Given two identical red balls, two identical black balls, and two identical white balls, in how many distinct ways can these balls be arranged in the cells shown above so that no two balls of the same color are placed next to each other?

Explanation

Let's analyze the problem in two parts: Case 1: Arrangements where two balls of the same color are adjacent and the other balls are of different colors. The number of ways to arrange four balls with two identical balls adjacent is calculated as 4! divided by (2! × 2!) = 6 ways. Since this can happen for each of the three colors, total arrangements here are 6 × 3 = 18. Case 2: Arrangements where balls of two colors alternate without any two identical balls touching. Selecting two colors out of three can be done in 3 ways. For each selection, the two balls of each color can be alternated in 2 ways. Hence, total arrangements in this case are 3 × 2 = 6. Adding both cases, the total number of valid arrangements is 18 + 6 = 24.