I was wondering how many ways a regular icosahedron (below) constructed from 20 congruent equilateral triangles could be numbered from 1 to 20. I'm going to be referring to the solid in 'layers', so the top and bottom layers are made of 5 triangles forming a pentagonal pyramid, whilst the middle layer is made of 10 triangles.
So far my thoughts are:
- Choose 5 numbers for the top layer and permute as if arranging around a circle - $\binom{20}{5}\cdot4!$
- Choose 10 numbers for the middle and permute as if arranging around a circle - $\binom{15}{10}\cdot9!$
- Permute the remaining 5 numbers for the bottom layer - $4!$ Which gives $\binom{20}{5}\cdot\binom{15}{5}\cdot(4!)^{2}\cdot9!$ in total.
However, I am unsure how to (or if I need to) further account for any symmetry in the solid, especially since it doesn't matter where I put the first number. Could someone explain please in detail if my steps are somewhat correct or if I'm completely wrong and how I would account for that symmetry?
The answer is very simple: there are $20!$ ways to assign the 20 distinct numbers to faces, but the rotational group of the icosahedron has order 60, so we divide by 60 to obtain the final answer of $\frac{20!}{60}=4.05×10^{16}$. If you consider reflections to be the same, this number must be divided by a further two, giving $\frac{20!}{120}=2.03×10^{16}$ numberings.