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 | + | 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 | + | 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 | + | 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 22:22, 21 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 from
and so
has a different color from
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 from
and
has a different color from
,
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 from
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 .