In how many ways you can place 18 chess bishops on a 10x10 chequered Board without hitting each other?

Question

Calculate number of ways 18 chess bishops could be placed on a a 10x10 chequered Board without hitting each other.

Answer 1

Consider all the diagonals of the Board, including those that consist of a single cell. They are all parallel to one of the two directions, and there are exactly 19 diagonals for each of these directions. It follows that more than 19 bishops cannot be placed in the required way. Moreover, you can not place 19, since in this case each diagonal will be occupied by one Bishop, and then there will be an bishop on two corner diametrically opposite cells, but this is not possible.

So, 18 is the maximum possible number of bishops. From the above it follows that on the opposite corner cells (1,1) and (10,10), in the coordinate entry, the Bishop stands exactly one. (Recall that each of these cells is considered a diagonal.) Next, consider two diagonals parallel to the same direction, i.e. the cells (1,2), (2,1) and (9,10), (10,9). Each of them has exactly one Bishop (without this, you can't reach the number 18). Note that the centers of these four cells are located at the vertices of the rectangle. Let's assume that the Bishop is on (1,2). Then there is no Bishop on (2,1) or (9,10). This means that it must be present on (10,1) -- a cell that is centrally symmetrical (1,2). And if the Bishop was on (2,1) instead of (1,2), then the second Bishop in this rectangle will be on (10,9).

Next, we consider a rectangle formed by cells (1,3), (3,1), (10,8), (8,10). On the diagonal connecting the first two cells, there is an bishops, and it stands on one of these cells, since it is not on (2,2). This cell is under the Bishop's battle with (1,1) or (10,10). A similar conclusion is true for a pair of centrally symmetric cells. Therefore, there are exactly two possible cases: the bishops are located either on (1,3), (10,8), or on (3,1), (8,10).

Next, we think similarly. For cells (1,4), (4,1), (10,7), (7,10) consider the same rectangle. And here again, on the diagonal connecting the first two cells, there is one Bishop, and it must stand on the edge. On the square (2,3), it is hit by an already exposed Bishop from the field (1,2) or (9,10), and the same applies to the field (3,2). Here the bishops will be either (1,4), (10,7), or (4,1), (7,10).

This way, all the bishops are lined up on the Board border. In each of the rectangles, they can stand in two ways, and choosing one of these two ways does not affect anything. Among the rectangles we are considering, there are also two "degenerate" ones. The first of these was considered at the beginning, and it consists of a single diagonal running from (1,1) to (10,10). The second is the diagonal from (1.10) to (10.1). Here, too, there is a choice of two options - on which end we put the bishops.

Thus, the answer is $2^k$, where $k$ is the number of rectangles under consideration. (We have a choice of exactly two options for each such rectangle, including "degenerate" ones.) It is easy to understand that each rectangle contains exactly one cell of the lower horizontal, i.e. $k=10$.

Answer: $2^{10}=1024$ ways.