I'm trying to find how many different ways there are to colour the edges of a regular tetrahedron with n colours such that there are no monochromatic triangles.
Certainly for one triangle there must be n choose 3 ways but I'm not quite sure how to generalise this to a tetrahedron.
Any help would be much appreciated!
Best Answer
For one triangle there are actually $n^3-n$ edge colourings, including reflections and rotations.
There are $6$ edges in a tetrahedron, $n^6$ ways to colour the edges, $4$ triangles, $n^4$ colourings with any given triangle monocromatic, $6$ pairs of triangles, $n^2$ colourings with any given pair of triangles both monochromatic, 4 ways to pick $3$ triangles, $n$ colourings with $3$ triangles, hence all triangles, monochromatic, $1$ way to pick $4$ triangles and $n$ colourings with $4$ triangles monochromatic.
The number of $n$-colourings without monochromatic triangles is then $n^6-4n^4+6n^2-3n$.