TITLE: "Gray code" for building teams
QUESTION [7 upvotes]: Motivation. In a team of $n$ people, we had the task to build subteams of a fixed size $k<n$ such that every day, $1$ person of the subteam is replaced by another person in the team, but not in the subteam (so that the new subteam consisted of $k$ again). The question arose if and for what choices of $k$ and $n$ the subteam schedule can be built to contain each choice of $k$ people out of the team exactly once.
Formal version. For any set $X$ and positive integer $k$, let $[X]^k$ be the collection of subset of $X$ having $k$ elements. For $n\in\mathbb{N}$ let $[n] =\{1,\ldots,n\}$. For integers $1< k < n$ we define a graph $G(n, k)$ by $V(G(n,k)) = [[n]]^k$ and $$E(G(n,k)) = \big\{\{A, B\} : A, B \in [[n]]^k \land |A\cap B| = k-1\big\}.$$
For what choices of $1< k < n$ does $G(n,k)$ have a Hamiltonian path? Bonus question: Replace "path" by "cycle" in the original question. (The bonus question need not be answered for acceptance.)

REPLY [6 votes]: Theorem. The graph $G(n,k)$ is Hamiltonian if $n\ge3$ and $0\lt k\lt n$.
Proof. If $k=1$ or $k=n-1$ it's obvious, because $G(n,k)\cong K_n$ in those cases. Now consider the graph $G=G(n,k)$ where $2\le k\le n-2$. Let
$$S=\{A\in[[n]]^k:1\in A\},\quad S'=\{A\in[[n]]^k:1\notin A\}.$$
Since the induced subgraphs $G[S]$ and $G[S']$ are isomorphic to $G(n-1,k-1)$ and $G(n-1,k)$ respectively, they are Hamiltonian by the inductive hypothesis. Moreover, since the graphs are edge transitive, we can choose a Hamiltonian cycle $C$ in $G[S]$ containing the edge
$$\{A,B\}=\{\{1,\dots,k\},\ \{1,\dots,k-1,k+1\}\}$$
and a Hamiltonian cycle $C'$ in $G[S']$ containing the edge
$$\{A',B'\}=\{\{2,\dots,k,k+2\},\ \{2,\dots,k+1\}\}.$$
Now we get a Hamiltonian cycle in $G$ by removing the edges $\{A,B\}$ and $\{A',B'\}$ from $C\cup C'$, and replacing them with $\{A,A'\}$ and $\{B,B'\}$.