11. If numbers $a_{1}, a_{2}, a_{3}$ are taken in increasing 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 total number of different ways to choose such numbers is $\qquad$.
If $a_{1}, a_{2}, a_{3}$ meet the conditions, then $a_{1}^{\prime}=a_{1}, a_{2}^{\prime}=a_{2}-2, a_{3}^{\prime}=a_{3}-4$ are distinct, so $a_{1}^{\prime}, a_{2}^{\prime}, a_{3}^{\prime}$ are selected in ascending order from the numbers $1,2, \cdots, 10$. Conversely, this is also true.
Therefore, the number of different ways to choose them is $C_{10}^{3}=120$.