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!$ ?