Tessa picks three real numbers $x, y, z$ and computes the values of the eight expressions of the form $\pm x \pm y \pm z$. She notices that the eight values are all distinct, so she writes the expressions down in increasing order. For example, if $x=2, y=3, z=4$, then the order she writes them down is \[ -x-y-z,+x-y-z,-x+y-z,-x-y+z,+x+y-z,+x-y+z,-x+y+z,+x+y+z \] How many possible orders are there?
Solution: There are $2^{3}=8$ ways to choose the sign for each of $x, y$, and $z$. Furthermore, we can order $|x|,|y|$, and $|z|$ in $3 !=6$ different ways. Now assume without loss of generality that $0<x<y<z$. Then there are only two possible orders depending on the sign of $x+y-z$ : \[ \begin{aligned} & -x-y-z,+x-y-z,-x+y-z,-x-y+z, x+y-z, x-y+z,-x+y+z, x+y+z \\ & -x-y-z,+x-y-z,-x+y-z, x+y-z,-x-y+z, x-y+z,-x+y+z, x+y+z \end{aligned} \] Thus, the answer is $8 \cdot 6 \cdot 2=96$. $\fbox{96}$.