2017 AMC 12B Problem 13

Before accounting for symmetry, there are $\dfrac{6!}{3!2!}=60$ paintings. The two nonidentity rotations partition the disks into two $3$-cycles, so neither can fix a painting having color counts $3,2,1.$ Each of the $3$ reflections fixes $2$ disks and swaps the other $4$ in $2$ pairs. For a painting to be fixed, the lone green disk and one of the $3$ blue disks must occupy the two fixed positions, in $2$ orders. Of the two swapped pairs, either one can be the red pair, giving $2\cdot2=4$ fixed paintings per reflection. Burnside’s Lemma therefore gives $\dfrac{60+3(4)}{6}=12$ distinct paintings. Thus, the correct answer is D.