Rectangles in a modified chessboard What is the number of rectangles in a chessboard with the diagonal unit squares removed(of the 4 corner unit squares, two were removed which were diagonal)? I tried counting but it was becoming difficult. The rectangles which have the corner squares are difficult for me to count. Also, the source is asking for exact answer, where I think I make mistakes, please help. Thanks.
 A: The number of rectangles in an ordinary $8\times8$ board is
$$(1+2+\cdots+8)^2=36^2=1296$$
The number of such rectangles that include the upper lefthand corner square is
$$8\cdot8=64$$
and likewise for rectangles that include the bottom righthand corner square.  There is only one rectangle (namely the entire board) that includes both those squares.  Thus the number of rectangles that include neither of the two opposite corner squares is
$$1296-(64+64)+1=1169$$
A: Assuming that the removed squares are a8,b7,c6,d5,e4,f3,g2,h1 then the chessboard is partitioned in two connected components and any rectangle may lie just in the lower left or in the upper right triangle. So we count the number of rectangles in the lower left triangle accordingly to the position of the upper right corner of the rectangle. For instance, assuming that the upper right corner falls in a7, then we have $7$ choices for the rectangle. Assuming that the upper right corner falls in b6, then we have $2\cdot 6$ choices. This gives that the total number of rectangles is given by:
$$ 2\sum_{j=1}^{7}\sum_{k=1}^{8-j}jk=\sum_{j=1}^{7}j(9-j)(8-j)=\color{red}{420}. $$
