Consider the following mad postman scenario. The mad postman has $n$ parcels which should go to $n$ different destinations. However, the mad postman assigns destinations to the parcels randomly. What is the expected number of parcels that have a correct destination address assigned?
I think in the following way. Let $X$ be a random variable that assumes values $1,2,\dots,n$ - the number of correctly addressed parcels. First, I want to know the probability distribution of $X$. So I need $P(X=k)$, for $k=0,1,\dots,n$. Overall, we have $n!$ possible assignments of destinations to parcels. I want to calculate the number of assignments which have exactly $k$ parcels with a correct address assigned. $n \choose k$ is the number of all sets of parcels of cardinality $k$ which could potentially get a correct address. Given any set of $k$ parcels, we want to know how many assignments there are which send the remaining $(n-k)$ parcels to incorrect destinations. This number is equal $(n-k-1)!$. Overall:
$$P(X=k) = \frac{ {n \choose k} (n-k-1)!}{n!}$$
Something bad happens with this expression for $k=n$, but $P(X=n) = \frac{1}{n!}$.
$$E(X) = \Sigma_{k=1}^{n-1} k \frac{{n \choose k} (n-k-1)!}{n!} + n\frac{1}{n!} = \Sigma_{k=1}^{n-1} \frac{1}{(k-1)!(n-k)} + \frac{1}{(n-1)!}$$
Is it the right way of dealing with this?
Your question is ambiguous. I am assuming that the postman randomly permutes the packages (so every address gets a package).
Use Linearity of Expectation. Let $X_i$ be the indicator variable for the $i^{th}$ parcel (hence $X_i=1$ if the $i^{th}$ parcel is delivered to the correct address and $X_i=0$ otherwise). Then $E[X_i]=\frac 1n$, clearly. By linearity, the answer you want is then $$E=E\Big [ \sum X_i \Big ]=\sum E[X_i]=n \times \frac 1n=1$$
Note: if, instead, you meant that each parcel is randomly assigned one of the addresses independently of all the other packages (so one address could get all the packages) then the answer, and the computation, are the same.