also evident that the function $\phi$ is injective.

Now let $Q\in{{\mathcal{Q}}}_n$ and let $K=\{a<b<c<d\}$ be a block of $Q$. Since $Q$ is noncrossing, each of the sets $$\label{inbetween}
    [a+1,b-1],\quad
    [b+1,c-1],\quad
    [c+1,d-1],\quad
    [d+1,4n-1]\cup[0,a-1]$$ must be a union of blocks, hence must have cardinality divisible by 4. In particular, $a\equiv c\not\equiv b\equiv d \pmod{2}$. It follows that replacing $K$ with the two pairs $\{a,c\},\{b,d\}$, and doing the same for every other block of $Q$, yields a $n$-basketball $B$ such that $\phi(B)=K$. Therefore $\phi$ is surjective.

The numbers $\frac{1}{3n+1}\binom{4n}{n}$ form sequence A002293 in the On-Line Encyclopedia of Integer Sequences [@EIS]. Other combinatorial objects enumerated in the same way include

-   quaternary trees with $n$ internal nodes;

-   dissections of a $(3n+2)$-gon into $n$ pentagons; and

-   rooted plane maps (see [@LW]).

Let $B$ be a basketball. An [*ear*]{} of $B$ is a quartet $\{i,j\},\{k,\ell\}$ such that the numbers $i,j,k,\ell$ are consecutive (in some order) modulo $4n$.

For instance, the basketball shown in Figure \[bijection-figure\] has two ears: $\{4,6\},\{5,7\}$ and $\{15,17\},\{16,18\}$.

\[thm:ear-thm\] Every basketball $B$ of order $n\geq 2$ has at least two ears.

We proceed by induction on $n$. The base case $n=2$ is amenable to proof by inspection, as there are only 