30. Answer: 165
Supposing there are $x_{1}$ numbers smaller than $a_{1}, x_{2}$ numbers between $a_{1}$ and $a_{2}, x_{3}$ numbers between $a_{2}$ and $a_{3}$ and $x_{4}$ numbers greater than $a_{3}$.
Finding the number of possible subsets of $\mathrm{A}$ is equivalent to finding the number of integer solutions of the equation $x_{1}+x_{2}+x_{3}+x_{4}=12$ with the conditions that $x_{1} \geq 0, x_{2} \geq 2, x_{3} \geq 2$ and $x_{4} \geq 0$.
The number of solution of this latter equation is equivalent to the number of solutions of the equation $y_{1}+y_{2}+y_{3}+y_{4}=8$, where $y_{1}, y_{2}, y_{3}$ and $y_{4}$ are nonnegative integers.
Hence the answer is $\binom{11}{3}=165$.
30. $\mathrm{P}$ is the set $\{1,2,3, \ldots, 14,15\}$. If $\mathrm{A}=\left\{a_{1}, a_{2}, a_{3}\right\}$ is a subset of $\mathrm{P}$ where $a_{1}<a_{2}<a_{3}$ such that $a_{1}+6 \leq a_{2}+3 \leq a_{3}$. How many such subsets are there of P?