Choose the Center Color

There are 5 colors available, and we can choose any one of them for the center region. This gives us 5 choices. We will consider these choices at the end.

Color the Outer Ring

The outer ring consists of 4 regions, and we need to ensure that no two adjacent regions share the same color. We will consider different cases based on how many colors are used in the outer ring .

Case A

Case B

Case C

Case A: Use All 4 Colors

If all four outer segments are colored differently, there’s no chance of having two identical neighbors. Each arrangement of 4 distinct colors is valid. The number of ways to arrange 4 distinct colors is given by the factorial of 4:

4! = 24

Case B: Use Exactly 3 of the 4 Colors

In this case, one color is excluded, and among the 3 chosen colors, one is used twice, and the other two are used once each. Let's count the possibilities:

Choose which color to exclude: There are 4 colors available, so there are 4 ways to pick the one we do not use.
Choose the color that will appear twice: From the remaining 3 colors, there are 3 ways to choose the color that will be repeated.
Arrange these four slots (2, 3, 4, 5) with a pattern like A, A, B, C: To avoid the two identical colors (A’s) being adjacent, they must be placed in opposite positions in the circle. In a 4-region circle, there are exactly 2 ways to choose opposite pairs for the repeated color: either (2 & 4) or (3 & 5).
Place the remaining two different colors in the 2 leftover spots: There are 2 ways to arrange these two distinct colors.

Multiply these together:

4 \text{ (exclude color)} \times 3 \text{ (choose repeated color)} \times 2 \text{ (opposite pair)} \times 2 \text{ (arrange last two colors)} = 48

Case C: Use Exactly 2 of the 4 Colors

We can pick 2 colors and arrange them alternately to avoid identical neighbors, such as A, B, A, B. This is the only way to avoid having two identical neighbors in a 4-cycle if we have only 2 colors.

Choose which 2 colors out of the 4: There are \binom{4}{2} = 6 ways to choose 2 colors .
Arrange them alternately around the circle: For chosen colors A and B, we can have A, B, A, B or B, A, B, A. That’s 2 ways per choice of the pair.

Total for 2-color usage:

6 \times 2 = 12

Step 3: Add Them Up

Now, we sum the number of ways for each case:

Using 4 colors: 24 ways
Using 3 colors: 48 ways
Using 2 colors: 12 ways

Total for the outer ring:

24 + 48 + 12 = 84 \text{ ways}

Step 4: Multiply by the Center Choices

We had 5 choices for the center color. Each choice leads to 84 ways to color the ring:

5 \times 84 = 420 \text{ ways}

Counting & Binomials