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

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- $C(n,r)=\large\frac{n!}{r!(n-r)!}$

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.

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