Question

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.

Answer 1

There are 5 black and 6 red balls in the bag.
2 black balls can be selected out of 5 black balls in 5C2 ways and 3 red balls can be selected out of 6 red balls in 6C3 ways.

Thus, by multiplication principle, required number of ways of selecting 2 black and 3 red balls
$=$ $^5C_2\times\space^6C_3$

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

$=\frac{5\times4}{2}\times\frac{6\times5\times4}{3\times2\times1}$
$=10\times20$
$=200$