplet.

Can anyone help me with this?

I suppose the inclusion-exclusion thing is needed here, but I'm not sure how to apply that here.


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