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 ${}^6{C}_3$ ways.

$3$ balls can be selected from $5$ white balls in ${}^5{C}_3$ ways.

$3$ balls can be selected from $5$ blue balls in ${}^5{C}_3$ ways.

Thus, by multiplication principle, required number of ways of selecting $9$ balls

${}^6{C}_3{\times }^5{C}_3{\times }^5{C}_3=\frac{6!}{3!3!}\times \frac{5!}{3!2!}\times \frac{5!}{3!2!}$
$=\frac{6\times 5\times 4\times 3!}{3!\times 3\times 2\times 1}\times \frac{5\times 4\times 3!}{3!\times 2\times 1}\times \frac{5\times 4\times 3!}{3!\times 2\times 1}$
$=20\times 10\times 10$

$=2000$