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

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:

$$\binom{n+3}{3} + 7\binom{n+2}{3}+\binom{n+1}{3}=\frac{(n+1)(3n^2+6n+2)}{2}$$

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 $$\sum_{x=0}^{8}\binom{25-3x+3-1}{2}$$

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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}