Finding the no. of non-negative integral solutions to $x+y+2z=33$. 
Finding the no. of non negative integral solutions to $x+y+2z=33$.

I tried many pure combinatorial approaches (I don't like making individual cases, too long). But they went in vain. I was just pondering, when randomly, a thought came to my mind.

$$x+y+2z=33$$
$$x+2y+z=33$$
$$2x+y+z=33$$
All would have the same number of solutions (integral). So if we add them up, then also, the solutions would remain the same. 
$$4x+4y+4z=99$$
$$x+y+z=24.75$$
Taking the floor of the right side (random), we get $x+y+z=24$. 
Then, the number of solutions would be ${26\choose2}$. But it is incorrect. So, what can I employ.
 A: A little nitpicking: if you are looking for integral solutions, the answer is infinite. I assume you are looking for solutions with $x, y, z \in \mathbb{N}_0 $ (where $\mathbb{N}_0 = \{ 0, 1, 2, ... \}$).
Observe that the solutions $(x, y) \in \mathbb{N_0}^2$ to $x + y = n$ are $(0, n), (1, n - 1), ..., (n, 0)$. So, there are $n + 1$ solutions.
So, if we fix $z$ we know how many solutions there are ($34 - 2z$, since we can rewrite the original equation to $x + y = 33 - 2z$). We also know that we can choose $z$ in the range $[0, 16]$ (since if we pick a negative number or 17 or higher for $z$ it would cause at least one of $x, y, z$ to be negative). So there are
$$ \sum_{z = 0}^{16} 34 - 2z = 306 $$
solutions.
A: For $z$ to be integral and positive, $x+y$ is odd, so it is $x+y=1,x+y=3,\dots,x+y=33$ and each has corresponding solution to $2z$ which is integer (positive), so we have fixed $z$, so the number of solutions are an arithmetic progression of $2,4,6,\dots,34$. Now summation of AP=$\frac{n}{2}(2a+(n-1)d)$ so plug in $a=2,n=17,d=2$ you get ,
Summation= $34\times 9=306$ and that are all possible solutions. Your book has a typo, I suppose, also suggested by Ian Miller. And if it's that $x, y, z$ belong to natural numbers rather than whole, then its $306-16\times6=210$ 
A: Clearly the number of non-negative integral solutions to the equations $x + y + 2z = 33$ is same as the number of similar solutions to $i + j + 2k = 33$. This is obviously equal to the coefficient of $x^{33}$ in the expansion $$(1 + x + x^{2} + \cdots + x^{i} + \cdots)(1 + x + x^{2} + \cdots + x^{j} + \cdots)(1 + x^{2} + \cdots + x^{2k} + \cdots)$$ which is the same as the coefficient of $x^{33}$ in $$\frac{1}{(1 - x)^{2}}\cdot\frac{1}{1 - x^{2}} = (1 + 2x + 3x^{2} + 4x^{3} + \cdots)(1 + x^{2} + x^{4} + x^{6} + \cdots)$$ and this coefficient is given by $$2 + 4 + 6 + \cdots + 32 + 34 = ((2 + 34)/2) \cdot 17 = 306$$
