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?
 A: 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 $$\frac{595+48+5}{6}=108$$ 
A: Your "positive unordered integral solutions" are called partitions (of $36$) into three parts, or $3$-partitions for short.
Since $a$, $b$, $c$ have to be $\geq1$, by "stars and bars" there are ${(36-3)+2\choose 2}=595$ ordered admissible triples summing to $36$. Here  each partition of $36$ with three different parts occurs six times, the $16$ $3$-partitions with exactly two equal parts occur  three times, and the partition $(12,12,12)$ occurs one time. If we now form the number
$$595+ 3\cdot 16+5\cdot 1=648$$
then all $3$-partitions of $36$ have been counted exactly six times. It follows that the number of $3$-partitions of $36$  is ${648\over6}=108$.
A: The number 91 that you have is the number of unordered triples $\{a,b,c\}$ such that $a,b,c$ are all distinct. For the number of unordered triples, add the 16 and 1 to get 108.
A: If $P(n,k)$ denotes the number of $k$ partitions of $n$, then we have the recurrence relation $$P(n,k) = P(n-k,k) + P(n-1,k-1)$$ Also note that $P(2m,2) = m$ and $P(2m+1,2) = 2$. Using these, it is easy to show that $P(6n,3) = 3n^2$. Hence $P(36,3) = 3\cdot 6^2 = 108$.
