Example 6 If numbers $a_{1}, a_{2}, a_{3}$ are taken in ascending order from the set $1,2, \cdots, 14$, such that both $a_{2}-a_{1} \geqslant 3$ and $a_{3}-a_{2} \geqslant 3$ are satisfied. Then, the number of all different ways to select the numbers is $\qquad$ kinds.
Solution: Given $a_{1} \geqslant 1, a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3$, $14-a_{3} \geqslant 0$, let $a_{1}=x_{1}, a_{2}-a_{1}=x_{2}, a_{3}-a_{2}=x_{3}, 14-a_{3}=x_{4}$, then
$$
x_{1}+x_{2}+x_{3}+x_{4}=14 .
$$

Thus, the problem is transformed into finding the number of integer solutions to the equation under the conditions
$$
x_{1} \geqslant 1, x_{2} \geqslant 3, x_{3} \geqslant 3, x_{4} \geqslant 0
$$

Transform the equation into
$$
\begin{array}{l}
x_{1}+\left(x_{2}-2\right)+\left(x_{3}-2\right)+\left(x_{4}+1\right)=11 . \\
\text { Let } x_{1}=y_{1}, x_{2}-2=y_{2}, x_{3}-2=y_{3}, \\
x_{4}+1=y_{4} .
\end{array}
$$

The number of positive integer solutions to $y_{1}+y_{2}+y_{3}+y_{4}=11$ is $\mathrm{C}_{10}^{3}$, so the number of different ways to choose $a_{1}, a_{2}, a_{3}$ that meet the conditions is $\mathrm{C}_{10}^{3}=120$.