Placing $9$ black and $9$ white rooks on $ 6\times6$ board without any attacks between different colours

You can always permute rows and columns, so if there is any solution, there is one for which the black rooks occupy some initial $k$ rows and $l$ columns. One must have $kl\geq9$ and $(6-k)(6-l)\geq9$, and some simple algebra will tell you that $k=l=3$ is the only solution to this. So one must count the number of ways to select $3$ rows and $3$ columns for the black rooks (the remainder being for white), and that number is $$ \binom 63^2=400. $$.

If white rooks and black rooks are indistinguishable, See the configuration below and black and white could be interchanged. Srinivas pointed out that we could stack up in columns too. That put if white occupies three columns to the top, the remaining columns could be occupied by Black to the bottom. Number of ways, a set of three white could occupy 3 columns is ${6\choose3}$ to the top and the rest of the columns should be occipied by Black. The answer is $({6\choose3})^2$

Marc, Understood. You have the full credit.