11. From 30 people with distinct ages, select two groups, the first group consisting of 12 people and the second group consisting of 15 people, such that the oldest person in the first group is younger than the youngest person in the second group. How many different ways are there to do this?
11. Let's assume the ages of these 30 people, in ascending order, are: $a_{1}<a_{2}<a_{3}<\cdots<{ }_{29}<a_{30}$.

Let the first group of 12 people be $a_{i 1}<a_{i 2}<\cdots<a_{i 12}$, and the second group of 15 people be $a_{j 1}<a_{j 2}<\cdots<a_{j 15}$, which are the selected groups that meet the requirements. Thus, $a_{i 12}<a_{j 1}$.

In this way, this selection method corresponds one-to-one with $a_{i 1}<a_{i 2}<\cdots<a_{12}<a_{j 1}<a_{j 2}<\cdots<a_{j 15}$, which is a 27-element subset of $\left\{a_{1}, a_{2}, \cdots, a_{30}\right\}$.
Therefore, the number of selection methods is $C_{30}^{27}=C_{30}^{3}=4060$.