In an examination of nine papers, a candidate has to pass in more papers than the number of papers in which he fails in order to be successful. The number of ways in which he can be unsuccessful is

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 is ${ }^9{\mathrm{C}}_9{+}^9{\mathrm{C}}_8{+}^9{\mathrm{C}}_7{+}^9{\mathrm{C}}_6{+}^9{\mathrm{C}}_5{=}^9{\mathrm{C}}_0{+}^9{\mathrm{C}}_1{+}^9{\mathrm{C}}_2{+}^9{\mathrm{C}}_3{+}^9{\mathrm{C}}_4$ (recall the concept of half series) 

$\begin{array}{l}=\frac{1}{2}\left({ }^9{\mathrm{C}}_0{+}^9{\mathrm{C}}_1+\cdots {+}^9{\mathrm{C}}_9\right)\\ =\frac{1}{2}\times 2^9=2^8\end{array}$