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Note: I'm working in python on this.
For example, given a list:
list = ['a','b','c','d','e','f','g','h','i','j'] I want to generate a list of lists with all possible 3-item combinations:
['a','b','c'], ['a','b','d'], ['a','b','e'] The permutations should not use the same item twice in a permutation, but the order is important and represents distinct permutations that should be included, e.g.,
['a','b','c'], ['a','c','b'] Should both be included.
"3" is the magic length for the permutations I'm looking to generate, but I wouldn't look down on a solution for arbitrary length permutations.
Thanks for any help!
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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}.
length() : length() is a sympy Python library function that finds the number of integers moved by the permutation.
You take first element of an array (k=0) and exchange it with any element (i) of the array. Then you recursively apply permutation on array starting with second element. This way you get all permutations starting with i-th element.
itertools.permutations(my_list, 3) Assuming you're in python 2.6 or newer:
from itertools import permutations for i in permutations(your_list, 3): print i If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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