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.

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 $\frac{5^2}{6^3}$.
