There are $n$ people. Each one distributes one file and one photo into one of the $n$ envelopes independently and randomly. The files and the photos are unique. Every envelope has a name on it, which corresponds to one of the $n$ people. Moreover, every envelope has and only has one file and one photo which are not necessarily from the same person.
1) Calculate $P_0(n)$ the probability that none of the envelopes has neither the right file nor the right photo.
2) Calculate $P_r(n)$ the probability that there are $r$ envelopes have both the right file the right photo.
3) Prove :
$$\lim_{n \rightarrow \infty} P_0(n) = 1,$$
$$\lim_{n \rightarrow \infty} P_r(n) = 0, (1\le r \le n)$$
My answer is:
The cardinal of full derangement of the files is: $D = n!\left(\frac{(-1)^2}{2!}+ \ldots+\frac{(-1)^n}{n!}\right)$. Thus, its probability is $\left(\frac{(-1)^2}{2!}+ \ldots+\frac{(-1)^n}{n!}\right)$. Regarding to the independence between the files and the photos, the joint full derangement probability is: $$P_0(n) =\left(\frac{(-1)^2}{2!}+ \ldots+\frac{(-1)^n}{n!}\right)^2$$ which leads to $$\lim_{n \rightarrow \infty} P_0(n) = \frac{1}{e^2},$$
and
$$P_r(n) = \left(\frac{1}{r!}-\frac{1}{(r+1)!} +\ldots+\frac{(-1)^{n-r}}{n!}\right)^2 \frac{1}{C_n^r}$$
Apparently, I did not do it right. Please help me to correct it.
Define: $$ \exp\rvert_n (z) = \sum_{0 \le k \le n} \frac{z^k}{k!} $$ Then the number of derangements of $n$ elements is $D_n = n! \exp\rvert_n(-1)$, and the number of permutations where $r$ elements stay fixed is just the number of derangements of $n -r$ elements times the number of ways of selecting $r$ elements to stay fixed, i.e., $\binom{n}{r} D_{n - r}$. Also note: $$ \lim_{n \to \infty} \exp\rvert_n(z) = \mathrm{e}^z $$ and the approximation $\exp\rvert_n(-1) \approx \mathrm{e}^{-1}$ is very good even for small $n$.
The number of ways of getting both all files and all pictures wrong is $D_n^2$, from among $(n!)^2$ distributions in total: \begin{align} P_0(n) &= \frac{D_n^2}{(n!)^2} \\ &= (\exp\rvert_n (-1))^2 \\ \lim_{n \to \infty} P_0(n) &= \mathrm{e}^{-2} \end{align}
Having $r$ pictures right, and the corresponding files right, means selecting the $r$ to be right, and deranging the other pictures and files: \begin{align} P_r(n) &= \frac{\binom{n}{r} D_{n - r}^2}{(n!)^2} \\ &= \frac{n!}{(n - r)! r!} \cdot \frac{((n - r)!)^2}{(n!)^2} \exp\rvert_{n - r}(-1) \\ &= \frac{(n - r)!}{ n! r!} (\exp\rvert_{n - r}(-1))^2 \\ \lim_{n \to \infty} P_r(n) &= 0 \end{align}