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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