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There are 5 letters and 5 different envelopes.The number of ways in which all the letters can be put in wrong envelope is

$\begin{array}{1 1}(A)\;119\\(B)\;44\\(C)\;59\\(D)\;40\end{array} $

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  • $n!=n(n-1)(n-2)(n-3)....(3)(2)(1)$
Total no. of letters =5
Total no. of envelopes =5
Required numbers =$5!\big[1-\large\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}\big]$
$\Rightarrow 120\big[1-\large\frac{1}{1}+\frac{1}{2}-\frac{1}{6}+\frac{1}{24}-\frac{1}{120}\big]$
$\Rightarrow 44$
Hence (B) is the correct answer.
answered Jun 20, 2014 by sreemathi.v

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