Problem 652

11. Clearly, $A(n, 1)=n$.

When $k \in \mathbf{Z}_{+}, k \geqslant 2$, and $n<2 k-1$,
$$A(n, k)=0 \text {. }$$

When $k \in \mathbf{Z}_{+}, k \geqslant 2$, and $n \geqslant 2 k-1$, let $\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}$ be a $k$-element subset of $\{1,2, \cdots, n\}$ that does not contain consecutive integers, and $a_{1}<a_{2}<\cdots<a_{k}$.

Then $\left\{a_{1}, a_{2}-1, a_{3}-2, \cdots, a_{k}-(k-1)\right\}$ is a $k$-element subset of $\{1,2, \cdots, n-(k-1)\}$.

Since $\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}$ and $\left\{a_{1}, a_{2}-1, a_{3}-2, \cdots, a_{k}-(k-1)\right\}$ are in one-to-one correspondence, we have $A(n, k)=\mathrm{C}_{n-(k-1)}^{k}$.