Example 3 If numbers $1,2, \cdots, 14$ are taken in ascending order as $a_{1}, a_{2}, a_{3}$, such that $a_{2}-a_{1} \geqslant 3$, and $a_{3}-a_{2} \geqslant 3$, find the number of different ways to choose them.
From the given information, we have
$$
\begin{aligned}
a_{1} & \leqslant a_{2}-3 \leqslant a_{3}-6 \\
& \Rightarrow 1 \leqslant a_{1}<a_{2}-2<a_{3}-4 \leqslant 10 .
\end{aligned}
$$

Substitute $\left(a_{1}, a_{2}-2, a_{3}-4\right)=(x, y, z)$.
Then the number of tuples $\left(a_{1}, a_{2}, a_{3}\right)$ equals the number of ways to choose three different elements from $\{1,2, \cdots, 10\}$, which is $\mathrm{C}_{10}^{3}=120$.