Compute the number of distinguishable coloured wheels having 12 compartments that can be formed with $q$ colours.(Assume that only one colour can be used on a single compartment and that the same colour can be used on different compartments)
My attempt:
Let, X be the collection of all possible coloured wheels , then $|X|=q^{12}$
What I understood from the Question is that we have an action of $\Bbb Z_{12}$ on $X$ (that sends the $i$th compartment to the $(i+j)$th compartment $\forall 0 \le i \le 11$ , $\forall j \in \Bbb Z_{12}$).
Then, using Burnside lemma, I obtained the result to be $$\frac{q^{12}+4q+2q^2+2q^3+2q^4+q^6}{12}$$
Is my answer correct? Please point out mistakes if any.
Thanks in advance for help!
If the compartments are all on one side, then your answer is correct.
If the wheel is two-sided so that compartments also show up on the other side and we can flip the wheel, the modifications aren't that complicated. We have in addition six reflections whose lines of reflection cross two compartments, fixing $q^8$ wheels each, and six reflections crossing no compartments fixing $q^6$ elements each, giving the number of colourings as $$\frac{q^{12}+4q+2q^2+2q^3+2q^4+7q^6+6q^8}{24}$$