How to generate all possible X permutations of N numbers with repetition in Python

1 Answer

0 votes
from itertools import product
 
lst_N_numbers = [3, 0, 8, 1, 7] 
X = 3
 
for numbers_set in product(lst_N_numbers, repeat = X):
    print(numbers_set)
   
  
    
        
'''
run:
        
(3, 3, 3)
(3, 3, 0)
(3, 3, 8)
(3, 3, 1)
(3, 3, 7)
(3, 0, 3)
(3, 0, 0)
(3, 0, 8)
(3, 0, 1)
(3, 0, 7)
(3, 8, 3)
(3, 8, 0)
(3, 8, 8)
(3, 8, 1)
(3, 8, 7)
(3, 1, 3)
(3, 1, 0)
(3, 1, 8)
(3, 1, 1)
(3, 1, 7)
(3, 7, 3)
(3, 7, 0)
(3, 7, 8)
(3, 7, 1)
(3, 7, 7)
(0, 3, 3)
(0, 3, 0)
(0, 3, 8)
(0, 3, 1)
(0, 3, 7)
(0, 0, 3)
(0, 0, 0)
(0, 0, 8)
(0, 0, 1)
(0, 0, 7)
(0, 8, 3)
(0, 8, 0)
(0, 8, 8)
(0, 8, 1)
(0, 8, 7)
(0, 1, 3)
(0, 1, 0)
(0, 1, 8)
(0, 1, 1)
(0, 1, 7)
(0, 7, 3)
(0, 7, 0)
(0, 7, 8)
(0, 7, 1)
(0, 7, 7)
(8, 3, 3)
(8, 3, 0)
(8, 3, 8)
(8, 3, 1)
(8, 3, 7)
(8, 0, 3)
(8, 0, 0)
(8, 0, 8)
(8, 0, 1)
(8, 0, 7)
(8, 8, 3)
(8, 8, 0)
(8, 8, 8)
(8, 8, 1)
(8, 8, 7)
(8, 1, 3)
(8, 1, 0)
(8, 1, 8)
(8, 1, 1)
(8, 1, 7)
(8, 7, 3)
(8, 7, 0)
(8, 7, 8)
(8, 7, 1)
(8, 7, 7)
(1, 3, 3)
(1, 3, 0)
(1, 3, 8)
(1, 3, 1)
(1, 3, 7)
(1, 0, 3)
(1, 0, 0)
(1, 0, 8)
(1, 0, 1)
(1, 0, 7)
(1, 8, 3)
(1, 8, 0)
(1, 8, 8)
(1, 8, 1)
(1, 8, 7)
(1, 1, 3)
(1, 1, 0)
(1, 1, 8)
(1, 1, 1)
(1, 1, 7)
(1, 7, 3)
(1, 7, 0)
(1, 7, 8)
(1, 7, 1)
(1, 7, 7)
(7, 3, 3)
(7, 3, 0)
(7, 3, 8)
(7, 3, 1)
(7, 3, 7)
(7, 0, 3)
(7, 0, 0)
(7, 0, 8)
(7, 0, 1)
(7, 0, 7)
(7, 8, 3)
(7, 8, 0)
(7, 8, 8)
(7, 8, 1)
(7, 8, 7)
(7, 1, 3)
(7, 1, 0)
(7, 1, 8)
(7, 1, 1)
(7, 1, 7)
(7, 7, 3)
(7, 7, 0)
(7, 7, 8)
(7, 7, 1)
(7, 7, 7)
  
'''

 



answered Oct 2, 2021 by avibootz
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