diff --git a/arai60/permutations/phase1.py b/arai60/permutations/phase1.py
new file mode 100644
index 0000000..5403388
--- /dev/null
+++ b/arai60/permutations/phase1.py
@@ -0,0 +1,14 @@
+class Solution:
+    def permute(self, nums: List[int]) -> List[List[int]]:
+        if len(nums) == 0:
+            return [[]]
+        
+        adding_number = nums.pop()
+        added_permutations = self.permute(nums) # permutations without adding_number
+        permutations: List[List[int]] = [] # all the pattern of the permutations of nums
+        for added_permutation in added_permutations:
+            for i in range(len(added_permutation)+1):
+                permutations.append(added_permutation[:i] + [adding_number] + added_permutation[i:]) 
+                # insert adding_number as i-th number of added_permutations
+        
+        return permutations
\ No newline at end of file
diff --git a/arai60/permutations/phase2.py b/arai60/permutations/phase2.py
new file mode 100644
index 0000000..93df19f
--- /dev/null
+++ b/arai60/permutations/phase2.py
@@ -0,0 +1,53 @@
+"""
+Reference
+shining-aiさん: https://github.com/shining-ai/leetcode/pull/50/files itertoolsでの解答と再帰によるbacktrackを見ました。再帰でのbacktrackを練習してみます。
+hayashi-ayさん: https://github.com/hayashi-ay/leetcode/pull/57/files 
+
+今回は全ての数が等しい前提のもとでコードを書いたが, next_permutationを用いた方が重複がある場合にも対応できるのではないかと思った。
+Phase1で書いたものの時間計算量はO(Σ_{0->n}k!)という悲惨なものに, これは書いてはいけない。
+要素数10で比較してみた結果
+itertoolsを使った場合: 0.5979518890380859
+phase1解法: 2.563912868499756 
+再帰でbacktrackをする方法: 0.0003998279571533203 
+とにかく似たような書いたように見えていても本当にまずいことをしていることがわかった。再帰を書くときはnaiveなフィボナッチ実装のように再帰が積み上がってないかを確認
+"""
+
+class Solution:
+    def permute(self, nums: List[int]) -> List[List[int]]:
+        return list(itertools.permutations(nums, len(nums)))
+
+# phase1で書いたもの, 時間計算量が非常に高い, この回答はまずい気がする
+class Solution:
+    def permute(self, nums: List[int]) -> List[List[int]]:
+        if len(nums) == 0:
+            return [[]]
+        
+        adding_number = nums.pop()
+        added_permutations = self.permute(nums) # all the permutations of nums without adding_number
+        permutations: List[List[int]] = [] # all the permutations of nums
+        for added_permutation in added_permutations:
+            for i in range(len(added_permutation)+1):
+                permutations.append(added_permutation[:i] + [adding_number] + added_permutation[i:]) 
+                # insert adding_number as i-th number of added_permutations
+        
+        return permutations
+
+# 再帰でbacktrack
+# イメージ的には, ある要素を加える, その要素を加える前の状態に戻るを繰り返す感じだろうか
+
+class Solution:
+    def permute(self, nums: List[int]) -> List[List[int]]:
+        def make_permutations(current):
+            if len(current) == len(nums):
+                permutations.append(current[:])
+                return
+            for num in nums:
+                if num in current:
+                    continue
+                current.append(num)
+                make_permutations(current)
+                current.pop()
+
+        permutations = []
+        make_permutations([])
+        return permutations
diff --git a/arai60/permutations/phase3.py b/arai60/permutations/phase3.py
new file mode 100644
index 0000000..b661309
--- /dev/null
+++ b/arai60/permutations/phase3.py
@@ -0,0 +1,16 @@
+class Solution:
+    def permute(self, nums: List[int]) -> List[List[int]]:
+        all_permutations: List[List[int]] = []
+        def make_permutations(candidates: List[int]):
+            if len(candidates) == len(nums):
+                all_permutations.append(candidates.copy())
+                return
+            for num in nums:
+                if num in candidates:
+                    continue
+                candidates.append(num)
+                make_permutations(candidates)
+                candidates.pop() # backtrack
+        
+        make_permutations([])
+        return all_permutations
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