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Difference between revisions of "2011 AMC 10A Problems/Problem 22"

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There are <math>6</math> ways to assign a color to <math>A</math>. WLOG, give vertex <math>A</math> a color; we can multiply by <math>6</math> at the end. Since vertices <math>A</math> and <math>C</math> cannot have the same color, there are <math>5</math> ways to assign colors to vertex <math>C</math>. Using this same logic, there are <math>5</math> ways to assign a color to vertices <math>E</math>, <math>B</math>, and <math>D</math>, giving a total of <math>5^4=625</math> ways. However, vertex <math>D</math> cannot be the same color as vertex <math>A</math>. To use complementary counting, we need to find the amount of ways for <math>D</math> and <math>A</math> to have the same color. Notice that this is equivalent to the amount of ways to arrange colors among the vertices of a square. Using the same logic as above, there are <math>5^3=125</math> ways, except we must subtract the number of ways for a triangle. Each time, there is <math>1</math> less vertex, so <math>5</math> times less ways to color. This process stops when there are only <math>2</math> vertices left; in this case there are simply <math>5</math> ways to color this figure.  
 
There are <math>6</math> ways to assign a color to <math>A</math>. WLOG, give vertex <math>A</math> a color; we can multiply by <math>6</math> at the end. Since vertices <math>A</math> and <math>C</math> cannot have the same color, there are <math>5</math> ways to assign colors to vertex <math>C</math>. Using this same logic, there are <math>5</math> ways to assign a color to vertices <math>E</math>, <math>B</math>, and <math>D</math>, giving a total of <math>5^4=625</math> ways. However, vertex <math>D</math> cannot be the same color as vertex <math>A</math>. To use complementary counting, we need to find the amount of ways for <math>D</math> and <math>A</math> to have the same color. Notice that this is equivalent to the amount of ways to arrange colors among the vertices of a square. Using the same logic as above, there are <math>5^3=125</math> ways, except we must subtract the number of ways for a triangle. Each time, there is <math>1</math> less vertex, so <math>5</math> times less ways to color. This process stops when there are only <math>2</math> vertices left; in this case there are simply <math>5</math> ways to color this figure.  
 
So in conclusion, there are <math>6(5^4-(5^3-(5^2-(5))))=6(5^4-5^3+5^2-5)=\boxed{3120 \ \mathbf{(C)}}</math> ways.
 
So in conclusion, there are <math>6(5^4-(5^3-(5^2-(5))))=6(5^4-5^3+5^2-5)=\boxed{3120 \ \mathbf{(C)}}</math> ways.
 +
 +
~Anonymous
  
 
==Solution 4==
 
==Solution 4==

Revision as of 16:41, 8 July 2020

Problem 22

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 1

Let vertex $A$ be any vertex, then vertex $B$ be one of the diagonal vertices to $A$, $C$ be one of the diagonal vertices to $B$, and so on. We consider cases for this problem.

In the case that $C$ has the same color as $A$, $D$ has a different color from $A$ and so $E$ has a different color from $A$ and $D$. In this case, $A$ has $6$ choices, $B$ has $5$ choices (any color but the color of $A$), $C$ has $1$ choice, $D$ has $5$ choices, and $E$ has $4$ choices, resulting in a possible of $6 \cdot 5 \cdot 1 \cdot 5 \cdot 4 = 600$ combinations.

In the case that $C$ has a different color from $A$ and $D$ has a different color from $A$, $A$ has $6$ choices, $B$ has $5$ choices, $C$ has $4$ choices (since $A$ and $B$ necessarily have different colors), $D$ has $4$ choices, and $E$ has $4$ choices, resulting in a possible of $6 \cdot 5 \cdot 4 \cdot 4 \cdot 4 = 1920$ combinations.

In the case that $C$ has a different color from $A$ and $D$ has the same color as $A$, $A$ has $6$ choices, $B$ has $5$ choices, $C$ has $4$ choices, $D$ has $1$ choice, and $E$ has $5$ choices, resulting in a possible of $6 \cdot 5 \cdot 4 \cdot 1 \cdot 5 = 600$ combinations.

