$p$ parallel lines are drawn in the plane and $q$ lines perpendicular to them are also drawn. How many rectangles are bounded by the lines?
1. To form a rectangle, we need two pairs of parallel lines: one pair from the set of $p$ parallel lines and one pair from the set of $q$ perpendicular lines.
2. The number of ways to choose 2 lines from $p$ parallel lines is given by the binomial coefficient $\binom{p}{2}$.
   \[
   \binom{p}{2} = \frac{p(p-1)}{2}
   \]
3. Similarly, the number of ways to choose 2 lines from $q$ perpendicular lines is given by the binomial coefficient $\binom{q}{2}$.
   \[
   \binom{q}{2} = \frac{q(q-1)}{2}
   \]
4. Each pair of parallel lines from the $p$ lines and each pair of perpendicular lines from the $q$ lines will form a unique rectangle. Therefore, the total number of rectangles is the product of these two binomial coefficients:
   \[
   \binom{p}{2} \times \binom{q}{2} = \left( \frac{p(p-1)}{2} \right) \times \left( \frac{q(q-1)}{2} \right) = \frac{p(p-1)}{2} \cdot \frac{q(q-1)}{2} = \frac{p(p-1)q(q-1)}{4}
   \]

The final answer is $\boxed{\frac{p(p-1)q(q-1)}{4}}$