$\begin{array}{1 1}(A)\;60\\(B)\;62\\(C)\;64\\(D)\;66\end{array} $

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

Total white balls =2

Total black balls =3

Total red balls =4

The possibilities to draw three balls with atleast one black ball =(1 black+2 non black)+(2 black+1 non black)+(3 black+0 non black)

$\Rightarrow 3C_1\times 6C_2+3C_2\times 6C_1+3C_3\times 6C_0$

$\Rightarrow 45+18+1$

$\Rightarrow 64$

Hence (C) is the correct answer.

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