Example 2 From the numbers $1,2, \cdots, 14$, select $a_{1}, a_{2}, a_{3}$ in ascending order, and $a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3$. How many different ways of selection are there that meet the conditions?
Solution: Notice that
$$
a_{1}+\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+\left(14-a_{3}\right)=14 \text {, }
$$

where $a_{1} \geqslant 1, a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3,14-a_{3}$ $\geqslant 0$, i.e.,
$$
\begin{array}{l}
a_{1}-1+\left(a_{2}-a_{1}-3\right)+\left(a_{3}-a_{2}-3\right)+ \\
\left(14-a_{3}\right)=7 .
\end{array}
$$
$$
\text { Let } x_{1}=a_{1}-1, x_{2}=a_{2}-a_{1}-3, x_{3}=
$$
$a_{3}-a_{2}-3, x_{4}=14-a_{3}$, then
$$
x_{1}+x_{2}+x_{3}+x_{4}=7 \text {, }
$$

where $x_{1} \geqslant 0, x_{2} \geqslant 0, x_{3} \geqslant 0, x_{4} \geqslant 0$.
By the theorem, the equation has $\mathrm{C}_{10}^{3}=120$ solutions. Therefore, there are $\mathrm{C}_{10}^{3}=120$ ways that meet the conditions.