There are two identical red, two identical black and two identical white balls. In how many different ways can the balls be placed in the cells (each cell to contain one ball) shown above such that balls of the same colour do not occupy any two consecutive cells?

To find the number of ways to arrange 2 red (R), 2 black (B), and 2 white (W) identical balls in 6 cells such that no same-colored balls are consecutive:

• Total arrangements (Linear): First, consider a linear row. The total permutations are 6! / (2! 2! 2!) = 90.
• Linear Constraints: Using the Inclusion-Exclusion Principle, the number of ways where no same-colored balls are adjacent in a row is 30.
• Circular Arrangement: The diagram in the original IAS 2008 exam showed these cells in a circle. In a circular layout, the first and last cells are also adjacent.
• We must subtract the linear cases where the first and last balls are the same color. For each color, there are 2 such cases (e.g., R-B-W-B-W-R and R-W-B-R-W-B), totaling 6 cases (2 per color × 3 colors).
• Result: 30 - 6 = 24.