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.
 A: 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 $25/6^3 = 25/216$.
A: 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.
A: We must find the number of solutions of the equation
$$x_1 + x_2 + x_3 = 9 \tag{1}$$
in the positive integers subject to the restrictions that $x_1, x_2, x_3 \leq 6$.
A particular solution of equation 1 in the positive integers corresponds to the placement of $3 - 1 = 2$ addition signs in the $9 - 1 = 8$ spaces between successive ones in a row of nine ones. 
$$1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1$$
For instance, placing addition signs in the third and seventh spaces yields
$$1 1 1 + 1 1 1 1 + 1 1$$
which corresponds to the solution $x_1 = 3$, $x_2 = 4$, $x_3 = 2$.  
The number of solutions of equation 1 in the positive integers is 
$$\binom{9 - 1}{3 - 1} = \binom{8}{2}$$
since we must choose which two of the eight spaces between successive ones to fill with addition signs.  
However, we have included solutions in which some $x_i > 6$.  Since the summands are positive integers, this can only occur if one of the $x_i$'s is $7$ and the other two $x_i$'s are $1$.  There are three choices for which $x_i$ is equal to $7$.  Hence, the number of ways the sum of the numbers on the three dice could add up to $9$ is 
$$\binom{8}{2} - \binom{3}{1}$$
We can formalize the argument about the restrictions as follows.  Suppose $x_1 > 6$.  Then $x_1' = x_1 - 6$ is a positive integer.  Substituting $x_1' + 6$ for $x_1$ in equation 1 yields
\begin{align*}
x_1' + 6 + x_2 + x_3 & = 9\\
x_1' + x_2 + x_2 & = 3 \tag{2}
\end{align*}
Equation 2 is an equation in the positive integers with 
$$\binom{3 - 1}{3 - 1} = \binom{2}{2}$$
solutions.  By symmetry, the number of solutions of equation 1 in which $x_1 > 6$ is equal to the number of solutions in which $x_2 > 6$ and to the number of solutions in which $x_3 > 6$.  Hence, 
$$\binom{3}{1}\binom{2}{2}$$
solutions of equation 1 violate the restrictions.  Therefore, the number of admissible solutions is 
$$\binom{8}{2} - \binom{3}{1}\binom{2}{2}$$
