HERE is THE RECURSION APPROACH you all are searching for👾
NOTE: MY FIRST SOlUTION I EVER POSTED!!\n# Intuition\n=> The code leverages factorials to compute the count of anagrams for each word efficiently.\n=> The calc 
NOTE: MY FIRST SOlUTION I EVER POSTED!!\n# Intuition\n=> The code leverages factorials to compute the count of anagrams for each word efficiently.\n=> The calc function recursively computes the product of factorials for character counts.\n=> The outer loop handles each word, and the inner loops calculate factorials and update the result.\n=> Modular arithmetic is used to prevent integer overflow.\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n=>The code splits the input string into words and processes each word independently.\n=>For each word, it calculates the count of each character and computes the product of factorials for the character counts.\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(n * m * k), where n is the total characters, m is the average word length, and k is the total unique characters in all words.\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n + k), where n is the total characters and k is the total unique characters in all words.\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nfrom math import factorial\n\n\nclass Solution:\n    def countAnagrams(self, s: str) -> int:\n        s=s.split(\' \')\n        prod=1\n        for word in s:\n            seen={}\n            for ch in word:\n                seen[ch]= seen.get(ch,0)+1\n            prod*=(factorial(len(word)))\n            for key,val in seen.items():\n                prod//= (factorial(val))\n        return int(prod)% (10**9 +7)\n\'\'\'\n\nclass Solution:\n    def countAnagrams(self, s: str) -> int:\n        \'\'\'\n```
['Python3']