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.

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 $$\frac{90 + 0 + 0 + 6 + 6 + 6}{6} = 18$$.