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> | ||
− | == Solution == | + | == Solution 1 == |
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. | ||
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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>. | ||
+ | ==Solution 2== | ||
+ | |||
+ | First, notice that there can be <math>3</math> cases. One with all vertices painted different colors, one with one pair of adjacent vertices painted the same color and the final one with two pairs of adjacent vertices painted two different colors. | ||
+ | |||
+ | Case <math>1</math>: There are <math>6!</math> ways of assigning each vertex a different color. <math>6! = 720</math> | ||
+ | |||
+ | Case <math>2</math>: There are <math>\frac {6!}{2!} * 5</math> ways. After picking four colors, we can rotate our hexagon in <math>5</math> ways to get different outcomes. <math>\frac {6!}{2} * 5 = 1800</math> | ||
+ | |||
+ | Case <math>3</math>: There are <math>\frac {\frac {6!}{3!} * 10}{2!}</math> ways of arranging the final case. We can pick <math>3</math> colors for our hexagon. There are <math>5</math> spots for the first pair of colors. Then, there are <math>2</math> possible ways we can put the final pair in the last <math>3</math> spaces. But because the two pairs are indistinguishable, we divide by <math>2!</math>. <math>\frac {\frac {6!}{3!} * 10}{2!} = 600</math> | ||
+ | |||
+ | Adding all the possibilities up, we get <math>720+1800+600=\boxed{3120 \ \mathbf{(C)}}</math> | ||
+ | |||
+ | ~ZericHang | ||
== See Also == | == See Also == |
Revision as of 21:42, 25 May 2018
Contents
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 1
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 .
Solution 2
First, notice that there can be cases. One with all vertices painted different colors, one with one pair of adjacent vertices painted the same color and the final one with two pairs of adjacent vertices painted two different colors.
Case : There are
ways of assigning each vertex a different color.
Case : There are
ways. After picking four colors, we can rotate our hexagon in
ways to get different outcomes.
Case : There are
ways of arranging the final case. We can pick
colors for our hexagon. There are
spots for the first pair of colors. Then, there are
possible ways we can put the final pair in the last
spaces. But because the two pairs are indistinguishable, we divide by
.
Adding all the possibilities up, we get
~ZericHang
See Also
2011 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.