Distribution of the random variable which counts the number of fixed points of functions on $n$ symbols

Question

Let $A_n$ denote the set of all functions from $\{1,...,n\}$ to itself and for $f\in A_n$, let $X_n(f)$ denote the number of fixed points of $f$. Then is it true that $P(X_n=k)={n \choose k}\big(\dfrac 1 n\big)^k \big(1- \dfrac 1 n\big)^{n-k} $ ? i.e. is it true that $X_n $ follows Binomial $(n, \dfrac 1n)$ ?

NOTE: I am equipping $A_n$ with uniform counting measure i.e. for a subset $S$ of $A_n$ the measure of $S$ is $|S|/n^n$ .

Answer 1

Yes, the number of fixed points of $f$ is distributed like $\text{Bin}(n,1/n)$. Since $f$ is uniformly distributed over $A_n$, it follows the values $f(1),f(2),\dots,f(n)$ are independent, so the number of $f(k)$ which are equal to $k$ will follow a binomial distribution.