# Difference between revisions of "2011 AMC 12A Problems/Problem 16"

## 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

We can do some casework when working our way around the pentagon from $A$ to $E$. At each stage, there will be a makeshift diagram.

1.) For $A$, we can choose any of the 6 colors.

        A : 6


2.) For $B$, we can either have the same color as $A$, or any of the other 5 colors. We do this because each vertex of the pentagon is affected by the 2 opposite vertices, and $D$ will be affected by both $A$ an $B$.

      A : 6
B:1        B:5


3.) For $C$, we cannot have the same color as $A$. Also, we can have the same color as $B$ ($E$ will be affected), or any of the other 4 colors. Because $C$ can't be the same as $A$, it can't be the same as $B$ if $B$ is the same as $A$, so it can be any of the 5 other colors.

      A : 6
B:1        B:5
C:5     C:4   C:1


4.) $D$ is affected by $A$ and $B$. If they are the same, then $D$ can be any of the other 5 colors. If they are different, then $D$ can be any of the (6-2)=4 colors.

      A : 6
B:1        B:5
C:5     C:4   C:1
D:5     D:4   D:4


5.) $E$ is affected by $B$ and $C$. If they are the same, then $E$ can be any of the other 5 colors. If they are different, then $E$ can be any of the (6-2)=4 colors.

      A : 6
B:1        B:5
C:5     C:4   C:1
D:5     D:4   D:4
E:4     E:4   E:5


6.) Now, we can multiply these three paths and add them: $(6\times1\times5\times5\times4)+(6\times5\times4\times4\times4)+(6\times5\times1\times4\times5)=600+1920+600=3120$

7.) Our answer is $C$!

## See also

 2011 AMC 12A (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions
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