# All permutations of range from $1$ to $n$ [duplicate]

Given a positive input $$\n\$$, output all permutations of either $$\\{0,1,\ldots,n-1\}\$$ or $$\\{1,2,\ldots,n\}\$$.

## Examples

Outputting permutations of $$\\{1,2,\ldots,n\}\$$.

Input Output
1 [(1)]
2 [(1, 2), (2, 1)]
4 [(1, 2, 3, 4), (1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (1, 4, 2, 3), (1, 4, 3, 2), (2, 1, 3, 4), (2, 1, 4, 3), (2, 3, 1, 4), (2, 3, 4, 1), (2, 4, 1, 3), (2, 4, 3, 1), (3, 1, 2, 4), (3, 1, 4, 2), (3, 2, 1, 4), (3, 2, 4, 1), (3, 4, 1, 2), (3, 4, 2, 1), (4, 1, 2, 3), (4, 1, 3, 2), (4, 2, 1, 3), (4, 2, 3, 1), (4, 3, 1, 2), (4, 3, 2, 1)]

Standard loopholes are forbidden. The shortest code wins.

• Wouldn't it make more sense for the output to be the permutations of either $\{0, 1, ..., n-1\}$ or $\{1, 2, ..., n\}$? Aug 17, 2022 at 19:05
• Duplicate? Aug 17, 2022 at 19:25

# Jelly, 2 bytes

Œ!


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# Jelly, 4 bytes

!Rœ?


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## How it works

!Rœ? - Main link. Takes n on the left
!    - Yield n!
R   - Range; [1, 2, ..., n!]
œ? - For each 1 ≤ i ≤ n!, get the ith permutation of [1, 2, ..., n]


# Python, 54 bytes

Outputs all permutations of $$\\{0,1,\ldots,n-1\}\$$.

lambda n:permutations(range(n))
from itertools import*


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## Print out -- Python 3, 58 bytes (@dingledooper)

def f(n,*r):[f(n,*r,x)for x in{*range(n)}-{*r}]or print(r)


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## Return list -- Python, 62 bytes

f=lambda n,*p:sum((f(n,*p,i)for i in{*range(n)}-{*p}),[])or[p]

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Ignoring the standard lib's itertools.permutations.

• @solid.py but they conveniently take a set as input which saves them the range and the conversion to set. Aug 17, 2022 at 19:47
• 58 bytes by printing the results Aug 17, 2022 at 21:26