Find the probability of solutions of an equation. Let $x+y+z=20$. What is the probability that all the solutions are distinct? (No two variables have the same value). Assuming that the solutions are only positive integers or zero.

I have tried- $n(S)=\frac{22!}{2!20!}$ 
EDIT:$n(S)=231$. Now suppose we choose any two variables $x,y,z$, which can be done in ${3 \choose 2}$ ways. Now no. of ways of getting the same value in the two chosen variables-$10$. Now ${3 \choose 2}*10=30$. So probability will be: $1-\frac{30}{231}=\frac{67}{77}$. So where did I commit the mistake? 
Edit: Sorry, the question says that solution can be zero too.

You have three distinct boxes, and want to count ways to put 20 indistinct balls into them. The total count of solutions is, as you have calculated:
$$\frac{22!}{2!20!} = 231$$
Now, to generate forbidden solutions you can choose 2 boxes, then put the same number of balls $n$ into each of them, and the remainder into the other. Since the number $n$ can vary from zero to ten, the ways to do this are:
$$\frac{3!}{2!1!}\times\sum_{n=0}^{10} 1 = 33$$
Noting that there's no way to fill three boxes all with the same number of balls (20 is not so divisible), we are done.
Assuming that each solution has equal probability of occurring, therefore the probability that no two terms in the solution are the same, is:
$$1-\frac{33}{231}=\frac{6}{7}$$
