Question 1183744
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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 


<A HREF =https://www.algebra.com/algebra/homework/Permutations/Stars-and-bars-method-for-Combinatorics-problems-2.lesson>Stars and bars method for Combinatorics problems</A> 


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


in this site.



<pre>
(a)  The number of different solutions is  {{{C[15-1]^(3-1)}}}    = {{{C[14]^2}}} = {{{(14*13)/2}}} = 7*13 = 91.     <U>ANSWER</U>


(b)  The number of different solutions is  {{{C[15+3-1]^(3-1)}}} = {{{C[17]^2}}} = {{{(17*16)/2}}} = 17*8 = 136.    <U>ANSWER</U>
</pre>