Three dice are rolled simultaneously. In how many different ways can the sum of the numbers appearing on the top faces of the dice be $9$? 
Three dice are rolled simultaneously. In how many different ways can the sum of the numbers appearing on the top faces of the dice be $9$?

What I did:
I know that the maximum value on the dice can be $6$. So, restricting the values of dice, $6-x+6-y+6-z=9$, $x+y+z=9$, $n=9$ and $r=3$.
Applying partition again:
$$\binom{9+3-1}{3-1}=\binom{11}2=55$$
But the answer is $25$. Please, someone explain this one.
This is a gmat exam question.

In these types of problems, it is not too hard to just consider case by case. Let's list out all the possibilities and how many ways to reorganize them:
$$1,2,6 \rightarrow 3!=6 \text{ ways}$$
$$1,3,5 \rightarrow 3!=6 \text{ ways}$$
$$1,4,4 \rightarrow 3!/2!=3 \text{ ways}$$
$$2,2,5 \rightarrow 3!/2!=3 \text{ ways}$$
$$2,3,4 \rightarrow 3!=6 \text{ ways}$$
$$3,3,3 \rightarrow 3!/3!=1 \text{ way}$$
So in total there are $25$ ways to get a sum of $9$. If you want the probability, just take this over the total number of possibilities and you get $25/6^3 = 25/216$.
