Let each of $m$ distinct points on the positive part of the $x$-axis be joined to $n$ distinct points on the positive part of the $y$-axis. Obtain a formula for the number of intersection points of these segments, assuming that no three of the segments are concurrent.
1. **First Method: Combinatorial Approach**

   To find the number of intersection points of the segments, we can use a combinatorial approach. Each intersection point is determined by choosing two distinct points on the $x$-axis and two distinct points on the $y$-axis. The segments joining these points will intersect exactly once.

   - Choose 2 points from $m$ points on the $x$-axis: $\binom{m}{2}$
   - Choose 2 points from $n$ points on the $y$-axis: $\binom{n}{2}$

   The total number of intersection points is given by the product of these two binomial coefficients:
   \[
   \binom{m}{2} \binom{n}{2} = \frac{m(m-1)}{2} \cdot \frac{n(n-1)}{2} = \frac{mn(m-1)(n-1)}{4}
   \]

2. **Second Method: Summation Approach**

   We can also derive the formula by considering the intersections incrementally.

   - Start by joining each point $x_i$ on the $x$-axis to one point $y_1$ on the $y$-axis.
   - For the second point $y_2$ on the $y$-axis, the number of intersections with segments from $y_1$ is $m-1$, and this decreases by one for each subsequent segment until $y_2x_m$ where there are no intersections. This gives:
     \[
     \sum_{i=0}^{m-1} i = \frac{m(m-1)}{2}
     \]
   - For the third point $y_3$ on the $y$-axis, the number of intersections is $2(m-1)$, contributed by $y_1$ and $y_2$. This pattern continues, and the total number of intersections for $y_3$ is:
     \[
     2 \cdot \frac{m(m-1)}{2} = (m-1)(m-1)
     \]
   - This pattern continues for each successive point on the $y$-axis. The total number of intersections for all points on the $y$-axis is:
     \[
     \left(\sum_{i=0}^{n-1} i\right) \left(\frac{m(m-1)}{2}\right) = \frac{n(n-1)}{2} \cdot \frac{m(m-1)}{2} = \frac{mn(m-1)(n-1)}{4}
     \]

Thus, both methods yield the same result.

The final answer is $\boxed{\frac{mn(m-1)(n-1)}{4}}$