$\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.

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