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# Three balls are drawn from a bag containing 5 red,4 white and 3 black balls.The number of ways in which this can be done if at least 2 are red is

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

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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.