30. Let $A=\{1,2,3,4,5,6,7,8,9,10,11\}, f$ be a mapping from $A$ to $A$, satisfying:
(1) For any $x \in A, f_{15}(x)=x$;
(2) For any positive integer $k(1 \leqslant k \leqslant 14)$, there exists at least one $a \in A$ such that $f_{k}(a) \neq a$.
Find the number of $f$ that satisfy the above conditions. Here, it is defined that $f_{1}(x)=f(x), f_{k+1}(x)=f\left(f_{k}(x)\right)(x \in A$, $k=1,2,3, \cdots)$.
30. From the analysis of Example 4 in §3, we know that $A$ contains only mapping cycles of $f$, and the least common multiple of the number of elements in each mapping cycle is 15, while the sum of the number of elements in each mapping cycle is 11. Noting that $15=1 \times 3 \times 5$ and $11=5+3+3$, and the least common multiple of $5,3,3$ is 15, or $11=5+3+1+1+1$ and the least common multiple of $5,3,1,1,1$ is 15. Therefore, $f$ can only be of the following 2 types: (1) $A$ contains one mapping cycle with 5 elements and two mapping cycles with 3 elements each. The number of such mappings $f$ is $C_{11}^{5} C_{6}^{3} C_{3}^{3} \cdot 4!\cdot(2!)^{2} = \frac{11!}{45}$; (2) $A$ contains one mapping cycle with 5 elements and one mapping cycle with 3 elements, while the remaining 3 elements are fixed points of $f$. The number of such $f$ is $C_{11}^{5} C_{6}^{3} \cdot 4!\cdot 2! = \frac{11!}{90}$. Therefore, the total number of $f$ is $\frac{11!}{45}+\frac{11!}{90}=\frac{11!}{30}$.