Let $n\in\mathbb{N}$ be a positive integer. We say that an $n\times n$-matrix $A$ with all entries in $\{0,1\}$ is $k$-regular for some $k\in \{0,\ldots,n\}$ if the sum of every row and the sum of every column of $A$ equals $k$. Let $M(n, k)$ be the number of $k$-regular $n\times n$-matrices with all entries in $\{0,1\}$.

It's easy to see that $M(n,1)=n!$. Moreover, a symmetry argument shows that $M(n,k) = M(n, n-k)$ for all $k\in \{0,\ldots,n\}$.

Question. Given $n>1$, is it true that for all $k\in\{0,\ldots, 1\}$ we have $M(n,k)\leq M(n, \lfloor\frac{n}{2}\rfloor)$?

Let $n\in\mathbb{N}$ be a positive integer. We say that an $n\times n$-matrix $A$ with all entries in $\{0,1\}$ is $k$-regular for some $k\in \{0,\ldots,n\}$ if the sum of every row and the sum of every column of $A$ equals $k$. Let $M(n, k)$ be the number of $k$-regular $n\times n$-matrices with all entries in $\{0,1\}$.

It's easy to see that $M(n,1)=n!$. Moreover, a symmetry argument shows that $M(n,k) = M(n, n-k)$ for all $k\in \{0,\ldots,n\}$.

Question. Given $n>1$, is it true that for all $k\in\{0,\ldots, n\}$ we have $M(n,k)\leq M(n, \lfloor\frac{n}{2}\rfloor)$?