Question

There are 12 balls of which 4 are red, 3 black and 5 white: In how many ways can you arrange the balls so that no two white balls may occupy consecutive positions if balls of the same colour are identical?

Answer 1

Balls of the same colour identical.

Let us arrange 4 red and 3 black balls in a row in$\frac{7!}{3!4!}=35$

Now there will be 8 gaps in which 5 identical white balls can be arranged in $\frac{1}{5!}{}^8{P}_5$ or in

${}^8{C}_5\cdot \frac{5!}{5!}=\frac{8!}{3!5!}=42$

Hence by fundamental theorem $35\times 42=1960$ ways.