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.