Difference between revisions of "2011 AMC 12A Problems/Problem 16"

(Solution: removed wrong solution)
(Casework)
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== Solution ==
 
== Solution ==
  
{{solution}}
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We can do some casework when working our way around the pentagon from <math>A</math> to <math>E</math>. At each stage, there will be a makeshift diagram.
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1.) For <math>A</math>, we can choose any of the 6 colors.
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        A : 6
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 +
2.) For <math>B</math>, we can either have the same color as <math>A</math>, or any of the other 5 colors. We do this because each vertex of the pentagon is affected by the 2 opposite vertices, and <math>D</math> will be affected by both <math>A</math> an <math>B</math>.
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      A : 6
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  B:1        B:5
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3.) For <math>C</math>, we cannot have the same color as <math>A</math>. Also, we can have the same color as <math>B</math> (<math>E</math> will be affected), or any of the other 4 colors. Because <math>C</math> can't be the same as <math>A</math>, it can't be the same as <math>B</math> if <math>B</math> is the same as <math>A</math>, so it can be any of the 5 other colors.
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      A : 6
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  B:1        B:5
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  C:5    C:4  C:1
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 +
4.) <math>D</math> is affected by <math>A</math> and <math>B</math>. If they are the same, then <math>D</math> can be any of the other 5 colors. If they are different, then <math>D</math> can be any of the (6-2)=4 colors.
 +
 
 +
      A : 6
 +
  B:1        B:5
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  C:5    C:4  C:1
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  D:5    D:4  D:4
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 +
5.) <math>E</math> is affected by <math>B</math> and <math>C</math>. If they are the same, then <math>E</math> can be any of the other 5 colors. If they are different, then <math>E</math> can be any of the (6-2)=4 colors.
 +
 
 +
      A : 6
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  B:1        B:5
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  C:5    C:4  C:1
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  D:5    D:4  D:4
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  E:4    E:4  E:5
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6.) Now, we can multiply these three paths and add them:
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<math>(6\times1\times5\times5\times4)+(6\times5\times4\times4\times4)+(6\times5\times1\times4\times5)=600+1920+600=3120</math>
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7.) Our answer is <math>C</math>!
  
 
== 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 20:19, 17 February 2012

Problem

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

We can do some casework when working our way around the pentagon from $A$ to $E$. At each stage, there will be a makeshift diagram.

1.) For $A$, we can choose any of the 6 colors.

        A : 6

2.) For $B$, we can either have the same color as $A$, or any of the other 5 colors. We do this because each vertex of the pentagon is affected by the 2 opposite vertices, and $D$ will be affected by both $A$ an $B$.

      A : 6
 B:1        B:5

3.) For $C$, we cannot have the same color as $A$. Also, we can have the same color as $B$ ($E$ will be affected), or any of the other 4 colors. Because $C$ can't be the same as $A$, it can't be the same as $B$ if $B$ is the same as $A$, so it can be any of the 5 other colors.

      A : 6
 B:1        B:5
 C:5     C:4   C:1

4.) $D$ is affected by $A$ and $B$. If they are the same, then $D$ can be any of the other 5 colors. If they are different, then $D$ can be any of the (6-2)=4 colors.

      A : 6
 B:1        B:5
 C:5     C:4   C:1
 D:5     D:4   D:4

5.) $E$ is affected by $B$ and $C$. If they are the same, then $E$ can be any of the other 5 colors. If they are different, then $E$ can be any of the (6-2)=4 colors.

      A : 6
 B:1        B:5
 C:5     C:4   C:1
 D:5     D:4   D:4
 E:4     E:4   E:5

6.) Now, we can multiply these three paths and add them: $(6\times1\times5\times5\times4)+(6\times5\times4\times4\times4)+(6\times5\times1\times4\times5)=600+1920+600=3120$

7.) Our answer is $C$!

See also

2011 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AMC 12 Problems and Solutions
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