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