4. In how many ways can 8 identical rooks be placed on an $8 \times 8$ board symmetrically with respect to the diagonal passing through the bottom-left corner?
Solution. On an $8 \times 8$ board, there are 8 diagonal and 56 non-diagonal cells, the latter of which can be divided into 28 pairs of cells symmetric relative to the diagonal. All placements of rooks will be divided into 5 non-overlapping classes - in the $m$-th class, we will include placements where $m$ pairs of rooks fall on the diagonal.

When $m=0$, no rook stands on the diagonal, and thus all rooks occupy 4 out of 28 pairs of symmetric non-diagonal cells. There will be $C_{28}^{4}=\frac{28 \cdot 27 \cdot 26 \cdot 25}{4 \cdot 3 \cdot 2 \cdot 2 \cdot 1}=20475$ such placements. When $m=1$, exactly two rooks are on the diagonal, which can be arranged in $C_{8}^{2}=$ 28 ways. The remaining 6 rooks occupy 3 out of 28 pairs of symmetric non-diagonal cells. Therefore, there will be $C_{8}^{2} \cdot C_{28}^{3}=91728$ such placements.

By similar reasoning, we find the number of required placements for $m=1$ and $m=3$, which are $C_{8}^{4} \cdot C_{28}^{2}=26460$ and $C_{8}^{6} \cdot C_{28}^{1}=784$ respectively. Finally, when $m=4$, there is only one placement where all rooks fall on the diagonal. The total number of symmetric placements is

$$
C_{28}^{4}+C_{8}^{2} \cdot C_{28}^{3}+C_{8}^{4} \cdot C_{28}^{2}+C_{8}^{6} \cdot C_{28}^{1}+1=139448
$$