14. Let $A=\{1,2, \cdots, 10\}, f$ be a one-to-one mapping from $A$ to $A$. Define
$$
f_{1}(x)=f(x), f_{n+1}(x)=f\left(f_{n}(x)\right) \text {. }
$$

The mapping $f$ also satisfies the conditions:
(1) For all $x \in A, f(x) \neq x$;
(2) For all $x \in A, f_{21}(x)=x$.

How many mappings $f$ satisfy the above conditions?
14. Since $f$ is a one-to-one mapping from $A$ to $A$, and $A$ is a finite set, for every $x \in A$, there is a positive integer $k(1 \leqslant k \leqslant 10)$ such that $f_{k}(x)=x$.
From condition (1), it follows that for any $x \in A$, $k \neq 1$;
From condition (2), it follows that for $x \in A$, $k=7$ or 3.
Since $10=7+3, 21=7 \times 3$, the set $A$ can be divided into two parts:
$$
A=\left\{a_{1}, a_{2}, \cdots, a_{7}\right\} \cup\left\{b_{1}, b_{2}, b_{3}\right\}.
$$

Construct the mapping $f, a_{1} \xrightarrow{f} a_{2} \xrightarrow{f} \cdots \xrightarrow{f} a_{7} \xrightarrow{f} a_{1}$ and $b_{1} \xrightarrow{f} b_{2} \xrightarrow{f} b_{3} \xrightarrow{f} b_{1}$, which satisfies the requirements.
Therefore, the total number of mappings $f$ that satisfy the given conditions is
$$
S=\mathrm{C}_{10}^{7} \frac{7!}{7} \cdot \mathrm{C}_{3}^{3} \frac{3!}{3}=\mathrm{C}_{10}^{7} 6! \cdot 2!.
$$