Count anagrams

It is obvious that the answer is the product of the number of unique permutations for each word in a sentence. The last one is just the number of permutations of all letters (treating same letters as distinct, i.e. n!) corrected (divided) by the number of permutations within each group of same letters.

class Solution:
    def countAnagrams(self, s: str) -> int:
        res = 1
        for word in s.split():
            res *= math.factorial(len(word))
            for ch, count in collections.Counter(word).items():
                res //= math.factorial(count)

        return res

Time complexity: O(n)
Space complexity: O(1)