【Question 12】
As shown in the figure, line $a$ is parallel to line $b$. Line $a$ has 10 points, which are $a_{1}, a_{2}, a_{3}, \cdots, a_{10}$, and line $b$ has 11 points, which are $b_{1}, b_{2}, b_{3}, \cdots, b_{11}$. Connecting each point on $a$ with each point on $b$, many line segments can be obtained. It is known that no three line segments intersect at a point outside lines $a$ and $b$. The total number of intersection points of these line segments (excluding $a_{1}, a_{2}, \cdots, a_{10}, b_{1}, b_{2}, \cdots, b_{11}$) is $\qquad$.
【Analysis and Solution】
Graph counting.
As shown in the figure, if we notice the following fact:
For any two points $M, N$ on line $a$ and any two points $P, Q$ on line $b$, a quadrilateral $MNQP$ can be formed, and the intersection point of the diagonals $NQ, NP$ of this quadrilateral is exactly the point we need to count.
At the same time, for any four points (two points are taken from each of the lines $a$ and $b$), such an intersection point can be generated. Therefore, the correspondence between the intersection points of the two line segments and the quadrilaterals is as follows:
This means that to count the number of intersection points,
we only need to determine how many quadrilaterals $MNQP$ satisfy the conditions in lines $a$ and $b$. Thus, the problem is transformed into: there are 10 points on line $a$ and 11 points on line $b$. How many quadrilaterals $MNQP$ are there? Among them, points $M, N$ are on line $a$, and points $P, Q$ are on line $b$.
This is a conventional combinatorial counting problem, which can be calculated in 2 steps using the multiplication principle:
Step 1, determine the line segment $MN$, there are $1+2+\cdots+9=(1+9) \times 9 \div 2=45$ or $C_{10}^{2}=\frac{10 \times 9}{2 \times 1}=45$ ways to choose;
Step 2, determine the line segment $PQ$, there are $1+2+\cdots+10=(1+10) \times 10 \div 2=55$ or $C_{11}^{2}=\frac{11 \times 10}{2 \times 1}=55$ ways to choose; according to the multiplication principle, a total of $45 \times 55=2475$ quadrilaterals can be generated.
Given that no three line segments intersect at a point outside lines $a$ and $b$, these line segments have a total of 2475 intersection points.