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
\(=\) \(^5C_2\times\space^6C_3\)

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

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