# Generate all permutation of a set in Python

• Difficulty Level : Medium
• Last Updated : 21 Jan, 2022

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}.

Recommended Practice

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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