In the figure below, 4 of the 6 disks are to be colored black and 2 are to be colored white. Two colorings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same.
[image]
There are only four such colorings for the given two colors, as shown in Figure 1. In how many ways can we color the 6 disks such that 2 are colored black, 2 are colored white, 2 are colored blue with the given identification condition?

The six disks form a triangular array whose symmetry group has 6 elements, namely the identity, two rotations and three reflections. Without symmetry there are $$\frac{6!}{2!2!2!} = 90$$ colorings; the rotations split the disks into two 3-cycles and fix none of these colorings, while each reflection fixes 6 of them, since its two 2-cycles must take two different colors and the two fixed disks then take the third. By Burnside's lemma the answer is $$\frac{90 + 0 + 0 + 6 + 6 + 6}{6} = 18$$.