Combinatorics Olympiad Problem 2

Here the group acting on the set of colorings is the $6$ -element group consisting of rotations by multiple of $60$ degrees. By Burnside’s Lemma, the answer is $\frac{1}{6} (A + 2B + 2C + D)$ , where A is the number of colorings that are invariant under $0$ degree rotation, B is both the number of colorings that are invariant under $60$ degree clockwise rotation and the number of colorings that are invariant under $60$ degree counterclockwise rotation, C is both the number of colorings that are invariant under $120$ degree clockwise rotation and the number of colorings that are invariant under $120$ degree counterclockwise rotation, and D is the number of colorings that are invariant under $180$ degree rotation. We have $A = 6!/2!2!2! = 90$ (these are just ordinary permutations of the multiset), $B = 0$ (the only way a circular 6-permutation involving R’s, W’s, and B’s can be invariant under $60$ degree rotation is if all six letters are the same), $C = 0$ (for the same sort of reason as $B = 0$ , but more complicated: a circular $6$ -permutation involving R’s, W’s, and B’s can be invariant under $120$ degree rotation only if either some color occurs six times or two of the colors occur three times each, contradicting our requirement that each of the three colors occurs twice), and $D = 3! = 6$ (there are $3!$ ways of assigning the three colors to the three pairs of diametrically opposite positions in the circular $6$ -permutation, and each of these corresponds to a way of arranging two R’s two B’s, and two W’s). So the number of orbits, i.e., the number of $6$ -permutations, is $\frac{1}{6}(90 + 6) = 16$ .

This solution is not mine.

I just like to write all permutations. You can check if there are the same amount of permutations in each category with Bs and Ws (considering cyclic-permutations)