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+## Step 1. Initial Solution
+
+- 一つずつ残っているリストから数字を取り出して末尾に入れる
+- 作成済みの順列と残りのリストをペアで保持して処理していく
+- なぜかエラー→next_remaining_nums=remaining_numsにすると参照先をremoveしてしまう
+    - copy.deepcopyで対応
+
+```python
+class Solution:
+    def permute(self, nums: List[int]) -> List[List[int]]:
+        result = []
+        prefix_to_remaining = [([], nums)]
+        while prefix_to_remaining:
+            prefix, remaining_nums = prefix_to_remaining.pop()
+            for num in remaining_nums:
+                next_prefix = prefix + [num]
+                next_remaining_nums = remaining_nums.copy()
+                next_remaining_nums.remove(num)
+                if not next_remaining_nums:
+                    result.append(next_prefix)
+                else:
+                    prefix_to_remaining.append((next_prefix, next_remaining_nums))
+        return result
+```
+
+### Complexity Analysis
+
+- 時間計算量：O(N*N!)
+- 空間計算量：O(N*N!)
+
+## Step 2. Alternatives
+
+- 再帰関数で先頭とそれ以外の並び替えを回す方法が多い
+    - https://github.com/tokuhirat/LeetCode/pull/50/files#diff-f980f45b4d4f94c41556977f060a371872f422650416425386f1322e52deba58R2
+        - ここは綺麗に書けていて分かりやすいと感じた
+            - https://github.com/tokuhirat/LeetCode/pull/50/files#diff-f980f45b4d4f94c41556977f060a371872f422650416425386f1322e52deba58R47-R58
+        - まだ入っていないものを追加していく方法
+            - https://github.com/tokuhirat/LeetCode/pull/50/files#r2278840164
+            - 時間計算量の話は1≤num.lengths≤6では線形の方が速そうというところ
+    - https://github.com/docto-rin/leetcode/pull/51/files#diff-47cd87d2eeba6169d55b484ea322c16f1a03278b0d29a5b0d00f771aaf816f8bR74-R78
+        - Backtrackingとも呼ぶらしい？がほぼ同じに見える
+- 再帰で書いてみる
+
+```python
+class Solution:
+    def permute(self, nums: List[int]) -> List[List[int]]:
+        if len(nums) == 1:
+            return [nums]
+
+        permutations = []
+        for i in range(len(nums)):
+            prefix = [nums[i]]
+            sub_permutations = self.permute(nums[:i]+nums[i+1:])
+            for sub_permutation in sub_permutations:
+                permutations.append(prefix+sub_permutation)
+        return permutations
+```
+
+- 既出を管理する方法の方がスライスしない分速いか
+
+```python
+class Solution:
+    def permute(self, nums: List[int]) -> List[List[int]]:
+        def permute_helper(prefix: List[int]) -> List[List[int]]:
+            if len(prefix) == len(nums):
+                return [prefix]
+            permutations = []
+            for num in nums:
+                if num in prefix:
+                    continue
+                permutations += permute_helper(prefix + [num])
+            return permutations
+
+        return permute_helper([])
+```
+
+## Step 3. Final Solution
+
+- ヘルパー関数を使う方法に帰着
+
+```python
+class Solution:
+    def permute(self, nums: List[int]) -> List[List[int]]:
+        def permute_helper(permuted_nums: List[int]) -> List[List[int]]:
+            permutations = []
+            for num in nums:
+                if num in permuted_nums:
+                    continue
+                permutations += permute_helper(permuted_nums + [num])
+            return permutations if permutations else [permuted_nums]
+        return permute_helper([])
+```