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

\(= \frac{6!}{3!3!}\times\frac{5!}{3!2!}\times\frac{5!}{3!2!}\)

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