Why does Python's itertools.permutations contain duplicates? (When the original list has duplicates)

Question

It is universally agreed that a list of n distinct symbols has n! permutations. However, when the symbols are not distinct, the most common convention, in mathemati

Answer 1

Maybe i am wrong but seems that reason for this is in 'Elements are treated as unique based on their position, not on their value. So if the input elements are unique, there will be no repeat values in each permutation.' You have specified (1,1,2) and from your point of view 1 at the 0 index and 1 at the 1 index are the same - but this in not so since permutations python implementation used indexes instead of values.

So if we take a look at the default python permutations implementation we will see that it uses indexes:

def permutations(iterable, r=None):
    pool = tuple(iterable)
    n = len(pool)
    r = n if r is None else r
    for indices in product(range(n), repeat=r):
        if len(set(indices)) == r:
            yield tuple(pool[i] for i in indices)

For example if you change your input to [1,2,3] you will get correct permutations([(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)]) since the values are unique.