This question was in my course book but without an answer.
From a chessboard, two boxes each located on an opposite corner, get cut away. So for example the most top-left and bottom-right box get cut away. Is it possible to cover the whole chess-board with Domino-stone's, when an half of the Domino equals the size of one box on the chess-board (1 domino-stone covers two boxes).
My solution (math): Possible
Amount of boxes on this chessboard? 64 - 2 = 62 (Two cut away)
Amount of Domino's needed to cover the whole board: 31 (62 / 2)
Amount of horizontal boxes: 7 + 7 + (8 * 6) - 14
=> #horizontalFirstRow + #horizontalLastRow + #allOtherBoxes - #verticalBoxes = 48
How many Domino's fit on 48 horizontal boxes? 48/2 = 24
31 - 24 = 7 (7 Domino's left)
We have 14 vertical boxes, so 14/2 = 7 (Used the 7 left domino's), so 0 domino's left and all boxes covered.
My solution (graphical): Impossible You will have to zig-zag the domino-stones, but in the 7th row you won't be able to do the zig-zag because the bottom-right corner is cut-away.
Question: Is it possible or not? And how to prove that mathematically?
The issue with your first approach is that it only counts the squares without taking into account where the dominoes are placed.
The simplest approach here is to notice that each domino must cover one black square and one white square, but the two squares removed are the same color, so we cannot cover the board with dominoes.