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?

$6$

$8$

$9$

$12$

$15$

Solution:

We first will calculate the number of ways when the green is at the top. This is rotationally symmetric with every other corner, so we wouldn't have to count those again. Then, we can multiply our count by $2$ since the number of cases when the green is in the inner $3$ disks is the same as if we made each corner an edge and each edge piece a corner.

Suppose the green is on the top. Then, there are $\binom{5}{2}=10$ places to put the two reds, of which $2$ are symmetric. Thus, the number of non symmetric configurations are $\frac12 \cdot (10-2)=4$ after dividing by $2$ to remove the duplicates, and $4+2=6$ when putting those cases back.

This makes the total $6\cdot 2=12.$

Thus, the correct answer is D .

2017 AMC 10B Problem 18

Solution:

Problem 18 in Other Years