Question

In an examination of $9$ papers, a candidate has to pass in more papers than the number of papers in which he fails in order to get the success. The number of ways in which he can be fail, is

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

Answer 1

Answer: (B)

The candidate will fail if he fails in $9$ or $8$ or $7$ or $6$ or $5$ papers
$\therefore$ Required number of ways
$={}^9{C}_9+{}^9{C}_8+{}^9{C}_7+{}^9{C}_6+{}^9{C}_5$
$={}^9{C}_0+{}^9{C}_1+{}^9{C}_2+{}^9{C}_3+{}^9{C}_4(\because {}^n{C}_r={}^n{C}_{n-r})$
$=\frac{1}{2}({}^9{C}_0+{}^9{C}_1+...+{}^9{C}_9)=\frac{1}{2}(2^9)$
$(\because (1+x{)}^9=C_0+C_{1}x+C_2{x}^2+...+C_9{x}^9$
Substitute $x=1$, we get $2^9=C_0+C_1+....+C_9)$
$=2^8=16\times 16=256$