itertools.permutations generates where its elements are treated as unique based on their position, not on their value. So basically I want to avoid duplicates like this:
I came up with a very suitable implementation using itertools.product in this case (this is an implementation where you want all combinations
unique_perm_list = [''.join(p) for p in itertools.product(['0', '1'], repeat = X) if ''.join(p).count() == somenumber]
this is essentially a combination (n over k) with n = X and somenumber = k itertools.product() iterates from k = 0 to k = X subsequent filtering with count ensures that just the permutations with the right number of ones are cast into a list. you can easily see that it works when you calculate n over k and compare it to the len(unique_perm_list)