$\begin{array}{1 1}(A)\;60\\(B)\;70\\(C)\;80\\(D)\;90\end{array} $

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

Total no of red balls =5

Total no of white balls =4

Total no of black balls =3

Total no of ways =$5C_2\times 7C_1+5C_3\times 7C_0$

$5C_2=\large\frac{5!}{2!3!}=\frac{5\times 4\times 3!}{2\times 3!}$

$\Rightarrow 10$

$7C_1=\large\frac{7!}{1!6!}=\frac{7\times 6!}{1\times 6!}$

$\Rightarrow 7$

$5C_3=\large\frac{5!}{3!2!}=\frac{5\times 4\times 3!}{3!\times 2}$

$\Rightarrow 10$

Total no of ways =$10\times 7+10$

$\Rightarrow 80$

Hence (C) is the correct answer.

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