10. From the numbers $1,2,3, \cdots, 14$, select $a_{1}, a_{2}$, $a_{3}$ in ascending order, and $a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3$, then the number of different ways to select that meet the conditions is $\qquad$ kinds.
10. 120 Obviously, $a_{1}+\left(a_{2}-a_{1}\right)+\left(a_{3}-a_{2}\right)+\left(14-a_{3}\right)=14$,

where $a_{1} \geqslant 1, a_{2}-a_{1} \geqslant 3, a_{3}-a_{2} \geqslant 3,14-a_{3} \geqslant 0$.
Transform the above equation to: $\left(a_{1}-1\right)+\left(a_{2}-a_{1}-3\right)+\left(a_{3}-a_{2}-3\right)+\left(14-a_{3}\right)=7$.
At this point, $a_{1}-1, a_{2}-a_{1}-3, a_{3}-a_{2}-3,14-a_{3} \geqslant 0$.
The indeterminate equation (*) has $C_{10}^{7}=C_{10}^{3}$ different non-negative integer solutions.
Therefore, the number of different ways that meet the requirements is $C_{10}^{3}=120$.
Method 2 Let $S=\{1,2, \cdots, 14\}, S^{\prime}=\{1,2, \cdots, 10\}$.
If $\left\{a_{1}, a_{2}, a_{3}\right\}$ is a three-element subset of $S$, set $a^{\prime}{ }_{1}=a_{1}<a^{\prime}{ }_{2}=a_{2}-2<a^{\prime}{ }_{3}=a_{3}-4$.
Then $\left\{a^{\prime}{ }_{1}, a^{\prime}{ }_{2}, a^{\prime}{ }_{3}\right\}$ is a three-element subset of $S^{\prime}$, and $\left(a_{1}, a_{2}, a_{3}\right)$ and $\left(a^{\prime}{ }_{1}, a^{\prime}{ }_{2}, a^{\prime}{ }_{3}\right)=\left(a_{1}, a_{2}-2, a_{3}-4\right)$ are in one-to-one correspondence.
Thus, the number of different ways to choose is equal to the number of ways to choose any three different numbers from $S^{\prime}$, which is $C_{10}^{3}=120$.