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 $m, 0 \leq m < n!$, 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 $4!$ ?
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