Number of nonnegative integral solutions of $3x+y+z \leq 25$ 
Find the number of nonnegative integral solutions of $$3x+y+z \leq 25$$

I can get the answer to $3x+y+z=25$ but I can't get the answer with inequality. Please help.
 A: The generating function approach is that this is the coefficient of $x^{25}$ in:
$$\frac{1}{(1-x)^3(1-x^3)}$$
Which can be rewritten as:
$$\frac{(1+x+x^2)^3}{(1-x^3)^4}=(x^6 + 3 x^5 + 6 x^4 + 7 x^3 + 6 x^2 + 3 x + 1)\sum_{j=0}^{\infty}\binom{j+3}{3}x^{3j}$$
So the coefficient of $x^{25}$ is:
$$3\binom{11}{3}+6\binom{10}{3}$$
More generally, the number of solutions to $3x+y+z\leq 3n-2$ is:
$$3\binom{n+2}{3}+6\binom{n+1}{3}=\frac{3n^2(n+1)}{2}$$
The number of solutions to $3x+y+z\leq 3n-1$ is:
$$6\binom{n+2}{3}+3\binom{n+1}{3}=\frac{3n(n+1)^2}{2}$$
The number of solutions to $3x+y+z\leq 3n$ is:
$$\binom{n+3}{3} + 7\binom{n+2}{3}+\binom{n+1}{3}=\frac{(n+1)(3n^2+6n+2)}{2}$$ 
A: Set $t\in\{0,1,2,\cdots\}$, 
$$3x+y+z+t=25$$
Since $x\in\{0,1,\cdots,8\}$ then
$y+z+t=25-3x$ and we have
$$\sum_{x=0}^{8}\binom{25-3x+3-1}{2}$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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The number of solutions $\ds{\,\mc{S}_{s}}$ with $\ds{3x + y + z = s}$  $\ds{\pars{~\mbox{with}\ x,y,z\ \in\ \mathbb{N}_{\geq 0}\ \mbox{and}\
s \geq 0~}}$ is given by: 

\begin{align}
\mc{S}_{s} & \equiv \bracks{t^{s}}\sum_{x = 0}^{\infty}t^{3x}
\sum_{y = 0}^{\infty}t^{y}\sum_{z = 0}^{\infty}t^{z} =
\bracks{t^{s}}{1 \over \pars{1 - t^{3}}\pars{1 - t}^{2}}
\\[5mm] & =
\bracks{t^{s}}\sum_{i = 0}^{\infty}t^{3i}
\sum_{j = 0}^{\infty}{-2 \choose j}\pars{-t}^{j} =
\bracks{t^{s}}\sum_{i = 0}^{\infty}\sum_{j = 0}^{\infty}
\pars{j + 1}\sum_{k = 0}^{\infty}\delta_{k,3i + j}\,t^{k}
\\[5mm] & =
\bracks{t^{s}}\sum_{k = 0}^{\infty}\bracks{\sum_{i = 0}^{\infty}\sum_{j = 0}^{\infty}\pars{j + 1}\delta_{k,3i + j}}t^{k} =
\sum_{i = 0}^{\infty}\sum_{j = 0}^{\infty}\pars{j + 1}\delta_{s,3i + j} =
\sum_{i = 0}^{\left\lfloor s/3 \right\rfloor}\pars{s - 3i + 1}
\end{align}

The number of solutions with $\ds{3x + y + z \leq 25}$
  $\ds{\pars{~\mbox{with}\ x,y,z\ \in\ \mathbb{N}_{\geq 0}~}}$ is given by:

\begin{align}
\sum_{s = 0}^{25}\mc{S}_{s} & =
\sum_{s = 0}^{25}\sum_{i = 0}^{\left\lfloor s/3\right\rfloor}\pars{s - 3i + 1} =
\bbx{\ds{1215}}
\end{align}
