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 $3$?

It is a bad question as they have not described how the sides have been chosen, but their answer almost reveals what they intended:

*

*There are $171$ compositions of $20$ into three positive integer parts (any order).  Of these,

*

*$144$ compositions have the three parts distinct

*$27$ compositions have two of the parts equal and the third distinct

*$0$ compositions have all three parts equal ($20$ is not divisible by $3$)



*So removing duplicates which are the same but in a different order, there are

*

*$\frac{144}{3!}=24$ partitions which have the three parts distinct

*$\frac{27}{3}=9$ partitions which have two of the parts equal and the third distinct

*$0$ partitions have all three parts equal ($20$ is not divisible by $3$)

*making $24+9+0=33$ for the number of partitions of $20$ into three positive integer parts



The $36 \to 8$ is the same but where the two smaller numbers add up to more than the largest number
