2011 AMC 12A Problems/Problem 16

Revision as of 17:32, 30 June 2011 by Esque (talk | contribs) (Problem)

Problem

Each vertex of convex pentagon $ABCDE$ is to be assigned a color. There are $6$ colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?

$\textbf{(A)}\ 2520 \qquad \textbf{(B)}\ 2880 \qquad \textbf{(C)}\ 3120 \qquad \textbf{(D)}\ 3250 \qquad \textbf{(E)}\ 3750$

Solution

There are three cases to consider altogether: all the vertices have different colors, two of the vertices have the same color while the rest have different colors, and three of the vertices have the same color while the rest are all different..

When two of the vertices are of the same color, they have to be two consecutive vertices. For three of the vertices to have the same color, two pairs of two consecutive vertices must be the same color while one is a different color.

In the first case, the number of ways to color the map is $6 \times 5 \times 4 \times 3 \times 2 = 720$.

In the second case, the number of ways of coloring is $5 \times (6 \times 5 \times 4 \times 3) = 1800$.

In the third case, the number of ways of coloring is $5 \times (6 \times 5 \times 4 ) = 600$.


Total number of colorings is thus $3120 \rightarrow \boxed{\textbf{C}}$

See also

2011 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AMC 12 Problems and Solutions
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