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Output all permutations using 0-9 of of n-size up to 25. [Python] [on hold]

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Solving a problem in which I need to generate all possible permutations using the elements [0-9] of a number of n-size up o 25; where repetition is allowed. I've been using Python with different algorithm types, but they using a list type approach; permutation [A,B]: would result in AB, BA.

The function I'm after would be to permutation all combinations of a n-char/digit length using the element set Z, where Z is [0-9].

I've calculated approx. 11 septillion+ unique numbers would be generated.

n = 1: 10 numbers
n = 2: 100 numbers
n = 3: 1000 numbers
...

permutations combinations

share|cite|improve this question

asked Jun 3 '15 at 16:35

EllxEllx

1

$endgroup$

put on hold as unclear what you're asking by Ross Millikan, Wrzlprmft, Cesareo, Learnmore, uniquesolution 2 days ago

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is wrong with range(pow(10,25))? You won't finish them before the sun goes supernova. Voting to close as not about mathematics
  $endgroup$
  – Ross Millikan
  2 days ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am voting to close this for lack of a question (as in: request for information).
  $endgroup$
  – Wrzlprmft
  2 days ago
  

  
  
  
  

add a comment | 

-1

$begingroup$

Solving a problem in which I need to generate all possible permutations using the elements [0-9] of a number of n-size up o 25; where repetition is allowed. I've been using Python with different algorithm types, but they using a list type approach; permutation [A,B]: would result in AB, BA.

The function I'm after would be to permutation all combinations of a n-char/digit length using the element set Z, where Z is [0-9].

I've calculated approx. 11 septillion+ unique numbers would be generated.

n = 1: 10 numbers
n = 2: 100 numbers
n = 3: 1000 numbers
...

permutations combinations

share|cite|improve this question

asked Jun 3 '15 at 16:35

EllxEllx

1

$endgroup$

put on hold as unclear what you're asking by Ross Millikan, Wrzlprmft, Cesareo, Learnmore, uniquesolution 2 days ago

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is wrong with range(pow(10,25))? You won't finish them before the sun goes supernova. Voting to close as not about mathematics
  $endgroup$
  – Ross Millikan
  2 days ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am voting to close this for lack of a question (as in: request for information).
  $endgroup$
  – Wrzlprmft
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

Solving a problem in which I need to generate all possible permutations using the elements [0-9] of a number of n-size up o 25; where repetition is allowed. I've been using Python with different algorithm types, but they using a list type approach; permutation [A,B]: would result in AB, BA.

The function I'm after would be to permutation all combinations of a n-char/digit length using the element set Z, where Z is [0-9].

I've calculated approx. 11 septillion+ unique numbers would be generated.

n = 1: 10 numbers
n = 2: 100 numbers
n = 3: 1000 numbers
...

permutations combinations

share|cite|improve this question

asked Jun 3 '15 at 16:35

EllxEllx

1

$endgroup$

share|cite|improve this question

asked Jun 3 '15 at 16:35

EllxEllx

1

share|cite|improve this question

asked Jun 3 '15 at 16:35

EllxEllx

1

asked Jun 3 '15 at 16:35

EllxEllx

1

asked Jun 3 '15 at 16:35

EllxEllx

1

1 Answer
1

active

oldest

votes

0

$begingroup$

When you want to look for all the possible permutations with the numbers from $0$ to $9$ when you take $n$ of them and they can repeat, basically what you are doing is taking all numbers from $0$ to $10^{n+1}-1$, and there are $10^{n+1}$ numbers there.

Still though, in case you want the formula for different number of elements (rather than from $0$ to $9$), for permutations with repetition of $k$ different elements when you take $n$, is
$$PR_k^n=k^n$$

share|cite|improve this answer

answered Jun 3 '15 at 19:28

MasclinsMasclins

994721

$endgroup$

add a comment | 

0

$begingroup$

When you want to look for all the possible permutations with the numbers from $0$ to $9$ when you take $n$ of them and they can repeat, basically what you are doing is taking all numbers from $0$ to $10^{n+1}-1$, and there are $10^{n+1}$ numbers there.

Still though, in case you want the formula for different number of elements (rather than from $0$ to $9$), for permutations with repetition of $k$ different elements when you take $n$, is
$$PR_k^n=k^n$$

share|cite|improve this answer

answered Jun 3 '15 at 19:28

MasclinsMasclins

994721

$endgroup$

add a comment | 

0

$begingroup$

When you want to look for all the possible permutations with the numbers from $0$ to $9$ when you take $n$ of them and they can repeat, basically what you are doing is taking all numbers from $0$ to $10^{n+1}-1$, and there are $10^{n+1}$ numbers there.

Still though, in case you want the formula for different number of elements (rather than from $0$ to $9$), for permutations with repetition of $k$ different elements when you take $n$, is
$$PR_k^n=k^n$$

share|cite|improve this answer

answered Jun 3 '15 at 19:28

MasclinsMasclins

994721

$endgroup$

add a comment | 

$begingroup$

When you want to look for all the possible permutations with the numbers from $0$ to $9$ when you take $n$ of them and they can repeat, basically what you are doing is taking all numbers from $0$ to $10^{n+1}-1$, and there are $10^{n+1}$ numbers there.

Still though, in case you want the formula for different number of elements (rather than from $0$ to $9$), for permutations with repetition of $k$ different elements when you take $n$, is
$$PR_k^n=k^n$$

share|cite|improve this answer

answered Jun 3 '15 at 19:28

MasclinsMasclins

994721

$endgroup$

share|cite|improve this answer

answered Jun 3 '15 at 19:28

MasclinsMasclins

994721

answered Jun 3 '15 at 19:28

MasclinsMasclins

994721

answered Jun 3 '15 at 19:28

MasclinsMasclins

994721