What is the limiting distribution of a random bijection from $\{1,2,\ldots,n\}$ to itself as $n \to \infty$? 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.
 A: 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 $\lambda = 1$.
