Difference between revisions of "2011 AMC 10A Problems/Problem 22"
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<math> \textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750 </math> | <math> \textbf{(A)}\ 2520\qquad\textbf{(B)}\ 2880\qquad\textbf{(C)}\ 3120\qquad\textbf{(D)}\ 3250\qquad\textbf{(E)}\ 3750 </math> | ||
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+ | == Solution == | ||
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+ | Let vertex <math>A</math> be any vertex, then vertex <math>B</math> be one of the diagonal vertices to <math>A</math>, <math>C</math> be one of the diagonal vertices to <math>B</math>, and so on. We consider cases for this problem. | ||
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+ | In the case that <math>C</math> has the same color as <math>A</math>, <math>D</math> has a different color than <math>A</math> and so <math>E</math> has a different color than <math>A</math> and <math>D</math>. In this case, <math>A</math> has <math>6</math> choices, <math>B</math> has <math>5</math> choices (any color but the color of <math>A</math>), <math>C</math> has <math>1</math> choice, <math>D</math> has <math>5</math> choices, and <math>E</math> has <math>4</math> choices, resulting in a possible of <math>6 \cdot 5 \cdot 1 \cdot 5 \cdot 4 = 600</math> combinations. | ||
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+ | In the case that <math>C</math> has a different color than <math>A</math> and <math>D</math> has a different color than <math>A</math>, <math>A</math> has <math>6</math> choices, <math>B</math> has <math>5</math> choices, <math>C</math> has <math>4</math> choices (since <math>A</math> and <math>B</math> necessarily have different colors), <math>D</math> has <math>4</math> choices, and <math>E</math> has <math>4</math> choices, resulting in a possible of <math>6 \cdot 5 \cdot 4 \cdot 4 \cdot 4 = 1920</math> combinations. | ||
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+ | In the case that <math>C</math> has a different color than <math>A</math> and <math>D</math> has the same color as <math>A</math>, <math>A</math> has <math>6</math> choices, <math>B</math> has <math>5</math> choices, <math>C</math> has <math>4</math> choices, <math>D</math> has <math>1</math> choice, and <math>E</math> has <math>5</math> choices, resulting in a possible of <math>6 \cdot 5 \cdot 4 \cdot 1 \cdot 5 = 600</math> combinations. | ||
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+ | Adding all those combinations up, we get <math>600+1920+600=\boxed{3120 \ \mathbf{(C)}}</math>. |
Revision as of 01:30, 16 February 2011
Problem 22
Each vertex of convex pentagon 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
Let vertex be any vertex, then vertex be one of the diagonal vertices to , be one of the diagonal vertices to , and so on. We consider cases for this problem.
In the case that has the same color as , has a different color than and so has a different color than and . In this case, has choices, has choices (any color but the color of ), has choice, has choices, and has choices, resulting in a possible of combinations.
In the case that has a different color than and has a different color than , has choices, has choices, has choices (since and necessarily have different colors), has choices, and has choices, resulting in a possible of combinations.
In the case that has a different color than and has the same color as , has choices, has choices, has choices, has choice, and has choices, resulting in a possible of combinations.
Adding all those combinations up, we get .