Without loss of generality let the vertices be $0,1,2,\dots,n-1$, and assume $0$ is one of the vertices. We need to select the other $k-1$ vertices. This can be done by selecting positive integers $a_1,a_2,\dots, a_{k-1}$ such that $a_i\geq 2$ and $a_1+a_2+\dots+a_{k-1}\leq n-2$ and selecting the vertices of the form $a_1+a_2+\dots+a_i$.
The number of ways to do this is equal to the number of solutions in integers greater than $2$ to $a_1+a_2+\dots+a_{k-1}\leq n-2$. This is clearly the number of solutions to $a_1+a_2+\dots+a_k=n$ with each $a_i\geq 2$. This is equal to the number of solutions to $a_1+a_2+\dots+a_k=n-2k$ with $a_i\geq 0$. By star and bars there are $\binom{n-k-1}{k-1}$ solutions to this.
Hence the final answer is $\frac{\binom{n-k-1}{k-1}}{\binom{n-1}{k-1}}$