Question about number of non equivalent colourings of corners of a regular tetrahedron with k colours

Question

Due to Covid -19 , in our university quizzes are held online and it's hard to ask questions.

3 Days back in my Combinatorics quiz this question was asked on which I am struck. I couldn't solve it in the time alloted and struggled to find a proper strategy.

Question is ->Determine the number of non equivalent colourings of the corners of regular tetrahedron with k different colours.

My attempt -> I am trying to solve it by Burnside Theorem ( Number of non equivalent colourings in C are given by N(G, C) = 1/ |G| $\sum_{f \epsilon G } | C(f) | $. [C(f) = set of all colourings in C that are fixed by f ]

Group of permutations is $S_4$ and all$ (k^4)$ will be fixed by identity . But I am not able to think how to find colourings fixed by each permutation caused due to rotations and reflections. I have done it for pentagon which was easy.

Can someone please tell a way on how to efficiently and elegentally compute the value of C(f) in case of rotations and reflections.

I will be really thankful for the ideas.

Answer 1

The symmetric group is $S_4$ with $4!=24$ elements:

• 6 with cycle structure $(abcd)$ then $C(f)=k$;
• 8 with cycle structure $(abc)(d)$ then $C(f)=k^2$;
• 3 with cycle structure $(ab)(cd)$ then $C(f)=k^2$;
• 6 with cycle structure $(ab)(c)(d)$ then $C(f)=k^3$;
• 1 with cycle structure $(a)(b)(c)(d)$ then $C(f)=k^4$.

It follows that $$N(G, C)=\frac{6k+(8+3)k^2+6k^3+k^4}{24}=\frac{(k+3)(k+2)(k+1)k}{4!}=\binom{k+3}{4}.$$

Answer 2

As the symmetry group of the tetrahedron is $S_4$, colourings are already equivalent if they use the same colours the same number of times. Thus we have

• $k$ colourings of type $(a,a,a,a)$,
• $k\cdot(k-1)$ colourings of type $(a,a,a,b)$,
• $k\choose 2$ colourings of type $(a,a,b,b)$,
• $k\cdot{k-1\choose 2}$ colourings of type $(a,a,b,c)$,
• $k\choose 4$ colourings of type $(a,b,c,d){}$.