HMMT February, 2020

Combinatorics Difficulty 5.6 AIME, harder Prove it United States

Problem:
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

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 $8 \cdot 6 \cdot 2=96$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.