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

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== Solution ==
 
== Solution ==
  
There are three cases to consider altogether. The cases are all vertices have different colours, two of the vertices sharing the same colour while the rest of the vertices are of different colours and lastly, three of the vertices sharing the same colour while the rest are of different colours.  
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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..  
  
When two of the vertices are of the same colour, they have to be two consecutive vertices. Likewise for three of the vertices having the same colour, they have to be three consecutive vertices.
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When two of the vertices are of the same color, they have to be two consecutive vertices. Similarly, for three of the vertices to have the same color, they have to be three consecutive vertices.
  
 +
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>.
  
When all vertices have different colours,  
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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>.
  
Number of ways of colouring <math>= 6 \times 5 \times 4 \times 3 \times 2 = 720 </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>.
  
  
When two vertices sharing the same colour while other vertices having different colours,
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Total number of colorings is thus <math>3120 \rightarrow \boxed{\textbf{C}}</math>
 
 
Number of ways of colouring <math>= 5 \times (6 \times 5 \times 4 \times 3) = 1800  </math>
 
 
 
 
 
When three vertices sharing the same colour while other vertices having different colours,
 
 
 
Number of ways of colouring <math>= 5 \times (6 \times 5 \times 4 ) = 600  </math>
 
 
 
 
 
Total number of colouring <math> = 3120 </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 22:20, 11 February 2011

Problem

Each vertex of convex polygon $ABCDE$ 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. Similarly, for three of the vertices to have the same color, they have to be three consecutive vertices.

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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