1. **Problem statement:** We have a circle divided into 6 sectors, each painted either red, green, or blue, with exactly 2 sectors of each color. We want to find the number of ways to color the sectors so that no two adjacent sectors share the same color. 2. **Key points:** - There are 6 sectors arranged in a circle. - Colors: red (R), green (G), blue (B). - Exactly 2 sectors of each color. - No two adjacent sectors can have the same color. 3. **Approach:** - Since the circle has 6 sectors, and each color appears exactly twice, the problem reduces to counting permutations of the multiset \{R,R,G,G,B,B\} arranged in a circle with no two identical colors adjacent. 4. **Step 1: Count linear arrangements with no two identical colors adjacent.** - Total permutations of \{R,R,G,G,B,B\} without restriction: $$\frac{6!}{2!2!2!} = 90$$ - We want to count permutations with no two identical colors adjacent in a line first. 5. **Step 2: Use Inclusion-Exclusion for linear arrangements:** - Let A = permutations with R's adjacent - B = permutations with G's adjacent - C = permutations with B's adjacent - Number of permutations with R's adjacent: Treat R's as a single entity \(R_2\), so we have \(R_2, G, G, B, B\) = 5 elements with G and B repeated twice. Number of permutations: $$\frac{5!}{2!2!} = 30$$ - Similarly, permutations with G's adjacent = 30, with B's adjacent = 30. - Number of permutations with R's and G's adjacent: Treat R's as \(R_2\), G's as \(G_2\), so elements are \(R_2, G_2, B, B\) = 4 elements with B repeated twice. Number of permutations: $$\frac{4!}{2!} = 12$$ - Similarly, permutations with R's and B's adjacent = 12, with G's and B's adjacent = 12. - Number of permutations with R's, G's, and B's adjacent: Treat R's as \(R_2\), G's as \(G_2\), B's as \(B_2\), so elements are \(R_2, G_2, B_2\) = 3 distinct elements. Number of permutations: $$3! = 6$$ 6. **Step 3: Apply Inclusion-Exclusion:** $$ \text{No adjacent identical colors} = 90 - (30+30+30) + (12+12+12) - 6 = 90 - 90 + 36 - 6 = 30 $$ 7. **Step 4: Adjust for circular arrangement:** - In a circle, the first and last sectors are adjacent. - Some linear arrangements counted above have the same color at the ends, which is not allowed in the circle. - Count linear arrangements with no adjacent identical colors but with the first and last colors the same. - Fix the first and last color as R (since there are 2 R's), then arrange the remaining 4 sectors \{R,G,G,B,B\} with no two identical adjacent. - Number of such linear arrangements is the number of permutations of \{G,G,B,B\} with no two identical adjacent. - Number of permutations of \{G,G,B,B\} is $$\frac{4!}{2!2!} = 6$$ - Count how many have no two identical adjacent: Possible permutations: GGBB, GBGB, GBBG, BGGB, BGBG, BBGG No two identical adjacent are GBGB and BGBG (2 permutations). - So, 2 linear arrangements have first and last color R and no two identical adjacent inside. - Similarly for G and B as first and last colors, total 2 + 2 + 2 = 6 such linear arrangements. 8. **Step 5: Final count for circular arrangements:** $$ \text{Valid circular arrangements} = \text{linear no-adjacent} - \text{linear no-adjacent but first=last} = 30 - 6 = 24 $$ **Answer: 24** **Therefore, the number of different colorings possible is 24, which corresponds to option (D).