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

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== Solution ==
 
== Solution ==
  
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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..
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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.
 
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In the first case, the number of ways to color the map is <math>6 \times 5 \times 4 \times 3 \times 2 = 720</math>.
 
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In the second case, the number of ways of coloring is <math>5 \times (6 \times 5 \times 4 \times 3) = 1800</math>.
 
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In the third case, the number of ways of coloring is <math>5 \times (6 \times 5 \times 4 ) = 600</math>.
 
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Total number of colorings is thus <math>3120 \rightarrow \boxed{\textbf{C}}</math>
 
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|num-b=15|num-a=17|ab=A}}
 
{{AMC12 box|year=2011|num-b=15|num-a=17|ab=A}}

Revision as of 12:24, 12 February 2012

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

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See also

2011 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 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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