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Partial solution

I'd start with placing the rooks. If they don't hit each other, it means they are in different rows and different columns. That is, wherever they are actually placed, they allow two knights to be placed in the intersections in the other three rows and columns. This gives an upper bound for the possible placements:
the first rook can be placed on any of the 25 cells (factor of 25);
the second rook on the 16 cells which are neither same row, nor same column (factor of 16);
the two rooks cannot be differentiated (factor of 1/2);
the first knight on one of the remaining 9 tiles (factor of 9);
the other knight on as well, but not on the same tile (factor of 8);
the knight can be swapped as well (factor of 1/2);
which gives 7200 configurations, but again, these contain the ones, where the knights can hit each other or the rooks, so the final answer will be somewhat lower.

I can also provide a lower bound by noticing that the knights cannot hit each other for sure if they are on same color cells of the chessboard. By the pigeonhole principle, out of the 9 cells not hit by the rooks, at least 5 will be of the same color. By placing the knights on two of these 5, it is guaranteed they don't hit each other. This gives a factor of 5x4, where it was 9x8 in the previous one, so this provides a lower bound of 2000. It can be easily shown, that instead of 5x4, 5x4+4x3 is a more strict lower bound, giving 3200 as a lower bound for the total number of configurations. An even more strict lower bound comes from realizing that a knights hits 8 cells most, and that those are placed in 4 different columns/rows, each containing 2 cells or less, none of them in the same column/row as the knight itself. Thus after placing two rooks which don' hit each other and the knight, at most 4 cells are hit by only the knight. Considering that the rooks have left 9 cells unhit, and one of them is taken by the knight, there are at least 4 cells left for the second knight. So we managed to replace the previous factor by 9x4, which results in a final lower bound of 3600.

Partial solution

I'd start with placing the rooks. If they don't hit each other, it means they are in different rows and different columns. That is, wherever they are actually placed, they allow two knights to be placed in the intersections in the other three rows and columns. This gives an upper bound for the possible placements:
the first rook can be placed on any of the 25 cells (factor of 25);
the second rook on the 16 cells which are neither same row, nor same column (factor of 16);
the two rooks cannot be differentiated (factor of 1/2);
the first knight on one of the remaining 9 tiles (factor of 9);
the other knight on as well, but not on the same tile (factor of 8);
the knight can be swapped as well (factor of 1/2);
which gives 7200 configurations, but again, these contain the ones, where the knights can hit each other or the rooks, so the final answer will be somewhat lower.

EDIT: As Jaap Scherphuis pointed out, a lower bound I tried to give in a previous version of my answer had a flawed calculation.

Partial solution

I'd start with placing the rooks. If they don't hit each other, it means they are in different rows and different columns. That is, wherever they are actually placed, they allow two knights to be placed in the intersections in the other three rows and columns. This gives an upper bound for the possible placements:
the first rook can be placed on any of the 25 cells (factor of 25);
the second rook on the 16 cells which are neither same row, nor same column (factor of 16);
the two rooks cannot be differentiated (factor of 1/2);
the first knight on one of the remaining 9 tiles (factor of 9);
the other knight on as well, but not on the same tile (factor of 8);
the knight can be swapped as well (factor of 1/2);
which gives 7200 configurations, but again, these contain the ones, where the knights can hit each other or the rooks, so the final answer will be somewhat lower.

I can also provide a lower bound by noticing that the knights cannot hit each other for sure if they are on same color cells of the chessboard. By the pigeonhole principle, out of the 9 cells not hit by the rooks, at least 5 will be of the same color. By placing the knights on two of these 5, it is guaranteed they don't hit each other. This gives a factor of 5x4, where it was 9x8 in the previous one, so this provides a lower bound of 2000. It can be easily shown, that instead of 5x4, 5x4+4x3 is a more strict lower bound, giving 3200 as a lower bound for the total number of configurations.

