Consider the $[n]!_q = \prod\limits_{k = 1}^{n} \frac{q^k - 1}{q - 1} = \sum\limits_{k = 0}^{\binom n 2} c_k q^k$ and let $\{f_n\}_{n \in \mathbb{N}}$ be the sequence of the functions on $[0; 1]$ defined by the following $$f_n(x) = \frac{c_{\lfloor \binom n 2 x \rfloor}}{n!}$$ Is there a formula for $\lim\limits_{n \rightarrow \infty} f_n(x)$? Roughly speaking, what is the limit distribution of its coefficients?

## 2 Answers

Let $Z_n$ be the number of inversions of a random permutation in $S_n$. Then for all $x\in\mathbb{R}$, $$ \mathrm{Prob}\left(Z_n<\frac 14 n^2+\frac 16xn^{3/2}\right)\to \mathcal{N}(x), $$ the standard normal distribution. This goes back to Feller, 1945. See for instance Theorem 3.3.4 of https://www.routledgehandbooks.com/doi/10.1201/b18255-6.

Choose $\xi_i\in \{0,1,\dots,i-1\} $ uniformly at random. Then $c_k/n! $ is a probability that $\sum \xi_i=k$. The law of large numbers and central limit theorem work nicely for the distribution of $\sum \xi_i$. The limits you are talking about exist and are equal to zero.