Problem 718

12. 4060

Assume the ages of these 30 people, in ascending order, are $a_{1}<a_{2}<\cdots<a_{29}<a_{30}$. 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 $a_{j 1}<a_{j 2}<\cdots a_{111}$, which are the ages of the selected two groups that meet the requirements. Thus, $a_{t 12}<a_{11}$.

In this way, this selection method corresponds one-to-one with $a_{t 1}<a_{i 2}<\cdots<a_{i 12}<a_{j 1}<\cdots<a_{11}$. The latter is a 27-element subset of $\mid a_{1}, a_{2}, \cdots, a_{30}!$.
Therefore, the number of selection methods is $C_{30}^{27}=C_{30}^{3}=4060$.