Number of positive unordered integral solutions What are the number of positive unordered integral solutions for $a+b+c=36$
Solution given is $108.$.But I am getting $91$ as $$\frac{\binom{35}2-3\times16-1}{3!}.$$
$3\times16($ for $a=b$ cases and $1$ for $a=b=c$ cases)
please help me what is missing in this logic?

The idea of ${35\choose2}$ is that you are picking $a,a+b$. For $a<b<c$ you get the solution six times (corresponding to the permutations of $a,b,c$). But you only get the solution $12,12,12$ once by picking $12,24$. For a solution like $10,10,16$, you get it three times by picking $10,20$ or $10,26$ or $16,10$. So you have actually undercounted these cases. There are 17 cases $a,a,b$ including the case $12,12,12$ (we can take $a=1,2,\dots,17$). So before dividing by 6 we need to add back the 16 doubles three times each and the triple 5 times. Thus the answer is $$\frac{595+48+5}{6}=108$$ 
