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 $A$ and $B$ .

      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 $C$ !

Solution 2

Right off the bat, we can analyze three things:

1.) There can only be two of the same color on the pentagon.

2.) Any pair of the same color can only be next to each other on the pentagon.

3.) There can only be two different pairs of same colors on the pentagon at once.

Now that we know this, we can solve the problem by using three cases: no same color pairs, one same color pair, and two same color pairs.

1.) If there are no color pairs, it is a simple permutation: six different colors in five different spots. We count $6!=720$ cases. No rotation is necessary because all permutations are accounted for.

2.)If there is one color pair, we must count 6 possibilities for the pair(as one element), 5 for the third vertex, 4 for the fourth vertex, and 3 for the fifth vertex.

We get $6\times5\times4\times3=360$ .

However, there are 5 different locations the pair could be at. Therefore we get $360\times5=1800$ possibilities for one pair.

3.)If there are two color pairs, we must count 6 possibilities for the first pair(as one element), 5 possibilities for the next pair(as one element), and 4 possibilities for the final vertex.

We get $6\times5\times4=120$ .

Once again, there are 5 different rotations in the pentagon that we must account for. Therefore we get $120\times5=600$ possibilities for two pairs.

5.) If we add all of three cases together, we get $720+1800+600=3120$ . The answer is $C$ .

Solution by gsaelite

Solution 3

This solution essentially explains other ways of thinking about the cases stated in Solution 2.

Case 1:

${6\choose5} \cdot 5!$

5 colors need to be chosen from the group of 6. Those 5 colors have 5! distinct arrangements on the pentagon's vertices.

Case 2:

${6\choose4} \cdot4\cdot5\cdot3!$

4 colors need to be chosen from the group of 6. Out of those 4 colors, one needs to be chosen to form a pair of 2 identical colors. Then, for arranging this layout onto the pentagon, one of the five sides (same thing as pair of adjacent vertices) of the pentagon needs to be established for the pair. The remaining 3 unique colors can be arranged 3! different ways on the remaining 3 vertices.

Case 3:

${6\choose3} \cdot {3\choose2} \cdot5\cdot2$

3 colors need to be chosen from the group of 6. Out of those 3 colors, 2 need to be chosen to be the pairs. Then, for arranging this layout onto the pentagon, start out by thinking about the 1 color that is not part of a pair, as it makes things easier. It can be any one of the 5 vertices of the pentagon. The remaining 2 pairs of colors can only be arranged 2 ways on the remaining 4 vertices.

Solving each case and adding them up gets you 3120. $\boxed{C}$

Video Solution

https://youtu.be/FThly7dRBIE

~IceMatrix

2011 AMC 12A (Problems • Answer Key • Resources)
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