6. Given the set $A=\{1,2, \cdots, 10\}$. Define the function $f: A \rightarrow A$ such that:
(1) For any $x, y \in A$, if $x \leqslant y$, then $f(x) \leqslant f(y)$;
(2) $f(3)=4$.
The number of such functions $f$ is $\qquad$
6.17160 .

First consider the mapping from $\{a+1, a+2, \cdots, a+k\}$ to $\{b+1, b+2, \cdots, b+t\}$, which can be viewed as a mapping from $\{1,2, \cdots, k\}$ to $\{0,1, \cdots, t-1\}$. In fact, by arranging $k$ sticks and $t-1$ balls in a row, the number of balls to the left of the $k$-th stick can be taken as $f(k)$, and the number of such arrangements is $\mathrm{C}_{k+t-1}^{k}$. The original problem can be divided into two parts: one is $f(1)$ and $f(2)$, which can be viewed as a mapping from $\{1,2\}$ to $\{1,2,3,4\}$, with the number of such mappings being $\mathrm{C}_{5}^{2}=10$; the other is $f(4)$ and $f(10)$, which can be viewed as a mapping from $\{4,5, \cdots, 10\}$ to $\{4,5, \cdots, 10\}$, with the number of such mappings being $\mathrm{C}_{13}^{7}=1716$.
In summary, the total number is 17160.