(A domino is a $1 \times 2$ or $2 \times 1$ rectangle. Determine the number of ways to place $n^{2}$ dominos on a $2 n \times 2 n$ grid so that every $2 \times 2$ square of the grid has at least two unoccupied cells in the same row or column.)
Consider a layout of $n^{2}$ dominoes. We start by dividing the grid into $n^{2}$ squares of $2 \times 2$. Each can contain at most two occupied cells, as there are $n^{2}$ dominoes of two cells each, each square has two occupied cells that share a side. Suppose, for the sake of contradiction, that there is at least one square where these two cells belong to two different dominoes, and consider among these the one with the minimum y-coordinate, then the minimum x-coordinate. Thus, the squares below and to the left of this one (if they exist) contain exactly one domino. But with the hypothesis of the statement, the only way to have two dominoes in contact is for them to form four aligned cells, so the two dominoes having a cell in the considered square must extend to the square to the left or the one below, which is absurd.

Thus, each square has exactly one domino, which can be positioned to the left, top, right, or bottom of the square. We can therefore divide the squares into four categories $G, H, D, B$ respectively, according to the positioning of the domino. The hypothesis of the statement then implies that the square to the left of a $G$ square is a $G$ square, the one above an $H$ square is an $H$ square, etc.

In this way, we deduce that all squares with x-coordinate and y-coordinate less than those of a $G$ or $B$ square are always $G$ or $B$ squares, so the broken line starting from the top-left corner and arriving at the bottom-right corner, separating the $G$ or $B$ squares from the $D$ or $H$ squares, consists only of segments of length 2 downwards and to the right. Similarly, the one starting from the top-right corner and arriving at the bottom-left corner, separating the $G$ or $H$ squares from the $D$ or $B$ squares, consists only of segments of length 2 downwards and to the left.

Conversely, given two such broken lines, they separate the grid into four parts corresponding to the $G, H, B, D$ squares. For each broken line, there are $\binom{2 n}{n}$ choices, so the total number of configurations is

$$
\binom{2 n}{n}^{2}
$$