n letters to each of which corresponds an addressed envelope are placed in the envelopes at random. What is the probability that no letter is placed in the right envelope?

Let Ai denote the event that the ith letter is placed in the right envelope. Then the required probability is.

$\begin{array}{l}P\left(\overline{A_1}\cap \overline{A_2}\cap \dots \dots \cap \overline{A_n}\right)\\ =P\left(\overline{A_1\cup A_2\cup \dots \dots ...A_n}\right)[\mathrm{ByDe}-\mathrm{Morganlaw}]\\ =1-P\left(A_1\cup A_2\cup \dots ..\cup A_n\right)\\ =1-\left[\mathrm{\Sigma }P\left(A_j\right)-\mathrm{\Sigma }P\left(A_i\cap A_i\right)+\mathrm{\Sigma }P\left(A_i\cap A_i\cap A_2\right)-\dots \dots \right.\\ \left.+(-1{)}^{n-1}P\left(A_1\cap A_2\cap \dots \dots \cap A_N\right)\right]\\ i\ne j\ne k\end{array}$

Now, $P\left(A\right)=\frac{(n-1)!}{n!}$ as having placed $i^{th}$ letter in the I n! right envelope, the remaining letters can be placed in (n-1)! ways. 

Similarly $P\left(A_1\cap A_2\cap \dots \cap \cap A\right)=1$ Prob. of r particular

letters in right envelopes $=\frac{(n-r)!}{n!}$

$\begin{array}{l}\therefore \mathrm{\Sigma }P\left(A_1\cap A_2\cap \dots .\cap \cap A\right)\\ {=}^n{C}_r\cdot \frac{(n-r)!}{n!}=\frac{1}{r!}\end{array}$

where r=1,2,3,4,……..,n

$\begin{array}{l}\therefore \mathrm{\Sigma }\left({\overline{A}}_i\cap \overline{A_2}\cap \dots \cap \overline{A_n}\right)\\ =1-\left\{\frac{1}{1!}-\frac{1}{2!}+\frac{1}{3!}-\dots +(-1{)}^{n-1}\cdot \frac{1}{n!}\right\}\\ =\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\dots ..+(-1{)}^n\cdot \frac{1}{n!}\end{array}$

which is equal to first n- 2 terms in the expansion of $e^{-1}$