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 $9$?

Question

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.

Answer 1

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 $25/6^3 = 25/216$.

Answer 2

What you did also counts solutions where $x,\,y,\,z$ can be $0$ as well, but that can't happen with dice.

You should do this for $$(x-1)+(y-1) + (z-1) = 9 -3,$$ i.e. $$x'+y'+z' =6$$ because it is now ok to have $x', y'$ or $z'$ to be $0$.

Now we get $\binom{6+3-1}{3-1} = 28.$ But, this is wrong as well, because $x',y',z'\leq 5$, so if we discard the three solutions where one of the $x', y', z'$ is equal to $6$, we get exactly $25$ ways.