<s>[INST] 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.
 [/INST] 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 $z$.
$$\begin{align}
\sum_{y=2}^{6} 1 + \sum_{x=2}^6 \sum_{y=1}^{8-x} 1 & = 5 + \sum_{x=2}^{6}(8-x)
\\ & = 5+8\times 5 - \frac{6\times 7}{2} + 1   
\\ & = 25   
\end{align}$$
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