Let $A$ be an order-$n$ Latin square. A $k$-plex of $A$ is a set of entries , $k$ from each row and column and $k$ from each symbol.
My question is: Is there a Latin square with a large number of $k$-plexes?
More specifically, I am interested in this question for $k = \lambda n$ for some constant $0<\lambda\leq 1$ and for $n$ tending to infinity, and I want to show that the number of plexes is at least $$ e^{-o(n^2)} \cdot \binom{n^2}{\lambda n^2} $$
Initially, I thought that every Latin square should have many plexes. My intuition was that if each entry is chosen uniformly at random with probability $\lambda$, then the following events should be positively correlated.
$A$ - There are $\lambda n$ entries in each row.
$B$ - There are $\lambda n$ entries in each column.
$C$ - There are $\lambda n$ entries with each symbol.
Indeed, a paper by Ordentlich and Roth shows that $A$ and $B$ are positively correlated, and the same reasoning can show the same for $A,C$ and $B,C$. However, the paper "A Generalisation of Transversals for Latin Squares" by Wanless contains a construction of Latin squares with no $k$-plex when $k$ is odd, so the above argument can't hold for every Latin square.
So my question is: For a given $\lambda$, are there Latin squares (for every large enough $n$) with at least $e^{-o(n^2)} \cdot \binom{n^2}{\lambda n^2}$ $\lambda n$-plexes?
