2017 AMC 12B Problems/Problem 13

Problem 13

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?

$[asy] size(100); pair A, B, C, D, E, F; A = (0,0); B = (1,0); C = (2,0); D = rotate(60, A)*B; E = B + D; F = rotate(60, A)*C; draw(Circle(A, 0.5)); draw(Circle(B, 0.5)); draw(Circle(C, 0.5)); draw(Circle(D, 0.5)); draw(Circle(E, 0.5)); draw(Circle(F, 0.5)); [/asy]$

$\textbf{(A) } 6 \qquad \textbf{(B) } 8 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 15$

Solution

Looking at the answer choices, we see that the possibilities are indeed countable. Thus, we will utilize that approach in the form of two separate cases, as rotation and reflection take care of numerous possibilities. First, consider the case that the green ball is in a corner. This yields $6$ possible arrangements for the $3$ blue balls and $2$ red balls in the remaining available slots. Now, consider the case that the green ball is on an edge. This yields $6$ more possible arrangements for the $3$ blue balls and $2$ red balls in the remaining available slots. Thus, our answer is $6 + 6 = \boxed{\bold{(D)}\, 12}$

Solution by akaashp11

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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All AMC 12 Problems and Solutions

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2017 AMC 12B Problems/Problem 13

Problem 13

Solution

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