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.

`# 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 permuatation 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.

`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)]

This article is contributed by **Arpit Agarwal**. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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