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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. 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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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 all vertices have different colours,
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Number of ways of colouring <math>= 6 \times 5 \times 4 \times 3 \times 2 = 720 </math>.
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When two vertices sharing the same colour while other vertices having different colours,
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Number of ways of colouring <math>= 5 \times (6 \times 5 \times 4 \times 3) = 1800  </math>
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When three vertices sharing the same colour while other vertices having different colours,
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Number of ways of colouring <math>= 5 \times (6 \times 5 \times 4 ) = 600  </math>
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Total number of colouring <math> = 3120 </math>
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== 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 02:06, 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. 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.

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.


When all vertices have different colours,

Number of ways of colouring $= 6 \times 5 \times 4 \times 3 \times 2 = 720$.


When two vertices sharing the same colour while other vertices having different colours,

Number of ways of colouring $= 5 \times (6 \times 5 \times 4 \times 3) = 1800$


When three vertices sharing the same colour while other vertices having different colours,

Number of ways of colouring $= 5 \times (6 \times 5 \times 4 ) = 600$


Total number of colouring $= 3120$

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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