SOLUTION: Given the equation x + y + z = 15, how many different solutions are possible (i) If x, y, and z are positive integers? (ii) If x, y, and z are non-negative integers?

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Question 1183744: Given the equation x + y + z = 15, how many different solutions are possible
(i) If x, y, and z are positive integers?
(ii) If x, y, and z are non-negative integers?
Answer by ikleyn(52798) About Me  (Show Source):
You can put this solution on YOUR website!
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Given the equation x + y + z = 15, how many different solutions are possible
(a) If x, y, and z are positive integers?
(b) If x, y, and z are non-negative integers?
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This problem is on "Stars and bars method".


See this Wikipedia article
https://en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)

or my lesson

Stars and bars method for Combinatorics problems

https://www.algebra.com/algebra/homework/Permutations/Stars-and-bars-method-for-Combinatorics-problems-2.lesson

in this site. 


(a)  The number of different solutions is  C%5B15-1%5D%5E%283-1%29    = C%5B14%5D%5E2 = %2814%2A13%29%2F2 = 7*13 = 91.     ANSWER


(b)  The number of different solutions is  C%5B15%2B3-1%5D%5E%283-1%29 = C%5B17%5D%5E2 = %2817%2A16%29%2F2 = 17*8 = 136.    ANSWER