$\begin{array}{1 1}(A)\;74\\(B)\;84\\(C)\;64\\(D)\;20\end{array} $

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

The possibilities of choosing at least one block ball are :-

1 black +2 non-black (or) 2black +1 non-black (or) 3 black +0 non-black

$\therefore$No of ways of choosing 3 balls with at least one black ball =$(3C_1\times 6C_2)+(3C_2\times 6C_1)+(3C_3\times 6C_0)$

$3C_1=\large\frac{3!}{1!2!}=\frac{3\times 2\times 1}{2\times 1}$

$\Rightarrow 3$

$6C_2=\large\frac{6!}{2!4!}=\frac{6\times 5\times 4!}{2\times 4!}$

$\Rightarrow 15$

$3C_2=\large\frac{3!}{2!1!}=\frac{3\times 2!}{2\times 1}$

$\Rightarrow 3$

$6C_1=\large\frac{6!}{1!\times 5!}$

$\Rightarrow 6$

$3C_3=\large\frac{3!}{3!\times 0!}$

$\Rightarrow 1$

$\Rightarrow (3\times 15)+(3\times 6)+1$

$\Rightarrow 45+18+1$

$\Rightarrow 64$

Hence (C) is the correct answer.

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