Q. In an examination of 9 papers a candidate has to pass in more papers, then the number of papers in which he fails in order to be successful. The number of ways in which he can be unsuccessful, is
A) 255
B) 256
C) 193
D) 319
Explanation
(b)
$\because$ The candidate is unsuccessful, if he fails in 9 or 8 or 7 or 6 or 5 papers.
$\therefore$ Numbers of ways to be unsuccessful
$= {\ ^{9}C}_{9} +^{9}C_{8} +^{9}C_{7} +^{9}C_{6} +^{9}C_{5}$
$= \ ^{9}C_{0} + \ ^{9}C_{1} +^{9}C_{2} +^{9}C_{3} +^{9}C_{4}$
$= \frac{1}{2}(\ ^{9}C_{0} +^{9}C_{1} + .\ldots.. + \ ^{9}C_{9})$
$= \frac{1}{2}\left( 2^{9} \right) = 2^{8} = 256$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.