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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

Correct Answer: 24

Last Updated: February 07, 2020