Given $n$ unique items and an $m^{th}$ normalised value, compute $m^{th}$ permutation without factorial expansion
Well, from the clarifications made in the comments, then this could be a solution.
I haven't checked for correctness. Specifically the requirement of choosing the closest label to the one input. I think I am taking the one immediately below the number input. Probably that can be fixed by using closest integer, instead of floor below.
Let $L=L_1$ be the input label, which is a real number. Maybe we can just assume that it is in $[0,1]$.
We begin with the full collection of numbers say $C_1=\{1,2,3,...,n\}$ from where one will be eliminated at each step.
In the first step we compute $a_1=\lfloor L_1\cdot n\rfloor$. We pick the $a_1$-th element $p_1$ of $C_1$ to be the first element of the permutation and remove it form $C_1$. So, define $C_2=C_1\setminus\{p_1\}$. Define $L_2=L_1\cdot n-a_1$.
Continue in this way, in the $k$-th step, for $k=2,3,...,n$, put $a_k=\lfloor L_k\cdot (n-k+1)\rfloor$, take $p_k$ to be the $a_k$-th element of $C_k$ and define $C_{k+1}=C_k\setminus\{p_k\}$ and $L_{k+1}=L_k\cdot(n-k+1)-a_k$.
Finish when there are no more elements in $C_k$, or maybe when there is just one which will be forced to be the last element in the permutation.

See also factorial number system. Since this is essentially what we are doing.