Generate all permutation of a set in Python

Permutation is an arrangement of objects in a specific order. Order of arrangement of object is very important. The number of permutations on a set of n elements is given by  n!.  For example, there are 2! = 2*1 = 2 permutations of {1, 2}, namely {1, 2} and {2, 1}, and 3! = 3*2*1 = 6 permutations of {1, 2, 3}, namely {1, 2, 3}, {1, 3, 2}, {2, 1, 3}, {2, 3, 1}, {3, 1, 2} and {3, 2, 1}.

Method 1 (Backtracking) 
We can use the backtracking based recursive solution discussed here.
Method 2 
The idea is to one by one extract all elements, place them at first position and recur for remaining list.

Python3

# Python function to print permutations of a given list
def permutation(lst):

    # If lst is empty then there are no permutations
    if len(lst) == 0:
        return []

    # If there is only one element in lst then, only
    # one permutation is possible
    if len(lst) == 1:
        return [lst]

    # Find the permutations for lst if there are
    # more than 1 characters

    l = [] # empty list that will store current permutation

    # Iterate the input(lst) and calculate the permutation
    for i in range(len(lst)):
       m = lst[i]

       # Extract lst[i] or m from the list.  remLst is
       # remaining list
       remLst = lst[:i] + lst[i+1:]

       # Generating all permutations where m is first
       # element
       for p in permutation(remLst):
           l.append([m] + p)
    return l


# Driver program to test above function
data = list('123')
for p in permutation(data):
    print (p)

Output:

['1', '2', '3']
['1', '3', '2']
['2', '1', '3']
['2', '3', '1']
['3', '1', '2']
['3', '2', '1']

Method 3 (Direct Function) 
We can do it by simply using the built-in permutation function in itertools library. It is the shortest technique to find the permutation.

Python3

from itertools import permutations
l = list(permutations(range(1, 4)))
print(l)

Output:

[(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)] 

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