How to find the inverse of the method to find the nth lexicographic logical permutation
$begingroup$
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 1x9! 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?$
combinatorics permutations puzzle inverse-function
asked Jan 19 at 14:41
RicharCdRicharCd
54
$endgroup$
add a comment |
$begingroup$
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 1x9! 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?$
combinatorics permutations puzzle inverse-function
asked Jan 19 at 14:41
RicharCdRicharCd
54
$endgroup$
add a comment |
$begingroup$
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 1x9! 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?$
combinatorics permutations puzzle inverse-function
asked Jan 19 at 14:41
RicharCdRicharCd
54
$endgroup$
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 1x9! 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?$
combinatorics permutations puzzle inverse-function
combinatorics permutations puzzle inverse-function
asked Jan 19 at 14:41
RicharCdRicharCd
54
asked Jan 19 at 14:41
RicharCdRicharCd
54
asked Jan 19 at 14:41
RicharCdRicharCd
54
asked Jan 19 at 14:41
RicharCdRicharCd
54
asked Jan 19 at 14:41
RicharCdRicharCd
54
54
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
1469037825
1234567890
^----------> 0 * 9!
234567890
^--------> 2 * 8!
23567890
^-------> 3 * 7!
2357890
^-----> 5 * 6!
235780
^-----> 5 * 5!
23578
^---------> 1 * 4!
2578
^--------> 2 * 3!
258
^--------> 2 * 2!
25
^----------> 0 * 1!
$$0cdot9!+2cdot8!+3cdot7!+5cdot6!+5cdot5!+1cdot4!+2cdot3!+2cdot2!+0cdot1!+1=100001$$
This shows that $1469037825$ is the $100,001^text{st}$ lexicographic permutation of $1234567890$
1234567890
1469037825
^----------> 0 * 9!
469037825
^---> 7 * 8!
46903785
^------> 4 * 7!
4690785
^----------> 0 * 6!
690785
^-----> 5 * 5!
69078
^----------> 0 * 4!
9078
^--------> 2 * 3!
908
^--------> 2 * 2!
90
^----------> 0 * 1!
$$0cdot9!+7cdot8!+4cdot7!+0cdot6!+5cdot5!+0cdot4!+2cdot3!+2cdot2!+0cdot1!+1=303017$$
This shows that $1234567890$ is the $303,017^text{th}$ lexicographic permutation of $1469037825$
answered Jan 19 at 15:39
Daniel MathiasDaniel Mathias
1,15118
$endgroup$
$begingroup$
Thanks Daniel! This method works perfectly and has a very clean working.
$endgroup$
– RicharCd
Jan 21 at 14:03
add a comment |
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1 Answer
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1 Answer
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oldest
votes
$begingroup$
1469037825
1234567890
^----------> 0 * 9!
234567890
^--------> 2 * 8!
23567890
^-------> 3 * 7!
2357890
^-----> 5 * 6!
235780
^-----> 5 * 5!
23578
^---------> 1 * 4!
2578
^--------> 2 * 3!
258
^--------> 2 * 2!
25
^----------> 0 * 1!
$$0cdot9!+2cdot8!+3cdot7!+5cdot6!+5cdot5!+1cdot4!+2cdot3!+2cdot2!+0cdot1!+1=100001$$
This shows that $1469037825$ is the $100,001^text{st}$ lexicographic permutation of $1234567890$
1234567890
1469037825
^----------> 0 * 9!
469037825
^---> 7 * 8!
46903785
^------> 4 * 7!
4690785
^----------> 0 * 6!
690785
^-----> 5 * 5!
69078
^----------> 0 * 4!
9078
^--------> 2 * 3!
908
^--------> 2 * 2!
90
^----------> 0 * 1!
$$0cdot9!+7cdot8!+4cdot7!+0cdot6!+5cdot5!+0cdot4!+2cdot3!+2cdot2!+0cdot1!+1=303017$$
This shows that $1234567890$ is the $303,017^text{th}$ lexicographic permutation of $1469037825$
answered Jan 19 at 15:39
Daniel MathiasDaniel Mathias
1,15118
$endgroup$
$begingroup$
Thanks Daniel! This method works perfectly and has a very clean working.
$endgroup$
– RicharCd
Jan 21 at 14:03
add a comment |
$begingroup$
1469037825
1234567890
^----------> 0 * 9!
234567890
^--------> 2 * 8!
23567890
^-------> 3 * 7!
2357890
^-----> 5 * 6!
235780
^-----> 5 * 5!
23578
^---------> 1 * 4!
2578
^--------> 2 * 3!
258
^--------> 2 * 2!
