Label the vertices of this $n$-polygon as $1$, $2$, $\cdots$, $n$ and the vertices of a qualified $k$-polygon be $A_1$, $A_2$, $\cdots$, $A_k$. Then there are two possible cases:

• $A_1 = 1$, $3\le A_2 < A_3 <\cdots < A_k \le (n-1)$, and $A_{i+1}-A_i\ge 2$ for all $2\le i \le (k-1)$
  
• $2\le A_1 < A_2 <\cdots < A_k\le n$, and $A_{i+1}-A_i\ge 2$ for all $1\le i \le (k-1)$

In the first scenario above, let's transform $\mathbb{A}=\{A_2,\ A_3,\ \cdots,\ A_{k}\}$ to $\mathbb{A'}=\{A_2-2,\ A_3 - 3,\ \cdots,\ A_{k}-k\}$. This transformation is bijective because every element in $\mathbb{A}$ has a corrsponding one in $\mathbb{A'}$, and vice vesa. Meanwhile, elements in $\mathbb{A'}$ satisfy $$1\le A_2-2 < A_3-3 < \cdots < A_{k}-k\le n-k - 1$$

A selection of $\mathbb{A'}$ is to choose $(k-1)$ distinct numbers from $\{1$, $2$, $\cdots$, $n-k-1\}$ which has $\binom{n-k-1}{k-1}$ ways.

In the latter case, let's transform $\mathbb{A}=\{A_1,\ A_2,\ A_3,\ \cdots,\ A_{k}\}$ to $\mathbb{A'}=\{A_1-1,\ A_2-2,\ A_3 - 3,\ \cdots,\ A_{k}-k\}$. This transformation is also bijective. At the same time, elements in $\mathbb{A'}$ satisfy $$1\le A_1-1< A_2-2 < A_3-3 < \cdots < A_{k}-k\le n-k $$

and form a selection of $k$ distinct numbers from $\{1$, $2$, $\cdots$, $n-k\}$. There are $\binom{n-k}{k}$ ways.

Hence, the final answer is $$\binom{n-k-1}{k-1} + \binom{n-k}{k} = \boxed{\frac{n}{k}\binom{n-k-1}{k-1}}$$