Find the number of arrangements of 4 beads (2 red, 2 green, 2 blue) in a circle such that the two red beads are not adjacent.
Solution
We divide this problem into cases based on the relative position of the two red beads: - They are adjacent. Then, there are 4 possible placements of the green and blue beads: GGBB, GBBG, GBGB, BGGB. - They are 1 bead apart. Then, there are two choices for the bead between then and 2 choices for the other bead of that color, for a total of 4. - They are opposite. Then, there are three choices for the placement of the green beads. This gives a total of 11 arrangements.
Want a route through all this instead of an archive? The track
puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.