I'm looking for a proper method to find the nth logical permutation of a particular sequence that is able to return the desired permutation. For example, for a sequence 1234567890, the 100,001st logical permutation is 1469037825, as calculated by:
To find the first digit,
Since there are 10 digits, fixing the first digit, there are 9! ways to permute the rest.
$\ 100 001 – 0(9!) = 100 001$ (As 1×9! is larger than 100001)
Hence the first digit is the 0th element in the list of digits 1234567890, i.e. $\ 1$
To find the second digit
$\ 100 001 – 2(8!) = 19361$
Hence the second digit is the 2nd element in the list of digits 234567890, i.e. $\ 4$ (counting from 0)
To find the third digit
$\ 19361 – 3(7!) = 4241$
Hence the third digit is the 3rd element in the list of digits 23567890, i.e. $\ 6$ (counting from 0)
And so forth
$\ 4241 – 5(6!) = 641 $ 5th in 2357890 is $\ 9$
$\ 641- 5(5!) = 41$ 5th in 235780 is $\ 0$
$\ 41 – 1(4!) = 17 $ 1st in 23578 is $\ 3$
$\ 17 – 2(3!) = 5$ 2nd in 2578 is $\ 7$
$\ 5 – 2(2!) = 1 $ 2nd in 258 is $\ 8$
$\ 1 – 0(1!) = 1 $ 0th in 25 is $\ 2$ (Remainder has to be more than 0)
Hence last digit remaining is $\ 5$
The 100,001st permutation of 1234567890 is 1469037825
How would I then be able to find the value of n, such than the nth permutation of $\ 1469037825$ is $\ 1234567890?$
Best Answer
$$0\cdot9!+2\cdot8!+3\cdot7!+5\cdot6!+5\cdot5!+1\cdot4!+2\cdot3!+2\cdot2!+0\cdot1!+1=100001$$ This shows that $1469037825$ is the $100,001^\text{st}$ lexicographic permutation of $1234567890$
$$0\cdot9!+7\cdot8!+4\cdot7!+0\cdot6!+5\cdot5!+0\cdot4!+2\cdot3!+2\cdot2!+0\cdot1!+1=303017$$ This shows that $1234567890$ is the $303,017^\text{th}$ lexicographic permutation of $1469037825$