Total number of ways of filling the 5 boxes numbered as (1, 2, 3, 4 and 5) with either blue or red balls = 25 = 32.
Now, we determine the number of ways of filling the boxes such that the adjacent boxes are filled with blue.
Two adjacent boxes with blue can be obtained in 4 ways, i.e., (12), (23), (34) and (45).
Three adjacent boxes with blue can be obtained in 3 ways, i.e., (123), (234) and (345).
Four adjacent boxes with blue can be obtained in 2 ways, i.e., (1234) and (2345).
And all 5 boxes can have blue in only 1 way.
Hence, the total number of ways of filling the boxes such that adjacent boxes have blue = (4 + 3 + 2 +1) = 10.
Hence, the number of ways of filling up the boxes such that no two adjacent boxes have blue = 32 - 10 = 22
The correct option is D.