use frozenset to do it
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```python3 []\nclass Solution:\n    def similarPairs(self, words: List[str]) -> int:\n        from collections import Counter\n        ans=0\n        set_list=[]\n        for word in words:\n            set_list.append(frozenset(word))\n        counter=Counter(set_list)\n        for key,val in counter.items():\n            if val>1:\n                ans+=val*(val-1)//2\n        return ans\n\n        \n        \n\n        \n```
['Python3']