2008 AMC 10B Problem 22

There are $\tfrac{6!}{3!\,2!}=60$ distinguishable orderings. The three reds must occupy non-adjacent positions, and the possible red placements are $\{1,3,5\},\{2,4,6\},\{1,3,6\},$ and $\{1,4,6\}.$ For $\{1,3,5\}$ and $\{2,4,6\},$ the remaining seats are mutually non-adjacent, so the blue bead can go in any of the $3,$ giving $3+3=6.$ For $\{1,3,6\}$ and $\{1,4,6\},$ two remaining seats are adjacent, so the blue must separate the whites, giving $2+2=4.$ That is $10$ valid orderings, so the probability is $\tfrac{10}{60}=\tfrac16.$ Thus, the correct answer is C.