Difference between revisions of "2011 AMC 12A Problems/Problem 16"
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== Solution == | == Solution == | ||
− | There are three cases to consider altogether | + | 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 | + | 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>. | ||
− | + | In the second case, the number of ways of coloring is <math>5 \times (6 \times 5 \times 4 \times 3) = 1800</math>. | |
− | + | In the third case, the number of ways of coloring is <math>5 \times (6 \times 5 \times 4 ) = 600</math>. | |
− | + | Total number of colorings is thus <math>3120 \rightarrow \boxed{\textbf{C}}</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 22:20, 11 February 2011
Problem
Each vertex of convex polygon is to be assigned a color. There are colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
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 .
In the second case, the number of ways of coloring is .
In the third case, the number of ways of coloring is .
Total number of colorings is thus
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 |