Let $\mathfrak{S}\_{2n}$ be the permutation group of the letters $[2n]=\{1,2,\dots,2n\}$. Call a permutation $\pi\in\mathfrak{S}\_{2n}$ has an *$n$-distant pair* if there is some $j\in [2n-1]$ such that $\vert\pi\_j-\pi\_{j+1}\vert=n$.


Define $A=$ set of $\pi\in\mathfrak{S}\_{2n}$ with exactly one $n$-distant pair and $B=$ set of permutations without $n$-distant pairs.



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> **Question.** Which inequality is true in general: $\vert A\vert>\vert B\vert$ or $\vert A\vert<\vert B\vert$? And your proof?
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$n$-distant permutations more than not
There is an injective mapping from $B$ to $A$, which can be constructed as follows:
for a permutation $\pi=(p\_1,p\_2,\dots,p\_{2n})\in B$, find the element $p\_k$ that pairs with $p\_1$ (i.e., $|p\_k-p\_1|=n$); clearly we have $k>2$. Map $\pi$ to the permutation
$\pi'=(p\_2,\dots,p\_{k-1},p\_1,p\_k,p\_{k+1},\dots,p\_{2n})$ (i.e., remove $p\_1$ from $\pi$ and insert it right before $p\_k$). 


It can be easily seen that $\pi'\in A$, and $\pi'$ has only one pre-image in $B$. Hence, $|B|\leq |A|$. 


Furthermore, for $n>2$, there exist permutations in $A$ that have no pre-images; namely, these are the permutations that start with an $n$-distant pair. It follows that $|B|<|A|$.


In fact, we can infer from the above argument that $|A\_n| = |B\_n| + 2n\cdot |B\_{n-1}|$.


**UPDATE.** Inclusion-exclusion gives
$$|B\_n| = \sum\_{j=0}^n (-1)^j \binom{n}{j} 2^j (2n-j)! ,$$
which matches sequence [A007060](https://oeis.org/A007060) and can be further used to derive an explicit formula for $|A\_n|$ (added as [A285850](https://oeis.org/A285850)).


