Can one replace a 2x2 plate with a 1x4 plate in a patio vice versa? I need a little help with my solution to the following problem. 

Lasse's rectangular patio is covered with concrete . Some of the plates in the form of 2 x 2 tiles, others have the shape of the 1 x 4 tiles. One of the plates has been broken, but as a possible replacement Lasse only have a single plate of the other variety. Lasse can replace the broken plate with the one he has, by possibly rearranging the tiles that are on the patio?

So my answer was the following: 

If we introduce a color grading simlar to a chessboard, we would notice the following, a 2x2 plate will cover 2 white and 2 black tiles and a 1x4 plate also would cover over 2 black and 2 white tiles, because we don't know how large the patio is, neither do we know how the plates are distributed around the patio. By first intuition it ought to be possible for Lasse to rearrange the tiles because it would not affect the coverage of white and black tiles, and also does not affect the total number of boxes for the patio.

But it can't be done, it is due to assume that a 2x2 plate has been broken. Then, replace it with a 1 x 4 plate would change the parity for the dimension in which a 1-width plate is placed instead of a 2-width, this can not be compensated by rearranging plates likewise vice versa."


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*Graders mark:  "I don't know what you mean by the parity for the dimensions, also look over the color grading as such you can make the different tiles cover different amount of tiles". 


How can I argue the parity change in a better way? (Or isn't this the way to go?) And what kind of color grading would fit? We've only been introduced to problems where you solve it by "chesscoloring". 
 A: Here is an argument that it is impossible, even for other shapes than rectangles. More precisely, it shows that for any shape that can be tiled with such tiles at all, the parity of the number of $4\times1$ tiles is always the same, which of course excludes the possibility of gaining or losing just one such tile. My argument does not establish or use a complete characterisation of the valid shapes.
For every row and column, count the parity of the number of squares it has in the shape. Then combine these counts for all rows whose index is in the same class modulo$~4$, and similarly for the columns, giving four row parities and four column parities in all. Clearly every vertical $4\times1$ tile contributies one to each of the row parities, and no other tile does, so a neccesary condition is that all these parities are equal, and then it gives the parity of the number of vertical tiles. Similarly the column parities give the parity of the number of horizontal $4\times1$ tiles. This proves the claim that the shape determines the parity of the number of all such tiles.
