Different types of octahedral dice

Question

How many types of octahedral dice exist?

We know that cubic dice have two types: one where the 1,2 and 3 are placed clockwise relatively to their common vertex and one where they are placed counterclockwise, given that the sum of the values between opposite faces is constant and equal to 7.

In an octahedral die the sum of opposite faces is 9. My answer is that there are 16 different types of octahedral dice, here are the steps I took:

-Take the triangle with the value of 1 as a reference: it has a side in common with 3 of the other 7 triangles, and only a vertex with other 3. The remaining face contains the 8. -Placing the number 2 (and the 7, consequentially) we have 2 choices: letting it have an edge in common with the 1 or not. -Now we allign the solid with the 7 under the 1 and the 2 over it. The 3 can be placed in one of the 4 spots remaining and the combination can be determined with no ambiguity or repetitions, so 4 different placements. -We have then only 2 combinations for the placement of the 4 and the 5

The total combinations are then 2x4x2=16, is it right?

Answer 1

We can also solve this using burnside's lemma. Since all the faces have distinct numbers, the formula is greatly simplified and easily intuitive.

Let's first solve the case of the 6 sided die. It is well known that there are $24$ elements of the rotation group on a cube (this is also non-coincidentally the same number as that for an octahedron). I.e. if we rotate a painted cube so that it is in the same orientation (but its faces do not necessarily have to be in the same position), then there are $24$ different orderings of the distinct faces.

Now we must find how many ways there are to place the numbers on the cubical die without considering indistinguishability from rotation. We will take a constructive approach.

There are $6$ places to put the 1-face. The position of the 6-face will then be decided. There are then $4$ places left for the 2-face, which will also decide the position of the 5-face. There are $2$ places left for the 3-face, which also decides the 4-face. Hence, there are $6!!=6\cdot 4\cdot 2=48$ ways to place the numbers on the die. It follows that there are $(2n)!!$ ways to place the numbers on a die with $2n$ faces.

By burnside's lemma, we have that the total number of possible valid dice when rotations are considered distinct is $$\frac{6!!}{24}$$ $$\frac{48}{24}$$ $$\boxed{2}$$

Going back to the octahedron problem. We again have that there are $24$ elements in the rotation group. However, there are $8!!$ ways to place the numbers disregarding indistinguishability from rotation.

Hence, by burnside's lemma, the number of possible valid octahedral die when rotations are considered distinct is $$\frac{8!!}{24}$$ $$\frac{8\cdot 48}{24}$$ $$\boxed{16}$$

P.S. while this may seem like a long approach, most of the text here was exposition to burnside's lemma. If you are already comfortable with burnside's lemma, this will be a quick, surefire method that can be easily extended to other types of dice.

Answer 2

If we drop the constraint about opposite faces having fixed sum, we arrive at a different answer: A priori, there are $8!$ ways to label eight faces with eight numbers. But some of these labellings are "the same" in that a rotation of the octahedron turns one into the other. In fact, if the number of possible rotations is $n$, then there are always sets of $n$ labellings that we identify, and the number of actually different labellings is $\frac{8!}n$. So what is $n$? We can always rotate the octahedron in such a way that an arbitrary of the $8$ faces is at the bottom. After that, we can rotate in $3$ ways (in steps of $120^\circ$) while fixing that bottom face. Hence $n=8\cdot 3=24$ and there are $\frac{8!}{24}=1680$ ways (or $840$ if we additionally consider reflections as symmetry).