Adding all those combinations up, we get $600+1920+600=\boxed{3120 \ \mathbf{(C)}}$.

Solution 2

First, notice that there can be $3$ 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 $1$: There are $6!$ ways of assigning each vertex a different color. $6! = 720$

Case $2$: There are $\frac {6!}{2!} * 5$ ways. After picking four colors, we can rotate our pentagon in $5$ ways to get different outcomes. $\frac {6!}{2} * 5 = 1800$

Case $3$: There are $\frac {\frac {6!}{3!} * 10}{2!}$ ways of arranging the final case. We can pick $3$ colors for our pentagon. There are $5$ spots for the first pair of colors. Then, there are $2$ possible ways we can put the final pair in the last $3$ spaces. But because the two pairs are indistinguishable, we divide by $2!$. $\frac {\frac {6!}{3!} * 10}{2!} = 600$

Adding all the possibilities up, we get $720+1800+600=\boxed{3120 \ \mathbf{(C)}}$

~ZericHang

Solution 3

There are $6$ ways to assign a color to $A$. WLOG, give vertex $A$ a color; we can multiply by $6$ at the end. Since vertices $A$ and $C$ cannot have the same color, there are $5$ ways to assign colors to vertex $C$. Using this same logic, there are $5$ ways to assign a color to vertices $E$, $B$, and $D$, giving a total of $5^4=625$ ways. However, vertex $D$ cannot be the same color as vertex $A$. To use complementary counting, we need to find the amount of ways for $D$ and $A$ to have the same color. Notice that this is equivalent to the amount of ways to arrange colors among the vertices of a square. Using the same logic as above, there are $5^3=125$ ways, except we must subtract the number of ways for a triangle. Each time, there is $1$ less vertex, so $5$ times less ways to color. This process stops when there are only $2$ vertices left; in this case there are simply $5$ ways to color this figure. So in conclusion, there are $6(5^4-(5^3-(5^2-(5))))=6(5^4-5^3+5^2-5)=\boxed{3120 \ \mathbf{(C)}}$ ways.

~Anonymous

Solution 4

This problem is a direct application of the chromatic polynomial of a graph, represented by $P_k(G)$, which returns a polynomial representing the number of ways to color a graph $G$ with $k$ colors such that no two adjacent vertices share the same color. In other words, it gives us a polynomial in terms of $k$ where we can plug in $k=6$ to get our answer.

Well, if we want to find $P_k$ for a graph $G$, we should probably first draw the graph, right? We can first draw five vertices (without any edges) in the shape of a pentagon. After connecting the diagonals, we get a star. We attempt to shift the vertices around to simplify the graph, which we quickly realize is isomorphic (can be turned into) the five-cycle, or the pentagon.

Finding $P_k$ on a graph with a cycle of length at most 3 is straightforward -- we first pick a vertex v. It has 0 colored (visited) neighbors, so we can color it in k ways. We then move on to the vertices adjacent to v, etc, and at the end we multiply all these together. For example, the chromatic polynomial of a triangle is $k(k-1)(k-2)$. However, cycles with length > 3 introduce ambiguity, and thus the above technique fails. Thus, we need to use the recursive formula $P_k(G) = P_k(G-e)-P_k(G*e)$, where $G-e$ represents removing e and $G*e$ represents contracting e, or collapsing the two endpoints of e into one. When we hit a graph where the longest cycle has length 3, we can use the first technique to quickly find $P_k$.

After about four iterations of the algorithm, we get that the chromatic polynomial is $k(k-1)^4 - k(k-1)^3 + k(k-1)(k-2)$. Plugging in 6 gives us $\boxed{3120 \ \mathbf{(C)}}$.

~Helloworld1

Video Solution

https://youtu.be/FThly7dRBIE

~IceMatrix

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

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