**Motivation.** Today my sons played a card game, in which a fixed number $n$ of cards was lying on the table. A move consists of adding an unused card to the cards on the table, and removing a card from the table, so that after the move, there are again $n$ cards on the table. Which made me ponder the following question.


**Formalization.** For any integer $n>1$ let $[\mathbb{N}]^n$ be the collection of subsets of $\mathbb{N}$ consisting of $n$ elements. For which integers $n>1$ is there a bijection $\varphi:\mathbb{N}\to [\mathbb{N}]^n$ such that for all $k\in\mathbb{N}$ we have $|\varphi(k) \cap \varphi(k+1)| = n-1$?



Inspired by a card game: finding a path through $[\mathbb{N}]^n$
$[\mathbb{N}]^n$, with edges between $a,b\in[\mathbb{N}]^n$ if $\#(a\cap b)=n-1$, is an infinite graph in which all vertices have infinite degree. Moreover, for any two vertices $a,b$ in $[\mathbb{N}]^n$ and any finite subset $E\subseteq[\mathbb{N}]^n$ not containing $a,b$, there is a path from $a$ to $b$ which does not pass through any vertices of $E$: to find it, one can consider some $N\in\mathbb{N}$ bigger than all natural numbers appearing in elements of $E$, and then you go from $a$ to $\{N+1,\dots,N+n\}$ and from $\{N+1,\dots,N+n\}$ to $b$ in $\leq 2n$ steps.


So one can create a Hamiltonian path in the following way:


* Number the vertices $u\_1,u\_2,\dots$ in $[\mathbb{N}]^n$.
* Start the path in the vertex $u\_1$.
* At each step, take the first vertex that you have not visited and visit it using a path that does not pass through any previously visited vertices.


