Permutation a list and Powerset is a list of all subsets OOP python

Question

As a beginner need help, how to implement logic to figure out this problems. I am trying to find Permutation is a list of all arrangements of elements and Powerset is a list of all subsets of elements including the empty set.

from functools import cmp_to_key

class E1_recursion:

    numbers = [9, 4, 8, 10, 2, 4, 8, 3, 14, 4, 8]    #[4, 12, 3, 8, 17, 12, 1, 3, 8, 7]

    def perm(self, _numbers) -> list:
        
        
        return []


    def pset(self, _numbers) -> list:
        

        return []


    run_choices = [
            1,      # Permutation
            2,      # Powerset
        ]

# Ignore this code that loads solution from file, if exists.
# The solution is not distributed.
try:
    _from, _import = 'E1_recursion_sol', 'E1_recursion'
    E1_recursion = getattr(__import__(_from, fromlist=[_import]), _import)
#
except ImportError:
    pass


if __name__ == '__main__':
    
    run_choices = E1_recursion.run_choices
    numbers = [9, 4, 8, 10, 2, 4, 8, 3, 14, 4, 8]
    n1 = E1_recursion(numbers)

    #  Permutation
    if 1 in run_choices:
        lst = [1, 2, 3]
        perm = n1.perm(lst)
        print(f'perm({lst}) -> {perm}')

    #  Powerset
    if 2 in run_choices:
        lst = [1, 2, 3]
        pset = n1.pset(lst)
        print(f'\npset({lst}) -> {pset}')

    lst = n1.numbers

Output context

perm([1, 2, 3]) -> [[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]

pset([1, 2, 3]) -> [[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]]