If $\mathrm n$ letters are placed at random in $\mathrm n$ envelopes, what is the probability that exactly $\text{2}$ letters will be placed in the correct envelopes?
I know the solution about this question.
Step 1:
The probability of at least one letter will be placed in the correct envelope is $$\mathrm P_n = 1 - \frac{1}{2!} + \frac{1}{3!} + \cdots + (-1)^{n+1}\frac{1}{n!}$$
Step 2:
The probability of exact $2$ envelopes are matching means other $\mathrm n-2$ envelops are mismatching.
If we select $2$ from $n$ envelopes, we have $\mathrm{C_{n}^{2}}$ choices. For each choice, we have $(1 - \mathrm P_{n-2}) \cdot (n-2)! $ events that all other envelopes are mismatching. \begin{align} P(\text{exact 2 envelopes matching}) &= \frac{\# \text{Events that exact 2 matching}}{\# \text{All Events}}\\ & = \frac{\frac{n(n-1)}{2}(n-2)![1 - (\mathrm P_{n-2})]}{n!}\\ & = \frac{1}{2}[\frac{1}{2!} - \frac{1}{3!} + \cdots + (-1)^{n-2}\frac{1}{(n-2)!} ] \end{align}