Tiling of rectangular floor by $2\times 2, 1\times 4$ and $4\times 1$ tiles A rectangular floor is covered by $2\times 2$ and $1\times 4$ tiles.

One tile got smashed. There is a tile of the other kind available.
Show that the floor cannot be covered by rearranging the tiles.

Today I was discussing that question with my friend and he told me that one can use coloring technique to solve this problem. By coloring he meant assigning to each cell of rectangular floor one of the four colors $\{1,2,3,4\}$ and he suggested to consider the following coloring such as: $$1 \quad 2 \quad 1 \quad 2 \quad 1 \quad 2 \quad \dots \quad 1 \quad 2 \quad 1 \quad 2 \quad $$
$$3 \quad 4 \quad 3 \quad 4 \quad 3 \quad 4 \quad \dots \quad 3 \quad 4 \quad 3 \quad 4 \quad $$
$$1 \quad 2 \quad 1 \quad 2 \quad 1 \quad 2 \quad \dots \quad 1 \quad 2 \quad 1 \quad 2 \quad $$
$$3 \quad 4 \quad 3 \quad 4 \quad 3 \quad 4 \quad \dots \quad 3 \quad 4 \quad 3 \quad 4 \quad $$
$$\vdots $$
$$1 \quad 2 \quad 1 \quad 2 \quad 1 \quad 2 \quad \dots \quad 1 \quad 2 \quad 1 \quad 2 \quad $$
$$3 \quad 4 \quad 3 \quad 4 \quad 3 \quad 4 \quad \dots \quad 3 \quad 4 \quad 3 \quad 4 \quad $$
In this example rectangular has even number of columns and rows but of course it could have size smth like $(2n+1)\times 2m$ or $2n\times (2m+1).$
This coloring has the property that each $2\times 2$ tile covers exactly $4$ different colors and $1\times4$ or $4\times 1$ covers eaxctly 2 colors.  Intuitively I understand it is not possible  to cover with rearrangement but I cannot prove it rigorously.
Can anyone explain to me please? Thank you so much for your attention!
EDIT: What does "Show that the floor cannot be covered by rearranging the tiles" mean? Is he asking to prove that the floor cannot be covered by replacing the smashed tile with the new tile and rearanging all of them?
 A: It is enough to keep track of just a single color: say, the color $1$. The key difference between the tiles is that:

*

*A single $2 \times 2$ tile always covers a single tile of this color.

*A single $1 \times 4$ tile always covers $0$ or $2$ tiles of this color: an even number.

Therefore:

*

*If there is an odd number of this color, an odd number of $2 \times 2$ tiles must be used when we cover the entire floor.

*If there is an even number of this color, an even number of $2 \times 2$ tiles must be used when we cover the entire floor.

Swapping out a $2\times 2$ tile for a $1\times 4$ tile, or vice versa, changes the parity of $2\times 2$ tiles (from even to odd, or vice versa). Since we started with the correct parity to cover the floor, we now have the incorrect parity, and we cannot do it.
A: Using your numbering, consider the squares numbered $1$. Every $2\times 2$ covers exactly one $1$, which is an odd number of $1$'s, while every $1\times 4$ or $4\times 1$ covers either zero or two squares labeled $1$, which is always even. Therefore, in every tiling of a rectangle, the parity of the number of $1$'s must match the parity of the number of $2\times 2$ tiles. However, if you smash one tile and replace it with the other, then you would change the parity of the number of $2\times 2$ squares.
For example, consider a $5\times 4$ rectangle. Using the numbering below, the number of $1$'s is even. We see that we can tile this rectangle using either $0$, $2$, or $4$ squares. However, in any of those tilings, if you broke one of the squares and replaced it with a skinny tile, or if you broke a skinny tile and replaced it with a square, the number of squares would be odd. Since there are six $1$'s in the rectangle, an odd number of squares and several skinny tiles would always cover an odd number of ones, so there would be at least one $1$ leftover, so you could not tile the board with the modified tile set.
1 2 1 2 1
3 4 3 4 3    # of 1's = 6
1 2 1 2 1
3 4 3 4 3

0 squares     2 squares     4 squares
A B C D E     A A C D E     A A C C E
A B C D E     A A C D E     A A C C E
A B C D E     B B C D E     B B D D E 
A B C D E     B B C D E     B B D D E

A: Take a correct tiling of your floor with a set of tiles. You will have covered $\frac{nm}{4}$ ones, twos, threes and fours.
If one tile is replaced, say a $2\times 2$, you will expose a $1,2,3,4$. The new tile,  $4\times 1$, can only cover two of these twice, and so there is no covering available.
