Question

Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.

Answer 1

There are a total of 6 red balls, 5 white balls, and 5 blue balls.
9 balls have to be selected in such a way that each selection consists of 3 balls of each colour.

Here, 3 balls can be selected from 6 red balls in $^6C_3$ ways.
3 balls can be selected from 5 white balls in $^5C_3$ ways.
3 balls can be selected from 5 blue balls in $^5C_3$ ways.

Thus, by multiplication principle, required number of ways of selecting 9 balls
= 6C3 $\times$5C3 $\times$ 5C3

$= \frac{6!}{3!3!}\times\frac{5!}{3!2!}\times\frac{5!}{3!2!}$

$= \frac{6\times5\times4\times3!}{3!\times3\times2}\times\frac{5\times4\times3!}{3!\times2\times1}\times\frac{5\times4\times3!}{3!\times2\times1}$
$= 20\times10\times10$
$= 2000$