How many essentially distinct hands of size $n$ can be dealt from a standard deck

card-gamescombinatorics

In many card games, suits are interchangeable, so that one would be indifferent between the (4-card) hands: $5\diamondsuit \ 6\spadesuit \ 7\heartsuit \ J\heartsuit$ and $5\spadesuit \ 6\diamondsuit \ 7\clubsuit \ J\clubsuit$. I'm interested in the number of $n$-card hands which are essentially different, that is, They are distinct under every possible relabeling of the suits. Put another way, two hands are essentially the same if there is some relabeling of the suits that makes them identical. Specifically, if it matters, I am most interested in the case $n=6$.

Lots of people have done similar calculations for $n=5$, due to the poker connection, but their analyses are always poker-centric. In particular, every general explication of $n$-card hands I've found assumes that one is going to make their best 5-card hand. I haven't been able to find any reference to this specific problem, and I'd really like to avoid having to do it from scratch.

Searching the broader internet and M.SE both turn up lots of specific homework-type problems looking for the number of one specific type of hand. Obviously one could enumerate all distinct 6-card hands ignoring suits and then count the ways to add suit-shapes to them; I'd really like to avoid that. Surely there's either a quick way to answer the question, or an answer out there somewhere already?

Best Answer

Based on a comment by @lulu, it appears that the answer is given by the coefficient of $x^6$ in the expansion of the $4$th Complete Bell Polynomial $$B_4(\{0!\cdot(1+x)^{13}, \ 1!\cdot(1+x^2)^{13}, \ 2!\cdot(1+x^3)^{13}, \ 3!\cdot(1+x^4)^{13}\})$$ divided by $6!$. Here is the Mathematica code I used to generate the polynomial, and its result:

y1 = (1 + x)^13;
y2 = (1 + x^2)^13;
y3 = 2 (1 + x^3)^13;
y4 = 6 (1 + x^4)^13;
p = BellY[4, 1, {y1, y2, y3, y4}] + BellY[4, 2, {y1, y2, y3}] + 
    BellY[4, 3, {y1, y2}] + BellY[4, 4, {y1}];
Expand[p]

24 + 312 x + 4056 x^2 + 42120 x^3 + 394368 x^4 + 3227016 x^5 + 23111712 x^6 + ...

Giving the solution 23,111,712 / 24 = 962,988 distinct hands.

This solution is based on the "Formula" comment found at the OEIS link here. I haven't been able to find another reference for it, but using a brute force algorithm, I was able to confirm the result.

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