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You are given a blank six sided die and 21 dot stickers. You can distribute the stickers on the die as you please, but you must use all 21 of them. How many distinct dice can be formed? Note that we do not care about the 3d geometry of the cube, for example, (1,2,3,4,5,6) is equivalent to (2,1,3,4,5,6) even though they are different geometrically.

We can overcount by considering the number of nonnegative solutions to $a_1 + a_2 + a_3 + a_4 + a_5 + a_6 = 21$, which is ${26}\choose{5}$. It is here that I am stuck, and don't know how to remove the double counted distributions.

ANSWER

Firstly, I apologise for being imprecise in my problem formulation. Nevertheless, I was indeed trying to count the number of functionally different dice, which only depends on the frequency of numbers on the sides. Also note that blank faces are allowed. This is given by the partition function described here courtesy of lulu.

Now, some simple code gives us our answer, which is part(21+6,6) = 331 (the part(n,k) function necessitates that each part is nonempty; we can partition $n+k$ into $k$ parts and subtract $1$ from each of the $k$ parts to allows empty parts and thus there is a 1-1 correspondence).

R.T.
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  • See this question Note: you didn't say if faces could be left blank or not, but the one follows from the other by summing cases. – lulu Jul 14 '24 at 11:21
  • @lulu Your comment is misleading. $p(n,k)$ is for when there is total permutation symmetry, but here we only have cube symmetry. The correct method here involves the-lemma-that-is-not-Burnside’s, using the group of 24 symmetries of the cube. – Mike Earnest Jul 14 '24 at 17:32
  • @MikeEarnest How do you interpret the OP's comment that "we do not care about the $3d-$geometry of the cube" ? I interpreted that to mean we just meant total permutations without regards to the symmetries of the cube (but of course my interpretation might be wrong). To be sure, your interpretation leads to a more interesting problem. – lulu Jul 14 '24 at 17:39
  • You’re right, I’m wrong. I think the question is poorly worded; if you don’t want the reader to think of cubical symmetry, then don’t make it a problem about labeling a cube. – Mike Earnest Jul 14 '24 at 17:44
  • @MikeEarnest Agreed. – lulu Jul 14 '24 at 17:44
  • @lulu thank you for the helpful link. It is exactly what I was looking for. I have mentioned you in the updated post, with the answer to my question. – R.T. Jul 14 '24 at 23:13
  • @MikeEarnest I apologise for the confusion. I hope my edit and answer clarifies what I was requesting. Nevertheless, thank you for your efforts. – R.T. Jul 14 '24 at 23:14
  • It sounds like you want the number of weak (strong) partitions of $21$ into $6$. They are weak if you allow $0$, strong if not. You are asking for the number of ways to have six numbers sum to $21$ when different orders of summands are considered equivalent. – Ross Millikan Jul 15 '24 at 00:22

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