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

$\begin{array}{1 1}(A)\;199\\(B)\;200\\(C)\;201\\(D)\;202\end{array}$

Toolbox:
• $C(n,r)=\large\frac{n!}{r!(n-r)!}$
Total number of black ball =5
Total number of red ball =6
No of ways of selecting 2 black balls =$5C_2$
$\Rightarrow \large\frac{5!}{2!3!}$
$\Rightarrow \large\frac{5\times 4\times 3!}{2\times 3!}$
$\Rightarrow 10$
No of ways of selecting 3 red balls =$6C_3$
$\Rightarrow \large\frac{6!}{3!3!}$
$\Rightarrow \large\frac{6\times 5\times 4\times 3!}{3\times 2\times 3!}$
$\Rightarrow 20$
Total no of ways of selecting 2 black and 3 red ball =$10\times 20$
$\Rightarrow 200$
Hence (B) is the correct answer.