1. **Problem Statement:** We have a circle divided into 6 sectors of different sizes. We want to paint 2 sectors red, 2 green, and 2 blue such that no two adjacent sectors share the same color. 2. **Understanding the Problem:** This is a combinatorial coloring problem on a circular arrangement with 6 sectors. The constraint is that no two neighboring sectors have the same color. 3. **Key Points:** - There are 6 sectors. - Colors: Red (R), Green (G), Blue (B). - Each color is used exactly twice. - Adjacent sectors cannot have the same color. 4. **Approach:** - Since the circle is divided into 6 sectors, and each color appears twice, the coloring must be arranged so that no two same colors are adjacent. - We can think of this as arranging the sequence of colors R, R, G, G, B, B around a circle with no two identical colors adjacent. 5. **Counting Valid Arrangements:** - First, consider the linear arrangements of the multiset \{R,R,G,G,B,B\} with no two identical colors adjacent. - Then, account for the circular condition by ensuring the first and last colors are also different. 6. **Step 1: Count linear arrangements with no two identical colors adjacent.** - Total permutations of \{R,R,G,G,B,B\} is \( \frac{6!}{2!2!2!} = 90 \). - Use inclusion-exclusion to count arrangements with no two identical colors adjacent. 7. **Step 2: Use Inclusion-Exclusion:** - Let A = arrangements with R's adjacent. - Let B = arrangements with G's adjacent. - Let C = arrangements with B's adjacent. - \(|A| = |B| = |C| = \frac{5!}{2!2!} = 30\) (treating the pair as one element). - \(|A \cap B| = |A \cap C| = |B \cap C| = 12\) (treating two pairs as single elements). - \(|A \cap B \cap C| = 4! = 24\) (all pairs treated as single elements). - Number of arrangements with no two identical colors adjacent: $$90 - (30+30+30) + (12+12+12) - 24 = 90 - 90 + 36 - 24 = 12$$ 8. **Step 3: Adjust for circular condition:** - In a circle, the first and last sectors are adjacent. - From the 12 linear arrangements, exclude those where the first and last colors are the same. - By checking, 6 of these 12 have first and last colors different. 9. **Final Answer:** - There are 6 valid ways to paint the circle with 2 red, 2 green, and 2 blue sectors so that no two adjacent sectors share the same color. **Summary:** - Total valid circular colorings = 6 This ensures the coloring meets the problem's constraints.