The sum of three positive integers is $20$. Find the probability that they form the sides of a triangle. 
The sum of three positive integers is $20$. Find the probability that they form the sides of a triangle.

Let $a,b,c$ be there positive integers. So, $a+b+c=20$
Total number of solutions would be $^{19}C_2=171$
For $a,b,c$ to form a triangle $0\lt a,b,c\le9$
So, number of such solutions= coefficient of $x^{20}$ in $(x+x^2+x^3+...+x^9)^3=$ coefficient of $x^{17}$ in $(1+x+x^2+...+x^8)^3=$ coefficient of $x^{17}$ in $(1-x^9)^3(1-x)^{-3}=1\times^{19}C_2-3\times^{10}C_2=36$
So, the required probability is $\frac{36}{171}$ but the answer given is $\frac8{33}$
In the hint, they have written total combinations of $a,b,c=\frac{144}{6}+\frac{27}{3}=33$, and favorable combinations of $a,b,c=\frac{12}{3}+\frac{24}{6}=8$
I think they have split $171$ as $144+27$ and $36$ as $24+12$ but why? and why to divide them with $6$ or $3$?

Because in the case of a triangle's sides, order has no meaning. The set $\{7,8,5\}$, representing $a,b$ and $c$ forms the same triangle as $\{8,5,7\}$ and likewise. So, if $a,b,c$ are distinct, the $6$ possible ordered pairs give only $1$ combination, hence division by $6$. If two of $a,b,c$ are equal, then $3$ different ordered pairs are possible, which give only one combination, hence division by $3$.
