Code-Golf: Permutations

Python, 52

Input is a set. Output is a list of lists.

f=lambda a:[p+[x]for x in a for p in f(a-{x})]or[[]]

This is shorter than the answer that does all the work with a builtin.

Longer than gnibbler's, but implements things from scratch.

p=lambda x:x and[[a]+b for a in x for b in p([c for c in x if c!=a])]or[[]]

(i.@!@#A.[)

Usage:

   (i.@!@#A.[) 1 3 5
1 3 5
1 5 3
3 1 5
3 5 1
5 1 3
5 3 1

Explanation:

i.@!@# uses three verbs to return a list from 0 to (!n)-1 where n is the number of items in the given list.

[ returns the list itself. In the example shown that gives 0 1 2 3 4 5 A. 1 3 5.

A. returns one possible permutation of the second list for each item in the first list (kind of - the proper explanation is given here).