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 = 6x3 =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) 
= 3x2 = 6ways 
 Hence total number of arrangements = 18+6 =24 ways