Consider a permutation $\sigma$ of $[1..n]$. An inversion is defined as a pair $(i, j)$ of indices such that $i < j$ and $\sigma(i) > \sigma(j)$.Define $A_k$ to be the number of permutations of $[1..n]$ with at most $k$ inversions. Question: What's the tight asymptotic bound for $A_k$?A related question was asked before: Number of permutations which have the same Kendall-Tau distanceBut the question above was regarding computing $A_k$. It can be computed using dynamic programming, since it satisfies the recurrence relation shown here: https://stackoverflow.com/questions/948341/dynamic-programming-number-of-ways-to-get-at-least-n-bubble-sort-swapsThe number of permutations with exactly $k$ inversions has also been studied and it can be expressed as a generating function: http://en.wikipedia.org/wiki/Permutation#InversionsBut I can't find a closed-form formula or an asymptotic bound.
Asymptotically, how many permutations of $[1..n]$ have at most $k$ inversions?
co.combinatorics;permutations
According to Wikipedia, the number of permutations in $S_n$ with exactly $k$ inversions is the coefficient of $X^k$ in$$1(1+X)(1+X+X^2)\cdots(1+X+\cdots+X^{n-1}).$$Denote this by $c(n,k)$. This shows that$$c(n+1,k) = \sum_{l=0}^k c(n,k-l).$$So the number of permutations in $S_n$ with at most $k$ inversions is equal to the number of permutations in $S_{n+1}$ with exactly $k$ inversions. This has a neat combinatorial proof as well (hint: take $\pi\in S_{n+1}$ and remove $n+1$).If we are interested only in the coefficient of $X^k$, then factors $X^m$ for $m > k$ don't make any difference. So for $n > k$, $c(n,k)$ is the coefficient of $X^k$ in$$\begin{align*}&1(1+X)\cdots(1+X+\cdots+X^{k-1}) (1+X+\cdots+X^k+\cdots)^{n-k} \\ =&1(1+X)\cdots(1+X+\cdots+X^{k-1}) \frac{1}{(1-X)^{n-k}} \\ =&1(1+X)\cdots(1+X+\cdots+X^{k-1}) \sum_{t=0}^\infty \binom{t+n-k-1}{t} X^t.\end{align*}$$This implies the formula$$c(n,k) = \sum_{t=0}^k \binom{n+t-k-1}{t} c(k,k-t), \quad n > k.$$When $k$ is constant, the asymptotically most important term is the one corresponding to $t = k$, and we have$$c(n,k) = \binom{n-1}{k} + O_k(n^{k-1}) = \frac{1}{k!} n^k + O_k(n^{k-1}).$$The same asymptotics work for $c(n+1,k)$, which is what you were after. For non-constant $k$, using the fact that $\binom{n+t-k-1}{t} = \binom{n+t-k-1}{n-k-1}$ is increasing in $t$ and $\sum_{t=0}^k c(k,t) \leq k!$, we get the bounds$$\binom{n-1}{k} \leq c(n,k) \leq k! \binom{n-1}{k}.$$Better bounds are surely possible, but I'll leave that to you.