I have been thinking about this question for a while and am getting hopelessly lost. We (I) know that the arrangements of n objects on a circle can be listed in (n-1)! ways and by keeping 1 fixed and permuting the other 2 through n objects in (n-1)! ways. How do I extend this to the 3D sphere? I am interested in both the number of ways as well as in the listing of the arrangements.

So why would this matter? Suppose we have a number of n rays (each of a different color) that projects from the center of the sphere are approximately equally spaced from each other (can only be done exactly for the five Platonic solids). The question is who many ways can we arrange these rays and what is the listing for these arrangements of the colors that are listed $x_1,x_2,\ldots,x_n$?

Many thanks for your thoughts!

  • 2
    $\begingroup$ Your question makes no sense. $\endgroup$
    – TonyK
    Sep 4, 2021 at 15:27
  • $\begingroup$ Since this arrangement only exists for the five Platonic solids, are you asking for each of $n = 4, 6, 8, 12, 20$ how many ways there are to arrange $n$ distinct objects on the vertices of the corresponding Platonic solid? $\endgroup$ Sep 4, 2021 at 15:39
  • $\begingroup$ When arranging $n$ objects on a circle, you are really arranging them on an $n$-gon, and saying that two arrangements are equal when they can be obtained from eachother by rotation. In the case of a sphere, $n$ points don't generally define some unique polytope or other combinatorial structure. The (a) correct generalization would be to start with some polytope and count arrangements on their vertices that are not related by some symmetry of the polytope. $\endgroup$
    – doetoe
    Sep 4, 2021 at 15:42
  • $\begingroup$ The question of finding $n$ positions on a sphere that minimize some potential (hence give you some kind of unique/canonical set of $n$ spots on a sphere) is called the Thomson problem. $\endgroup$
    – doetoe
    Sep 4, 2021 at 15:44
  • $\begingroup$ @doetoe, agreed and I think that I understand that, even though I am not a mathematician but I was wondering how I can get such a listing. Is there a formula, etc. $\endgroup$ Sep 4, 2021 at 16:37

1 Answer 1


The answer depends very unpredictably on the value of $n$, and on the specific arrangement chosen. (Sometimes there is a specific optimal arrangement; sometimes the optimal arrangement is not known.) In Wikipedia's table for the Thomson problem, you see the degree of symmetry in the Symmetry column (in Schönflies notation) for the best known arrangements.

There is no general formula known for the symmetry group, because there is no general pattern known for the most symmetric spherical arrangement.

Essentially, the number of arrangement is $n!/s$, where $s$ is the number of symmetries (the order of the symmetry group). For small symmetry groups, this is the best way to solve the problem. You'll have to decide whether reflections (if they exist) are allowed or not - but the answer of $(n-1)!$ for the circular arrangement assumes they are not allowed.

For very symmetric arrangements like the icosahedron, this is tricky, because I for one don't know how many symmetries the icosahedron has. So I would solve the problem differently for highly symmetric cases. Here's how I would approach the icosahedron, for example:

  1. Orient the icosahedron such that object $x_1$ is on the top.
  2. There are $\binom{11}{5}$ ways to choose the $5$ objects adjacent to $x_1$, and (by the 2D problem) $4!$ ways to arrange them around $x_1$.
  3. Once that is done, and those objects are fixed, there is no more symmetry left, and there are $6!$ ways to arrange the remaining $6$ objects.

This gives us $\binom{11}{5} \cdot 4! \cdot 6! = \frac{11!}{5}$ as the final answer. This is equal to $\frac{12!}{60}$, so in the process we've concluded that there are $60$ rotational symmetries, which we could also have gotten from Wikipedia's article on icosahedral symmetry.

  • $\begingroup$ I already know that you are looking for a solution for general $n$, but I am giving you an example of how I would approach a specific $n$. For general $n$, I have already stated the answer: $n!/s$. $\endgroup$ Sep 4, 2021 at 18:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .