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Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
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In an examination of a 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

$\begin{array}{1 1}(A)\;255\\(B)\;256\\(C)\;193\\(D)\;319\end{array} $

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  • $n(C,r)=\large\frac{n!}{r!(n-r)!}$
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 =$9C_9+9C_8+9C_7+9C_6+9C_5$
$9C_9=\large\frac{9!}{9!0!}=$$1$
$9C_8=\large\frac{9!}{8!1!}=$$9$
$9C_7=\large\frac{9!}{7!2!}=\frac{9\times 8\times 7!}{7!\times 2}=$$36$
$9C_6=\large\frac{9!}{6!3!}=\frac{9\times 8\times 7\times 6!}{3\times 2\times 6!}=$$84$
$9C_5=\large\frac{9!}{5!4!}=\frac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 5!}=$$126$
$\Rightarrow 1+9+36+84+126$
$\Rightarrow 256$
Hence (B) is the correct answer.
answered May 19, 2014 by sreemathi.v
 

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