25
^----------> 0 * 1!
$$0cdot9!+2cdot8!+3cdot7!+5cdot6!+5cdot5!+1cdot4!+2cdot3!+2cdot2!+0cdot1!+1=100001$$
This shows that $1469037825$ is the $100,001^text{st}$ lexicographic permutation of $1234567890$
1234567890
1469037825
^----------> 0 * 9!
469037825
^---> 7 * 8!
46903785
^------> 4 * 7!
4690785
^----------> 0 * 6!
690785
^-----> 5 * 5!
69078
^----------> 0 * 4!
9078
^--------> 2 * 3!
908
^--------> 2 * 2!
90
^----------> 0 * 1!
$$0cdot9!+7cdot8!+4cdot7!+0cdot6!+5cdot5!+0cdot4!+2cdot3!+2cdot2!+0cdot1!+1=303017$$
This shows that $1234567890$ is the $303,017^text{th}$ lexicographic permutation of $1469037825$
answered Jan 19 at 15:39
Daniel MathiasDaniel Mathias
1,15118
$endgroup$
$begingroup$
Thanks Daniel! This method works perfectly and has a very clean working.
$endgroup$
– RicharCd
Jan 21 at 14:03
add a comment |
$begingroup$
1469037825
1234567890
^----------> 0 * 9!
234567890
^--------> 2 * 8!
23567890
^-------> 3 * 7!
2357890
^-----> 5 * 6!
235780
^-----> 5 * 5!
23578
^---------> 1 * 4!
2578
^--------> 2 * 3!
258
^--------> 2 * 2!
25
^----------> 0 * 1!
$$0cdot9!+2cdot8!+3cdot7!+5cdot6!+5cdot5!+1cdot4!+2cdot3!+2cdot2!+0cdot1!+1=100001$$
This shows that $1469037825$ is the $100,001^text{st}$ lexicographic permutation of $1234567890$
1234567890
1469037825
^----------> 0 * 9!
469037825
^---> 7 * 8!
46903785
^------> 4 * 7!
4690785
^----------> 0 * 6!
690785
^-----> 5 * 5!
69078
^----------> 0 * 4!
9078
^--------> 2 * 3!
908
^--------> 2 * 2!
90
^----------> 0 * 1!
$$0cdot9!+7cdot8!+4cdot7!+0cdot6!+5cdot5!+0cdot4!+2cdot3!+2cdot2!+0cdot1!+1=303017$$
This shows that $1234567890$ is the $303,017^text{th}$ lexicographic permutation of $1469037825$
answered Jan 19 at 15:39
Daniel MathiasDaniel Mathias
1,15118
$endgroup$
1469037825
1234567890
^----------> 0 * 9!
234567890
^--------> 2 * 8!
23567890
^-------> 3 * 7!
2357890
^-----> 5 * 6!
235780
^-----> 5 * 5!
23578
^---------> 1 * 4!
2578
^--------> 2 * 3!
258
^--------> 2 * 2!
25
^----------> 0 * 1!
$$0cdot9!+2cdot8!+3cdot7!+5cdot6!+5cdot5!+1cdot4!+2cdot3!+2cdot2!+0cdot1!+1=100001$$
This shows that $1469037825$ is the $100,001^text{st}$ lexicographic permutation of $1234567890$
1234567890
1469037825
^----------> 0 * 9!
469037825
^---> 7 * 8!
46903785
^------> 4 * 7!
4690785
^----------> 0 * 6!
690785
^-----> 5 * 5!
69078
^----------> 0 * 4!
9078
^--------> 2 * 3!
908
^--------> 2 * 2!
90
^----------> 0 * 1!
$$0cdot9!+7cdot8!+4cdot7!+0cdot6!+5cdot5!+0cdot4!+2cdot3!+2cdot2!+0cdot1!+1=303017$$
This shows that $1234567890$ is the $303,017^text{th}$ lexicographic permutation of $1469037825$
answered Jan 19 at 15:39
Daniel MathiasDaniel Mathias
1,15118
answered Jan 19 at 15:39
Daniel MathiasDaniel Mathias
1,15118
answered Jan 19 at 15:39
Daniel MathiasDaniel Mathias
1,15118
answered Jan 19 at 15:39
Daniel MathiasDaniel Mathias
1,15118
1,15118
$begingroup$
Thanks Daniel! This method works perfectly and has a very clean working.
$endgroup$
– RicharCd
Jan 21 at 14:03
add a comment |
$begingroup$
Thanks Daniel! This method works perfectly and has a very clean working.
$endgroup$
– RicharCd
Jan 21 at 14:03
$begingroup$
Thanks Daniel! This method works perfectly and has a very clean working.
$endgroup$
– RicharCd
Jan 21 at 14:03
$begingroup$
Thanks Daniel! This method works perfectly and has a very clean working.
$endgroup$
– RicharCd
Jan 21 at 14:03
add a comment |
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