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

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== Solution ==
 
== Solution ==
  
There are three cases to consider altogether: all the vertices have different colors, two of the vertices have the same color while the rest have different colors, and three of the vertices have the same color while the rest are all different..  
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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.
  
When two of the vertices are of the same color, they have to be two consecutive vertices. For three of the vertices to have the same color, two pairs of two consecutive vertices must be the same color while one is a different color.
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1.) For <math>A</math>, we can choose any of the 6 colors.
  
In the first case, the number of ways to color the map is <math>6 \times 5 \times 4 \times 3 \times 2 = 720</math>.
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        A : 6
  
In the second case, the number of ways of coloring is <math>5 \times (6 \times 5 \times 4 \times 3) = 1800</math>.
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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> and <math>B</math>.
  
In the third case, the number of ways of coloring is <math>5 \times (6 \times 5 \times 4 ) = 600</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.
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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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  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.
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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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  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)
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=600+1920+600=3120</math>
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7.) Our answer is <math>C</math>!
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==Solution 2==
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Right off the bat, we can analyze three things:
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1.) There can only be two of the same color on the pentagon.
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2.) Any pair of the same color can only be next to each other on the pentagon.
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3.) There can only be two different pairs of same colors on the pentagon at once.
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Now that we know this, we can solve the problem by using three cases: no same color pairs,  one same color pair, and two same color pairs.
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1.) If there are no color pairs, it is a simple permutation: six different colors in five different spots. We count <math>6!=720</math> cases. No rotation is necessary because all permutations are accounted for.
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2.)If there is one color pair, we must count 6 possibilities for the pair(as one element), 5 for the third vertex, 4 for the fourth vertex, and 3 for the fifth vertex.
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We get <math>6\times5\times4\times3=360</math>.
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However, there are 5 different locations the pair could be at. Therefore we get <math>360\times5=1800</math> possibilities for one pair.
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3.)If there are two color pairs, we must count 6 possibilities for the first pair(as one element), 5 possibilities for the next pair(as one element), and 4 possibilities for the final vertex.
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We get <math>6\times5\times4=120</math>.
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Once again, there are 5 different rotations in the pentagon that we must account for. Therefore we get <math>120\times5=600</math>  possibilities for two pairs.
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5.) If we add all of three cases together, we get <math>720+1800+600=3120</math>. The answer is <math>C</math>.
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Solution by gsaelite
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==Video Solution==
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https://youtu.be/FThly7dRBIE
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~IceMatrix
  
Total number of colorings is thus <math>3120 \rightarrow \boxed{\textbf{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}}
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[[Category:Introductory Combinatorics Problems]]
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{{MAA Notice}}

Revision as of 14:40, 13 December 2020

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 $A$ and $B$.

      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 $C$!

Solution 2

Right off the bat, we can analyze three things:


1.) There can only be two of the same color on the pentagon.

2.) Any pair of the same color can only be next to each other on the pentagon.

3.) There can only be two different pairs of same colors on the pentagon at once.


Now that we know this, we can solve the problem by using three cases: no same color pairs, one same color pair, and two same color pairs.


1.) If there are no color pairs, it is a simple permutation: six different colors in five different spots. We count $6!=720$ cases. No rotation is necessary because all permutations are accounted for.


2.)If there is one color pair, we must count 6 possibilities for the pair(as one element), 5 for the third vertex, 4 for the fourth vertex, and 3 for the fifth vertex.

We get $6\times5\times4\times3=360$.

However, there are 5 different locations the pair could be at. Therefore we get $360\times5=1800$ possibilities for one pair.


3.)If there are two color pairs, we must count 6 possibilities for the first pair(as one element), 5 possibilities for the next pair(as one element), and 4 possibilities for the final vertex.

We get $6\times5\times4=120$.

Once again, there are 5 different rotations in the pentagon that we must account for. Therefore we get $120\times5=600$ possibilities for two pairs.


5.) If we add all of three cases together, we get $720+1800+600=3120$. The answer is $C$.

Solution by gsaelite

Video Solution

https://youtu.be/FThly7dRBIE

~IceMatrix


See also

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

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