how many different ways can 15 be written as the sum of 4 different counting numbers
(1, 0, 2, 12) , (1, 0, 3, 11) , (1, 0, 4, 10) , (1, 0, 5, 9) , (1, 0, 6, 8) , (2, 0, 3, 10) , (2, 0, 4, 9) , (2, 0, 5, 8) , (2, 0, 6, 7) , (3, 0, 4, 8) , (3, 0, 5, 7) , (4, 0, 5, 6) , (1, 2, 3, 9) , (1, 2, 4, 8) , (1, 2, 5, 7) , (1, 3, 4, 7) , (1, 3, 5, 6) , (2, 3, 4, 6) , Total = 18 such counting numbers.
Note: I assumed that order does NOT matter: such as 1, 2, 3, 9 is treated the same as 9, 3, 2, 1. if you wish to calculate ALL the permutations possible, then each group of 4 has 4! ==24 permutations, including some that will begin with zero "0".
