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

(Solution)
(Solution 3)
(15 intermediate revisions by 11 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
Each vertex of convex polygon <math>ABCDE</math> is to be assigned a color. There are <math>6</math> colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
+
Each vertex of convex pentagon <math>ABCDE</math> is to be assigned a color. There are <math>6</math> colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?
  
 
<math>
 
<math>
Line 11: Line 11:
 
== Solution ==
 
== Solution ==
  
There are three cases to consider altogether. The cases are all vertices have different colours, two of the vertices sharing the same colour while the rest of the vertices are of different colours and lastly, three of the vertices sharing the same colour while the rest are of different colours.  
+
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 colour, they have to be two consecutive vertices. Likewise for three of the vertices having the same colour, they have to be three consecutive vertices.
+
1.) For <math>A</math>, we can choose any of the 6 colors.
  
 +
        A : 6
  
When all vertices have different colours,  
+
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>.
  
Number of ways of colouring <math>= 6 \times 5 \times 4 \times 3 \times 2 = 720 </math>.
+
      A : 6
 +
  B:1        B:5
  
 +
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.
  
When two vertices sharing the same colour while other vertices having different colours,
+
      A : 6
 +
  B:1        B:5
 +
  C:5    C:4  C:1
  
Number of ways of colouring <math>= 5 \times (6 \times 5 \times 4 \times 3) = 1800  </math>
+
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
 +
  C:5    C:4  C:1
 +
  D:5    D:4  D:4
  
When three vertices sharing the same colour while other vertices having different colours,  
+
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.
  
Number of ways of colouring <math>= 5 \times (6 \times 5 \times 4 ) = 600  </math>
+
      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:
 +
<math>(6\times1\times5\times5\times4)+(6\times5\times4\times4\times4)+(6\times5\times1\times4\times5)
 +
=600+1920+600=3120</math>
 +
 +
7.) Our answer is <math>C</math>!
 +
 +
==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 <math>6!=720</math> 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 <math>6\times5\times4\times3=360</math>.
 +
 +
However, there are 5 different locations the pair could be at. Therefore we get <math>360\times5=1800</math> 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 <math>6\times5\times4=120</math>.
 +
 +
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.
 +
 +
 +
5.) If we add all of three cases together, we get <math>720+1800+600=3120</math>. The answer is <math>C</math>.
 +
 +
Solution by gsaelite
 +
 +
==Video Solution==
 +
https://youtu.be/FThly7dRBIE
 +
 +
~IceMatrix
  
Total number of colouring <math> = 3120 </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}}
 +
 +
[[Category:Introductory Combinatorics Problems]]
 +
{{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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS