Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 19"
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We look at the number of possibilities by considering each dot. If we start with <math>A</math>, there are <math>4</math> distinct colors for it. Then, there are <math>3!</math> ways to color the other three dots since they are all distinct. Thus, the number of ways to color the graph is <math>3! \cdot 4 \cdot 4=96</math>. However, <math>A,B,C,D</math> are distinct, or in other words, the graphs must be distinct which can occur one way for each pair of starting dots we choose. Thus, the total number is <math>96-{4\choose 2}=84</math>. | We look at the number of possibilities by considering each dot. If we start with <math>A</math>, there are <math>4</math> distinct colors for it. Then, there are <math>3!</math> ways to color the other three dots since they are all distinct. Thus, the number of ways to color the graph is <math>3! \cdot 4 \cdot 4=96</math>. However, <math>A,B,C,D</math> are distinct, or in other words, the graphs must be distinct which can occur one way for each pair of starting dots we choose. Thus, the total number is <math>96-{4\choose 2}=84</math>. | ||
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Revision as of 12:13, 23 July 2006
Problem
In the figure below, there are 4 distinct dots and , joined by edges. Each dot is to be colored either red, blue, green, or yellow. No two dots joined by an edge are to be colored with the same color. How many completed colorings are possible?
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
Solution
We look at the number of possibilities by considering each dot. If we start with , there are distinct colors for it. Then, there are ways to color the other three dots since they are all distinct. Thus, the number of ways to color the graph is . However, are distinct, or in other words, the graphs must be distinct which can occur one way for each pair of starting dots we choose. Thus, the total number is .