6 Set $A=\{0,1,2,3,4,5,9\}, a, b \in A$, and $a \neq b$. The number of functions of the form $y=-a x^{2}+(4-b) x$ whose graph vertex is in the first quadrant is $\qquad$ .
6 21. From the problem, we have
$$
y=-a\left(x-\frac{4-b}{2 a}\right)^{2}+\frac{(4-b)^{2}}{4 a},
$$

According to the problem, we have
$$
\left\{\begin{array}{l}
\frac{4-b}{2 a}>0, \\
\frac{(4-b)^{2}}{4 a}>0,
\end{array}\right.
$$

Clearly, $a>0, b<4$. Also, since
$$
A=\{0,1,2,3,4,5,9\},
$$

and $a, b \in A$, with $a \neq b$, when $b=0$, there are $C_{6}^{1}$ ways to choose $a$; when $b \neq 0$, there are $3C_{1}^{1}$ ways to choose $a$ and $b$, so in total, we have
$$
C_{6}^{1}+3 C_{1}^{1}=21 \text{ (ways). }
$$