imc

We will use complementary counting and do casework on the equations. There are $8$ possible equations: Equation 1: $0=0$ Equation 2: $x=0$ Equation 3: $y=0$ Equation 4: $z=0$ Equation 5: $x+y=0$ Equation 6: $x+z=0$ Equation 7: $y+z=0$ Equation 8: $x+y+z=0$ We will continue to refer to the equations by their number on this list. $8^3=512$ total systems. Note that no two equations by themselves can force $x=y=z=0$. Therefore no system with Equation 1 or with repeated equations can force $x=y=z=0$. Case 1: Equation 8 ($x+y+z=0$) is present. Case 1a: Equation 8, and two equations from $\{5,6,7\}$. There are $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$ ways to choose two equations from $\{5,6,7\}$ and $3!=6$ ways to arrange each case. The number of options that force $x=y=z=0$ is $3\cdot 3!=18$. Case 1b: Equation 8, one equation from $\{5,6,7\}$, and one equation from $\{2,3,4\}$. There are $\left(\genfrac{}{}{0ex}{}{3}{1}\right)=3$ ways to choose one equation from $\{5,6,7\}$. WLOG let us choose Equation 7. Given $x+y+z=0$ and $y+z=0$, we conclude that $x=0$. The third equation can be either $y=0$ or $z=0$. There are $3!$ ways to arrange each case. The number of options that force $x=y=z=0$ is $3\cdot 2\cdot 3!=36$. Case 1c: Equation 8, and two equations from $\{2,3,4\}$. There are $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$ ways to choose two equations from $\{2,3,4\}$ and $3!=6$ ways to arrange each case. Each of these cases forces $x=y=z=0$. $3\cdot 3!=18$ total options. Case 2: Equation 8 is $\text{not}$ present, at least one equation from $\{5,6,7\}$ is present. Case 2a: Equations $\{5,6,7\}$ are all present. There are $3!$ ways to arrange the three equations. $6$ options. Case 2b: Two equations from $\{5,6,7\}$ are present. One equation from $\{2,3,4\}$ is present. There are $\left(\genfrac{}{}{0ex}{}{3}{2}\right)$ ways to choose two equations from $\{5,6,7\}$. WLOG let Equations 5 and 6 be in our system: $x+y=0$ and $x+z=0$. Any equation from $\{2,3,4\}$ will force $x=y=z=0$. There are $3!$ ways to arrange the equations. The number of options that force $x=y=z=0$ is $\left(\genfrac{}{}{0ex}{}{3}{2}\right)\cdot \left(\genfrac{}{}{0ex}{}{3}{1}\right)\cdot 3!=54$. Case 2c: One equation from $\{5,6,7\}$ is present. Two equations from $\{2,3,4\}$ are present. There are $\left(\genfrac{}{}{0ex}{}{3}{1}\right)$ ways to choose one equation from $\{5,6,7\}$. WLOG let Equation 5 ($x+y=0$) be present. One of the two equations from $\{2,3,4\}$ must be Equation 4, $z=0$, since it is the only equation that restricts $z$. The last equation can be either 2 or 3. There are $3!$ ways to arrange the equations. The number of options that force $x=y=z=0$ is $\left(\genfrac{}{}{0ex}{}{3}{1}\right)\cdot \left(\genfrac{}{}{0ex}{}{2}{1}\right)\cdot 3!=36$. Case 3: Only equations $\{2,3,4\}$ are present. There are $3!$ ways to arrange the three equations. $6$ options. We add up the cases: $18+36+18+6+54+36+6=174$ total systems force $x=y=z=0$. Thus $512-174=\boxed{338}$ do not.