List 3. From $1,2,3, \cdots, 14$, select $a_{1}, a_{2}, a_{3}$ in ascending order, such that $a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3$. How many different ways are there to select the numbers that meet the above requirements? (1389, National High School League)
Solution: Obviously, $a_{1}+\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+(14-$ $\left.a_{3}\right)=14$, where $a_{1} \geqslant 1, a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3,14-$ $a_{3} \geqslant 0$.
Transform the above equation to
$$
\left(a_{1}-1\right)+\left(a_{2}-a_{1}-3\right)+\left(a_{3}-a_{2}-3\right)+(14
$$
$\left.-a_{3}\right)=7$.

At this point, $a_{1}-1, a_{2}-a_{1}-3, a_{3}-a_{2}-3,14-c_{3}$ are all non-negative integers. It is known that this indeterminate equation has $C_{i+4-1}^{\gamma}=$ $C_{10}^{?}$ different sets of non-negative integer solutions, therefore, the number of different ways that meet the requirements is $C_{10}^{3}=120$ (using the combination formula).