Let $A_n$ be the number of fixed points of a random bijection form $\{1,2,\ldots,n\}$ to itself. What is the distribution of $A_n$ as $n \to \infty$? What is the expected value?
Here is my idea:
I firstly calculated pmf of $A_n$ $$P(A_n = k) = {n \choose k} \frac{1}{k!}\left(\frac{1}{2!}-\frac{1}{3!}+\cdots+\frac{1}{n-k}\right)$$
I am not sure if I derive the expression correctly. If not, what is the correct one? If yes, how to proceed and calculate the limit? Thanks in advance.
The expected value is much easier to compute: linearity of expectation gives that it's $n$ times the probability that a particular point is fixed, which is $\frac{1}{n}$. So the expected number of fixed points is $1$, for all $n \ge 1$.
As for the distribution, your calculation is not correct, and for $n$ sufficiently large your "probabilities" will be greater than $1$. The probability that there are $k$ fixed points is ${n \choose k}$ times the probability that some particular subset of size $k$ is fixed (and the rest isn't fixed). This is $\frac{1}{n(n-1) \dots (n-(k-1))}$ times the probability that the rest of the permutation is a derangement (your $\frac{1}{k!}$ here is incorrect), which cancels most of the binomial coefficient, and we get that overall
$$\mathbb{P}(A_n = k) = \frac{1}{k!} \left( \sum_{i=0}^{n-k} \frac{(-1)^i}{i!} \right).$$
As $n \to \infty$ the derangement probability converges rapidly to $e^{-1}$, so we get
$$\lim_{n \to \infty} \mathbb{P}(A_n = k) = e^{-1} \frac{1}{k!}$$
which is the pdf of a Poisson distribution with $\lambda = 1$.