If a fair die is thrown three times, what is the probability that the sum of the faces is 9? 
If a fair die is thrown thrice, what is the probability that the sum of the faces is 9?

I did like this.
The total number of cases is $6^3=216$
Now,the number of solutions of the equation $x + y + z = 9$ with each of $x,y,z$ greater than equal to $1$ is ${8 \choose 2}$.
But am not sure about my answer. Please help.
 A: I get that there are only $25$ ways of writing $9$ as a sum of three integers in $[1,6]$, since:
$$[x^9](x+x^2+x^3+x^4+x^5+x^6)^3 = 25.$$
Hence the probability is $\frac{5^2}{6^3}$.
A: I solved it as such.
For each first die roll, I find how many solutions sum to the rest. That is:
For die roll 1: The next 2 dice must sum to 8. There are 5 ways to do this.
For die roll 2: The next 2 dice must sum to 7. There are 6 ways to do this.
For die roll 3: The next 2 dice must sum to 6. There are 5 ways to do this.
For die roll 4: The next 2 dice must sum to 5. There are 4 ways to do this.
For die roll 5: The next 2 dice must sum to 4. There are 3 ways to do this.
For die roll 6: The next 2 dice must sum to 3. There are 2 ways to do this.  
You add these up to get $25/216$
Finding the numbers for rolling 2 dice is pretty simple. It's small enough to be enumerated.
This is less elegant than the other solutions here. But I like its simplicity.
A: Alternatively: given that $x, y, z \in \{1,..6\}, x+y+z=9$
If $x=1$ then $y\in \{2,..6\}$, else if $x\in\{2,..6\}$ then $y\in\{1,.. 8-x\}$.  For each such pairing there is one value of $z$.
$$\begin{align}
\sum_{y=2}^{6} 1 + \sum_{x=2}^6 \sum_{y=1}^{8-x} 1 & = 5 + \sum_{x=2}^{6}(8-x)
\\ & = 5+8\times 5 - \frac{6\times 7}{2} + 1   
\\ & = 25   
\end{align}$$
A: You have included the solutions $7+1+1$ and relatives, which cannot occur with dice.  So your procedure is fine,  except that for the "favourables" we must use $\binom{8}{2}-3$.
