2. From $1,2,3, \cdots, 14$ select $a_{1}, a_{2}, a_{3}$ in ascending order, such that $a_{2}-a_{1} \geqslant 3$, $a_{3}-a_{2} \geqslant 3$ are simultaneously satisfied. The number of all different selection methods that meet the above requirements is $\qquad$.
2. 120 Detailed Explanation: $\because a_{1}-1 \geqslant 0, a_{2}-a_{1}-3 \geqslant 0, a_{3}-a_{2}-3 \geqslant 0,14-a_{3} \geqslant 0$, adding them up gives $a_{1}-1+a_{2}-a_{1}-3$ $+a_{3}-a_{2}-3+14-a_{3}=7$, thus the number of ways to choose $a_{1}, a_{2}, a_{3}$ is the number of non-negative integer solutions to the equation $x_{1}+x_{2}+x_{3}+x_{4}=7$, which is $C_{7+4-1}^{4-1}=C_{10}^{3}=120$.