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Chapter 16: Problem 2

If a regular pentagon is free to move in space and we can color its vertices with red, white, and blue paint, how many nonequivalent configurations have exactly three red vertices? How many have two red, one white, and two blue vertices?

Short Answer

Expert verified
There are 40 nonequivalent configurations with exactly three red vertices and 30 nonequivalent configurations with two red, one white, and two blue vertices.

Step by step solution

01

Configurations with exactly three red vertices

The task is to paint exactly 3 vertices of the pentagon red. The remaining 2 vertices can be either white or blue, so we have 2 ways of coloring each of the remaining vertices. So, the total number of nonequivalent configurations of painting 3 vertices red would be \( ^5C_3 \times 2^2 = 10 \times 4 = 40 \).
02

Configurations with two red, one white, and two blue vertices

Now, we need to paint 2 vertices red, 1 white, and 2 blue. First, choose 2 out of 5 vertices to paint red which can be done in \( ^5C_2 \) ways. After painting 2 vertices red, we are left with 3 vertices. Now, choose 1 out of these 3 vertices to paint white which can be done in \( ^3C_1 \) ways. The 2 remaining vertices can then be painted blue. So, the total number of nonequivalent configurations of painting 2 vertices red, 1 white, and 2 blue would be \( ^5C_2 \times ^3C_1 = 10 \times 3 = 30 \).

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