Question

In order to get through in an examination of nine papers, a candidate has to pass in more papers than the number of papers in which he fails. The number of ways in which he can fail, in this examination, is :

• A. $9\times \left(8\right)!$
• B. $128$
• C. $255$
• D. $256$

Answer 1

Answer: (D)

$256$

We know that,

$$

The candidate is unsuccessful if he falls in $9$ or $8$ or $7$ or $6$ or $5$ papers.

$\therefore$ The number of ways to be unsuccessful =${}^9{C}_9{+}^9{C}_8{+}^9{C}_7{+}^9{C}_6{+}^9{C}_5$

$\begin{array}{c}{}^9{C}_9=\frac{9!}{9!10!}=1\\ {=}^9{C}_8=\frac{9!}{8!1!}=9\\ {}^9{C}_7=\frac{9!}{7!2!}=\frac{9\times 8\times 7}{7!\times 2}=36\\ {}^9{C}_6=\frac{9!}{6!3!}=\frac{96\times 8\times 7\times 6!}{3\times 2\times 6!}=84\\ {}^9{C}_5=\frac{9!}{5!4!}=\frac{9\times 8\times 7\times 6}{4\times 3\times 2\times 5!}=126\\ \Rightarrow 1+9+36+84+126=256\end{array}$

Hence,

Option $D$ is correct answer.