A. 15
B. 18
C. 24
D. 30
Explanation:
Case I : 2 balls of the same colour and two balls are a different colour are arranged.
Two balls of the same colour and two balls of different colours can be arranged together in which two balls of the same colour are adjacent =4!/2!x2! = 6 ways
Therefore, Total number of arrangements = 6×3 =18 ways
Case II : Two colours out of 3 can be selected in = 3C1 = 3ways
Now 2 balls of each colour can be arranged alternatively in 2 ways
Thus 4 balls can be arranged(two of each colours)
= 3×2 = 6ways
Hence total number of arrangements = 18+6 =24 ways
