four basketballs of order two.

Suppose now that $n>2$. Let $B$ be a basketball and let $T=\{i,j\},\{k,\ell\}$ be a quartet with $i<j<k<\ell$.. If $T$ is an ear, then $B-T$ is a basketball on the ground set $[\ell+1,4n-1]\cup[0,i-1]$. By induction, $B-T$ contains at least two ears $E,F$. Without loss of generality, we have $\ell+1\not\in E$. Then $E$ is an ear of $B$.

Now suppose that $T$ is not an ear. Then at least two of the four sets of are nonempty. By a similar argument, each of those two sets contains an ear that is also an ear of $B$.

We point out several combinatorial problems related to bimatchings and basketballs.

1.  The set of $n$-basketballs is invariant under the operation of *rotation*: replacing each quartet $\{i,j\},\{k,\ell\}$ with $\{i-2,j-2\},\{k-2,\ell-2\}$, with all indices taken modulo $4n$. Let $r(n)$ be the number of $n$-basketballs up to rotation, so that $(r(1),r(2),\dots)=(1,2,6,22,103,614,3872,\dots)$. These numbers occur as sequence A103941 in [@EIS], enumerating unrooted loopless plane maps with $n$ edges [@LW Theorem 4.4]. There does not seem to be an obvious relationship between plane maps and basketballs.

2.  One might instead seek to count equivalence classes of basketballs up to *half-rotation*, replacing $\{i,j\},\{k,\ell\}$ with $\{k-1,\ell-1\},\{i-1,j-1\}$, or up 