Question: “Jeffrey has a set of tetrahedral dice. Each has a distinct arrangement of exactly six dots painted on the four faces, and no two dice can be rotated to appear the same. Faces can have any number of dots on each face including zero. How many dice are in a set that has each distinct arrangement of six dots exactly once?”
This question has puzzled me for a while now and I haven’t made much progress. Please help if you find a way to solve it.
Let's look for distinct configurations of the dots. You are looking for solutions to the Diophantine equation:
$$x_1+x_2+x_3+x_4=6 \\ 0\le x_1 \le x_2 \le x_3 \le x_4$$
This is easily enumerated:
$$0,0,0,6 \\ 0,0,1,5 \\ 0,0,2,4 \\ 0,0,3,3 \\ 0,1,1,4 \\ 0,1,2,3 \\ 0,2,2,2 \\ 1,1,1,3 \\ 1,1,2,2$$
These are all configurations. Next, for a single configuration of dots, let's determine if it is rotationally unique. For $x,x,x,y$, this is clearly the same as $x,x,y,x$ and $x,y,x,x$ and $y,x,x,x$. Simply turn the die until the one side that is different faces down.
Similarly for $x,x,y,y$ and $x,x,y,z$. But, I think $w,x,y,z$ is distinct from $w,z,y,x$. One would have the values arranged to increment clockwise while the other would have the increment be counterclockwise. So, I suspect the correct answer is $10$, one for each distinct configuration, plus one for the additional rotational arrangement of $0,1,2,3$.