A bag contains $5$ black and $6$ red balls. Determine the number of ways in which $2$ black and $3$ red balls can be selected.

There are $5$ black and $6$ red balls in the bag.
$2$ black balls can be selected out of $5$ black ball in ${}^5{C}_2$ ways and $3$ red ball can be selected out of $6$ red balls in ${}^6{C}_3$ ways.
Thus, by multiplication principle, required number of ways of selecting $2$ black and $3$ red balls is  ${}^5{C}_2{\times }^6{C}_3$

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

$=\frac{5\times 4}{2}\times \frac{6\times 5\times 4}{3\times 2\times 1}=10\times 20=200$