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.

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