2017 AMC 12B Problem 13

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

Small hint:

Count the labeled colorings first, then apply Burnside’s Lemma to the

6

symmetries of the triangle

Big hint:

A nonidentity rotation fixes none; under a reflection, the odd green and blue counts must use the two fixed disks

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.