2017 AMC 10B Problems/Problem 18

Revision as of 19:02, 16 February 2017 by Naman12 (talk | contribs) (Solution)

Problem

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?

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

Solution

Solution 1

First we figure out the number of ways to put the $3$ blue disks. Denote the spots to put the disks as $1-6$ from left to right, top to bottom. The cases to put the blue disks are $(1,2,3),(1,2,4),(1,2,5),(1,2,6),(2,3,5),(1,4,6)$. For each of those cases we can easily figure out the number of ways for each case, so the total amount is $2+2+3+3+1+1 = \boxed{\textbf{(D) } 12}$.

Solution 2

Denote the $6$ discs as in the first solution. Ignoring reflections or rotations, there are $\binom{6}{3} * \binom{3}{2} = 60$ colorings. Now we need to count the number of fixed points under possible transformations:

1. The identify transformation. Since this doesn't change anything, there are $60$ fixed points

2. Reflect about a line of symmetry. There are $3$ lines of reflections. Take the line of reflection going through the centers of circles $1$ and $5$. Then, the colors of circles $2$ and $3$ must be the same, and the colors of circles $4$ and $6$ must be the same. This gives us $4$ fixed points per line of reflection

3. Rotate by $120^\circ$ counter clockwise or clockwise with respect to the center of the diagram. Take the clockwise case for example. There will be a fixed point in this case if the colors of circles $1$, $4$, and $6$ will be the same. Similarly, the colors of circles $2$, $3$, and $5$ will be the same. This is impossible, so this case gives us $0$ fixed points per rotation.

By Burnside's lemma, the total number of colorings is $(1*60+3*4+2*0)/(1+3+2) = \boxed{\textbf{(D) } 12}$.

See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
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All AMC 10 Problems and Solutions

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