$\begin{array}{1 1}(A)\;255\\(B)\;256\\(C)\;193\\(D)\;319\end{array} $

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

The candidate is unsuccessful if he fails in 9 or 8 or 7 or 6 or 5 papers

$\therefore$ The number of ways to be unsuccessful =$9C_9+9C_8+9C_7+9C_6+9C_5$

$9C_9=\large\frac{9!}{9!0!}=$$1$

$9C_8=\large\frac{9!}{8!1!}=$$9$

$9C_7=\large\frac{9!}{7!2!}=\frac{9\times 8\times 7!}{7!\times 2}=$$36$

$9C_6=\large\frac{9!}{6!3!}=\frac{9\times 8\times 7\times 6!}{3\times 2\times 6!}=$$84$

$9C_5=\large\frac{9!}{5!4!}=\frac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 5!}=$$126$

$\Rightarrow 1+9+36+84+126$

$\Rightarrow 256$

Hence (B) is the correct answer.

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