Showing all possible permutations from a set
As noted by 200_success in the comments, the m == len(nlist) check is a bug that can be fixed by simply removing the condition.

Also, if m &gt; len(nlist): return is almost redundant because you would get an empty list as the result anyway in that case. (The deepest levels of recursion give empty results when the list runs out and the emptiness propagates all the way up)

Turning the function into a generator simplifies it and speeds it up at the same time: from 10.9 ms to 6.6 ms for the case perm(range(6), 6). (Of course I wrapped that in list() when timing the generator version)

def perm(nlist, m):
    if m == 0:
        yield []
        return

    for list_i in nlist:
        temp = list(nlist)          
        temp.remove(list_i)
        for p in perm(temp, m-1):
            yield [list_i] + p