Question

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

• A. $128$
• B. $255$
• C. $256$
• D. $9\times 8!$

Answer 1

The correct option is C $256$

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

The number of ways to be $=9c_9+9c_8+9c_7+9c_6+9c_5$

$\begin{array}{c}9c_9=\frac{9!}{9!0!}=1\\ 9c_8=\frac{9!}{8!1!}=9\left[\begin{array}{c}usingformula\\ {}^n{c}_r=\frac{n!}{r!\left(n-r\right)!}\end{array}\right]\\ 9c_7=\frac{9!}{7!2!}=36\\ 9c_6=\frac{9!}{6!3!}=84\\ 9c_5=\frac{9!}{5!4!}=126\\ \Rightarrow 1+9+36+84+126=256.\end{array}$