Partial solution

I'd start with placing the rooks. If they don't hit each other, it means they are in different rows and different columns. That is, wherever they are actually placed, they allow two knights to be placed in the intersections in the other three rows and columns. This gives an upper bound for the possible placements:
the first rook can be placed on any of the 25 cells (factor of 25);
the second rook on the 16 cells which are neither same row, nor same column (factor of 16);
the two rooks cannot be differentiated (factor of 1/2);
the first knight on one of the remaining 9 tiles (factor of 9);
the other knight on as well, but not on the same tile (factor of 8);
the knight can be swapped as well (factor of 1/2);
which gives 7200 configurations, but again, these contain the ones, where the knights can hit each other or the rooks, so the final answer will be somewhat lower.

I can also provide a lower bound by noticing that the knights cannot hit each other for sure if they are on same color cells of the chessboard. By the pigeonhole principle, out of the 9 cells not hit by the rooks, at least 5 will be of the same color. By placing the knights on two of these 5, it is guaranteed they don't hit each other. This gives a factor of 5x4, where it was 9x8 in the previous one, so this provides a lower bound of 2000. It can be easily shown, that instead of 5x4, 5x4+4x3 is a more strict lower bound, giving 3200 as a lower bound for the total number of configurations. An even more strict lower bound comes from realizing that a knights hits 8 cells most, and that those are placed in 4 different columns/rows, each containing 2 cells or less, none of them in the same column/row as the knight itself. Thus after placing two rooks which don' hit each other and the knight, at most 4 cells are hit by only the knight. Considering that the rooks have left 9 cells unhit, and one of them is taken by the knight, there are at least 4 cells left for the second knight. So we managed to replace the previous factor by 9x4, which results in a final lower bound of 3600.

Partial solution

I'd start with placing the rooks. If they don't hit each other, it means they are in different rows and different columns. That is, wherever they are actually placed, they allow two knights to be placed in the intersections in the other three rows and columns. This gives an upper bound for the possible placements:
the first rook can be placed on any of the 25 cells (factor of 25);
the second rook on the 16 cells which are neither same row, nor same column (factor of 16);
the two rooks cannot be differentiated (factor of 1/2);
the first knight on one of the remaining 9 tiles (factor of 9);
the other knight on as well, but not on the same tile (factor of 8);
the knight can be swapped as well (factor of 1/2);
which gives 7200 configurations, but again, these contatin the ones, where the knights can hit each other, so the final answer will be somewhat lower.

I can also provide a lower bound by noticing that the knights cannot hit each other for sure if they are on same color cells of the chessboard. By the pigeonhole principle, out of the 9 cells not hit by the rooks, at least 5 will be of the same color. By placing the knights on two of these 5, it is guaranteed they don't hit each other. This gives a factor of 5x4, where it was 9x8 in the previous one, so this provides a lower bound of 2000. It can be easily shown, that instead of 5x4, 5x4+4x3 is a more strict lower bound, giving 3200 as a lower bound.

Partial solution

I'd start with placing the rooks. If they don't hit each other, it means they are in different rows and different columns. That is, wherever they are actually placed, they allow two knights to be placed in the intersections in the other three rows and columns. This gives an upper bound for the possible placements:
the first rook can be placed on any of the 25 cells (factor of 25);
the second rook on the 16 cells which are neither same row, nor same column (factor of 16);
the two rooks cannot be differentiated (factor of 1/2);
the first knight on one of the remaining 9 tiles (factor of 9);
the other knight on as well, but not on the same tile (factor of 8);
the knight can be swapped as well (factor of 1/2);
which gives 7200 configurations, but again, these contain the ones, where the knights can hit each other or the rooks, so the final answer will be somewhat lower.

I can also provide a lower bound by noticing that the knights cannot hit each other for sure if they are on same color cells of the chessboard. By the pigeonhole principle, out of the 9 cells not hit by the rooks, at least 5 will be of the same color. By placing the knights on two of these 5, it is guaranteed they don't hit each other. This gives a factor of 5x4, where it was 9x8 in the previous one, so this provides a lower bound of 2000. It can be easily shown, that instead of 5x4, 5x4+4x3 is a more strict lower bound, giving 3200 as a lower bound for the total number of configurations.