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Difference between revisions of "2011 AMC 10A Problems/Problem 22"
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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.
 
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.
  
−
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.
+
In the case that <math>C</math> has the same color as <math>A</math>, <math>D</math> has a different color from <math>A</math> and so <math>E</math> has a different color from <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.
  
−
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.
+
In the case that <math>C</math> has a different color from <math>A</math> and <math>D</math> has a different color from <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.
 
   
 
   
−
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.
+
In the case that <math>C</math> has a different color from <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.
  
 
Adding all those combinations up, we get <math>600+1920+600=\boxed{3120 \ \mathbf{(C)}}</math>.
 
Adding all those combinations up, we get <math>600+1920+600=\boxed{3120 \ \mathbf{(C)}}</math>.

Revision as of 21:22, 21 February 2011

Problem 22

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

Let vertex $A$ be any vertex, then vertex $B$ be one of the diagonal vertices to $A$, $C$ be one of the diagonal vertices to $B$, and so on. We consider cases for this problem.

In the case that $C$ has the same color as $A$, $D$ has a different color from $A$ and so $E$ has a different color from $A$ and $D$. In this case, $A$ has $6$ choices, $B$ has $5$ choices (any color but the color of $A$), $C$ has $1$ choice, $D$ has $5$ choices, and $E$ has $4$ choices, resulting in a possible of $6 \cdot 5 \cdot 1 \cdot 5 \cdot 4 = 600$ combinations.

In the case that $C$ has a different color from $A$ and $D$ has a different color from $A$, $A$ has $6$ choices, $B$ has $5$ choices, $C$ has $4$ choices (since $A$ and $B$ necessarily have different colors), $D$ has $4$ choices, and $E$ has $4$ choices, resulting in a possible of $6 \cdot 5 \cdot 4 \cdot 4 \cdot 4 = 1920$ combinations.

In the case that $C$ has a different color from $A$ and $D$ has the same color as $A$, $A$ has $6$ choices, $B$ has $5$ choices, $C$ has $4$ choices, $D$ has $1$ choice, and $E$ has $5$ choices, resulting in a possible of $6 \cdot 5 \cdot 4 \cdot 1 \cdot 5 = 600$ combinations.

Adding all those combinations up, we get $600+1920+600=\boxed{3120 \ \mathbf{(C)}}$.

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