Extremal combinatorics

I am trying to prove the following claim, or at least find something related to it. Assume we have a complete bipartite graph with partitions of size $n, m$. We then have a family of matchings of this graph such that each matching is of size $k$ and any two matchings of the family are incompatible, i.e. their union is not a matching. Assuming $n \geq m$, is it true that the maximal size of such a family is $(n)! / (n - k)!$, which corresponds to fixing $k$ elements of the partition of size $m$ and producing all possible matchings of size $k$ covering the fixed vertices?

Call the family of $k$-matchings in which no two are compatible $\mathcal F$.

Let $M$ be a matching of size $m$ in $K_{m,n}$. There are $\binom mk$ sub-matchings of $M$ with size $k$, and only one of them can appear in $\mathcal F$, because any two of them are compatible.

Averaging over all choices of $M$, we see that at most $1$ in $\binom mk$ of the matchings of size $k$ can appear in $\mathcal F$.

In total, there are $k! \binom nk \binom mk$ matchings of size $k$, so if at most $1$ in $\binom mk$ of them appear in $\mathcal F$, then $|\mathcal F| \le k! \binom nk = \frac{n!}{(n-k)!}$.

(This is essentially the same idea as the one in Katona's proof of the Erdős–Ko–Rado theorem.)