<s>[INST] 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.
 [/INST] 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 $25/6^3 = 25/216$.
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