Recursion on lists : Append to a list vs passing subset

I am trying to solve the leetcode 46 problem (https://leetcode.com/problems/permutations/) and I came up with 2 solutions and both appear identical to me, however one of them just returns empty output.

Incorrect Solution

class Solution:
def permute(self, nums: List[int]) -> List[List[int]]:
    self.ans = list()
    self.final_length = len(nums)
    available_items = nums.copy()
    current_items = []
    self.solve(available_items,current_items)
    return self.ans
    
def solve(self,available_items,current_items ):
    if len(current_items) == self.final_length:
        self.ans.append(current_items)
    for i in range(len(available_items)):
        current_items.append(available_items.pop(i))
        self.solve(available_items,current_items)
        available_items.insert(i,current_items.pop())

For input [1,2,3] I get [[],[],[],[],[],[],[],[]]

Correct Solution

class Solution:
def permute(self, nums: List[int]) -> List[List[int]]:
    self.ans = list()
    self.final_length = len(nums)
    available_items = nums.copy()
    current_items = []
    self.solve(available_items,current_items)
    return self.ans
    
def solve(self,available_items,current_items ):
    if len(current_items) == self.final_length:
        self.ans.append(current_items)
    for i in range(len(available_items)):
        self.solve(available_items[:i]+available_items[i+1:],current_items+[available_items[I]])

For input [1,2,3] I get [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]

I do not understand what's the difference here in terms of storing answers. Am I missing some basic concept here in Python?