How many integer solutions to$ x+y+z \le 6$ where $x,y,z$ are non-negative? I understand the steps towards working this question out if $x + y + z = 6$, but what are the steps for an inequality?

$x$ can be anything from $0$ to $6$.
$y$ can be anything from $0$ to $6 - x$.
$z$ can be anything from $0$ to $6 - (x+y)$
$$\begin{align} 
\text{total} &= \sum_{x=0}^6 \quad \sum_{y=0}^{6-x}\quad \sum_{z=0}^{6-(x+y)} 1
\\ &\quad \text{...and the rest is just algebra}
\\ &= \sum_{x=0}^6 \quad \sum_{y=0}^{6-x} 6-(x+y) + 1
\\ &= \sum_{x=0}^6 \left(\sum_{y=0}^{6-x} 7 - x\right) - \left(\sum_{y=0}^{6-x}y\right)
\\ &= \sum_{x=0}^6  (6 - x + 1)(7 - x) - \left(\frac{(6-x)^2 + (6-x)}2\right)
\\ &= \frac 12 \sum_{x=0}^6 x^2 - 15x + 56
\\ &= \frac 12\left( \sum_{x=0}^6 x^2 - 15\sum_{x=0}^6x + \sum_{x=0}^6 56 \right)
\\ &= \frac 12\left( \frac{6(6+1)(2\cdot 6 + 1)}{6} - 15 \frac{6(6+1)}2 + 56\cdot(6+1) \right)
\\ &= 84
\end{align}$$
