Question

In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?

$6$

$8$

$9$

$12$

$15$

Answer: D

Concepts:casework symmetry

Difficulty rating: 1660

Solution:

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.

2017 AMC 12B Problem 13

Answer: D

Difficulty rating: 1660

Solution:

Problem 13 in Other Years