Given $n$ unique items and an $m^{th}$ normalised value, compute $m^{th}$ permutation without factorial expansion We know that the number of permutations possible for $n$ unique items is $n!$.
We can uniquely label each permutation with a number from $0$ to $(n!-1)$.
Suppose if $n=4$, the possible permutations with their labels are,
0:  1234
1:  1243
2:  1324
3:  1342
4:  1432
5:  1423
6:  2134
7:  2143
8:  2314
9:  2341
10: 2431
11: 2413
12: 3214
13: 3241
14: 3124
15: 3142
16: 3412
17: 3421
18: 4231
19: 4213
20: 4321
21: 4312
22: 4132
23: 4123

With any well defined labelling scheme, given a number $m, 0 \leq m &lt; n!$, we can get back the permutation sequence. Further, these labels can be normalised to be between $0$ and $1$. The above labels can be transformed into,
0:       1234
0.0434:  1243
0.0869:  1324
0.1304:  1342
0.1739:  1432
0.2173:  1423
0.2608:  2134
0.3043:  2143
0.3478:  2314
0.3913:  2341
0.4347:  2431
0.4782:  2413
0.5217:  3214
0.5652:  3241
0.6086:  3124
0.6521:  3142
0.6956:  3412
0.7391:  3421
0.7826:  4231
0.8260:  4213
0.8695:  4321
0.9130:  4312
0.9565:  4132
1:       4123

Now, given $n$ and $m^{th}$ normalised label, can we get the $m^{th}$ permutation while avoiding the expansion of $n!$ ?
For example, in the above set of permutations, if we were given the $m^{th}$ normalised label to be $0.9$, is it possible to get the closest sequence 4312 as the answer without computing $4!$ ?
Math SE crosspost
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.