Untested solution : We can see that for each $$$i$$$ we can go to some $$$j$$$ or $$$i-1$$$ using the handles. Now let $$$dp_{i,j}$$$ denote a way to position the first $$$i$$$ handles in a way such that you can reach $$$j$$$. Notice that if $$$j$$$ is reachable, all values below $$$j$$$ are also reachable.

This means at least one handle is at the $$$jth$$$ position, and none are at a larger position. Now we can see that $$$dp_{i,j} = dp_{i-1,j}\times j + \sum_{k=0}^{j-1} dp_{i-1,k}$$$.

The first term is for $$$j$$$ being reachable from before and the second term is setting the $$$ith$$$ handle to $$$j$$$ so that $$$j$$$ is now reachable. This can be computed in $$$O(n^2)$$$ by using prefix sums.

The base term is $$$dp_{0,1} = 1$$$ and the rest are $$$0$$$.

:0