$7 \cdot 34$ Given 7 points $A_{1}, A_{2}, \cdots, A_{7}$ in the plane, where no three points are collinear and their convex hull is $\triangle A_{1} A_{2} A_{3}$. How many convex quadrilaterals can be constructed using 4 of these points as vertices?
[Solution] We discuss the problem in two cases based on whether the convex hull of $\{A_{4}, A_{5}, A_{6}, A_{7}\}$ is a triangle or a quadrilateral. For convenience, we refer to the first 3 points as outer points and the last 4 points as inner points.
(1) Suppose the convex hull of the 4 inner points is $\triangle A_{4} A_{5} A_{6}$. In this case, the 4 inner points can form 6 different pairs of points, each pair determining a line that intersects exactly two sides of $\triangle A_{1} A_{2} A_{3}$. Thus, the two inner points on the line and the two endpoints of the side not intersected by the line together form a convex quadrilateral. Therefore, the number of convex quadrilaterals formed by two inner points and two outer points is 6.

Next, we count the number of convex quadrilaterals formed by 3 inner points and 1 outer point. For this, we introduce the concept of "angle capacity." Extending the two sides of a triangle's interior angle, if there is a given point in the region $F$ bounded by the extensions of the two sides and the third side of the triangle, then this point and the three vertices of the triangle can serve as the four vertices of a convex quadrilateral. Thus, the number of points in region $F$ is called the capacity of the angle. In this way, the capacity of each angle is exactly the number of convex quadrilaterals formed by the three vertices of the triangle and the points inside the angle but outside the triangle.
The 4 inner points form 4 triangles, with a total of 12 interior angles. It is easy to verify that the sum of their capacities is at most 9, so the number of convex quadrilaterals formed by 3 inner points and 1 outer point is at most 9. Therefore, when the convex hull of the 4 inner points is a triangle, there are at most 15 convex quadrilaterals.
(2) Suppose the quadrilateral $A_{4} A_{5} A_{6} A_{7}$ is a convex quadrilateral. We first state a lemma:
Lemma In the quadrilateral $A_{4} A_{5} A_{6} A_{7}$,
(i) If the capacities of a pair of adjacent angles are both 2, then the capacities of the other two angles are both zero;
(ii) The capacities of opposite angles cannot both be 2.
The proof of the lemma is straightforward and is omitted here. According to this lemma, the sum of the capacities of the 4 interior angles of the quadrilateral $A_{4} A_{5} A_{6} A_{7}$ is at most 5. In the figure shown below, the sum of the capacities of the 4 interior angles of the quadrilateral $A_{4} A_{5} A_{6} A_{7}$ is exactly 5. Therefore, the number of convex quadrilaterals formed by 3 inner points and 1 outer point is at most 10. As in (1), the number of convex quadrilaterals formed by two inner points and two outer points is always 6, plus the quadrilateral $A_{4} A_{5} A_{6} A_{7}$, there are a total of 17 convex quadrilaterals.
In summary, there are at most 17 convex quadrilaterals.