Setting $y=A x^{2}+B=C x^{2}+D$, we find that $A x^{2}-C x^{2}=x^{2}(A-C)=D-B$, so $x^{2}=\frac{D-B}{A-C} \geq 0$ by the trivial inequality. This implies that $D-B$ and $A-C$ must both be positive or negative. If two distinct values are chosen for $(A, C)$ and $(B, D)$ respectively, there are 2 ways to order them so that both the numerator and denominator are positive/negative (increasing and decreasing). We must divide by 2 at the end, however, since the 2 curves aren\textquotesingle t considered distinct. Calculating, we get $$ \frac{1}{2} \cdot\left(\begin{array}{l} 6 \textbackslash\textbackslash{} 2 \end{array}\right)\left(\begin{array}{l} 4 \textbackslash\textbackslash{} 2 \end{array}\right) \cdot 2=(\text { C) } 90 $$ 
Values for $A, B, C$, and $D$ are to be selected from $\textbackslash{1,2,3,4,5,6\textbackslash}$ without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves $y=A x^{2}+B$ and $y=C x^{2}+D$ intersect? (The order in which the curves are listed does not matter; for example, the choices $A=3, B=2, C=4, D=1$ is considered the same as the choices $ A=4, B=1, C=3, D=2 $.) 
  * $$30$$ 
  * $$60$$ 
  * $$90$$ 
  * $$180$$ 
  * $$360$$ 