Count number of pair of sequences Lets say you have two sequences of non negative integers each of length $n$.
ie $(a_1,a_2,...,a_n)$ and $(b_1,b_2,...,b_n)$ such that
$\max(a_i) < k$
and
$\max(b_i) < k$
Game rule:
You can edit both sequence with $\mathrm{swap}(a_i, b_i)$ for $1 ≤ i≤ n$,
Goal:
$a_i ≤ a_j$ for all $i ≤ j$
and
$b_i ≤b_j$ for all $i ≤ j$
But not all initial sequence $a$ and $b$ can be solved. For example $(2,0)$ and $(0,1)$ is a pair of sequence that can't be solved.
Now given $n$ and $k$, count number of different pair of initial sequence $(a,b)$ that can be solved with game described above.
Example:
for $n=1$,$k=2$:
These are the cases: ${[(0),(0)],[(0),(1)],[(1),(0)],[(1),(1)]}$.
Hence answer would be $4$.
 A: Not an answer but an elaboration on P. Quinton's remark.
A pair of sequences $a$ and $b$ is solvable if and only if the pair of sequences
$$(\min (a_1,b_1),\ldots, \min(a_n,b_n))\quad\mbox{and}\quad (\max (a_1,b_1),\ldots,\max(a_n,b_n)) $$
is solved.
Proof sketch. $\Rightarrow$ Say $\min (a_1,b_1) > \min (a_2,b_2)$, then $a_1,b_1 > \min (a_2,b_2)$ contradicting solvability. Similar argument for the maximums. $\Leftarrow$ Winning strategy is clear.
Call a pair of sequences a pair of min/max sequences if it is invariant with respect to the min/max strategy. A pair of min/max sequences can be scrambled in $2^n$ different ways (for each index either swap or no swap). So the number of all solvable pairs of sequences is bounded above by $2^nP(n,k)$, where $P(n,k)$ is the number of pairs of min/max sequences of length $n$ and bound $k$.
There is a caveat. A pair of constant and equal sequences, for instance, is invariant with respect to scrambling. I don't know how to efficiently account for the $a_i=b_i$ cases that will otherwise lead to considerable overestimation.
