The question is a bit hard to phrase. If I have a set of $n$ unique numbers $N=\{1, 2, 3, ... , n\}$, then the number of permutations where no position is repeated would naturally be the derangement of n.
Let's say I pick a single permutation from the derangement $!N$. I thus have my original set, $N$, and a single permutation $K_{1}$. No element in $N$ and $K_{1}$ share a common position. Now how many permutations of $N$ can I find, such that no positions are repeated in both $N$ and $K_{1}$?
Further extending this, if I have a set $N$, and $k$ permutations $K_{1}$, $K_{2}$,...,$K_{k}$ such that none of them have any positions in common with each other or $N$, how many permutations of $N$ are there which do not repeat any positions from $N$ or any $K$? An empirical example would be:
For $\{1, 2, 3, 4, 5\}$ there are $44$ possible permutations which don't share any position in common (derangement of 5).
For k=1, that is, $\{1, 2, 3, 4, 5\}$ and $\{2, 1, 4, 5, 3\}$, there are $12$ possible permutations which don't share any position in common.
For k=2, that is, $\{1, 2, 3, 4, 5\}$, $\{2, 1, 4, 5, 3\}$, and $\{3, 4, 5, 1, 2\}$, there are only 2 possible permutations.
For k=3, that is, $\{1, 2, 3, 4, 5\}$, $\{2, 1, 4, 5, 3\}$, $\{3, 4, 5, 1, 2\}$, and $\{5, 3, 1, 2, 4\}$ there would be just one possible permutation left: $\{4, 5, 2, 3, 1\}$.
How to formulate this in terms of $n$ and $k$?
Interestingly, it seems that depending on the choice of the additional permutation in question, the answer can change. For example, for k=1, if the existing permutation is $\{2, 4, 1, 5, 3\}$, instead of $\{2, 1, 4, 5, 3\}$ there seem to be $13$ possible permutations instead of $12$.
As such, it may not be possible to directly formulate a solution for this in terms of n and k only.