Fix the set of $n$ green squares (one in each row), and consider how many ways there are to color $n$ blue squares of the remaining $n(n-1)$ white squares such that (a) there is one in each column and (b) there is one in each row: (b) is easy; there are $n-1$ choices per row left, so the answer is $B = (n-1)^n$. (a) depends; if there are $c_i$ white squares left in the $i^\textrm{th}$ column, then the answer is $A = \prod_i c_i$. But $\sum_i c_i = n(n-1)$ (that's how many white squares are left) and by AM-GM, \begin{align*} \frac{\sum_i c_i}{n} &\geq \sqrt[n]{\prod_i c_i} \\ n-1 &\geq \sqrt[n]{A} \\ B = (n-1)^n &\geq A \end{align*} so there are more blue square selections with one in each row. Furthermore, the inequality is strict for any set of green squares that is not one in each column, of which many clearly exist. So, simply by summing across all initial choices of green squares, we conclude that there are more colourings for which there is exactly one green square and exactly one blue square in each row, than colourings for which there is exactly one green square in each row and exactly one blue square in each column.