8. given an arbitrary floor plan composed of $n$ unit squares. Albert and Berta want to cover this floor with tiles, each tile having the shape of a $1 \times 2$ domino or a T-tetromino. Albert only has tiles of one color available, while Berta has dominoes of two colors and tetrominoes of four colors. Albert can cover this floor plan with tiles in $a$ ways, Berta in $b$ ways. Assuming that $a \neq 0$ applies, determine the ratio $b / a$.

## 1st solution
We call Albert's covers colorless and Berta's colored. Let $A$ and $B$ be the set of colorless and colored overlaps, respectively. By "forgetting the color" we obtain a mapping $\varphi: B \rightarrow A$ and we claim that this mapping is $2^{\frac{n}{2}}: 1$. So we have to show that every colorless cover can be colored in exactly $2^{\frac{n}{2}}$ ways. To do this, weight each unit square of the floor plan with the value $\sqrt{2}$. Each colorless tile can now be colored in exactly as many ways as the product of the weights that the tile covers (namely 2 or 4 depending on the tile type). According to the product rule, the number of ways to color a fixed colorless overlay is equal to the product of the number of ways to color each of the tiles individually. However, this is simply the product of the weights in all unit squares of the floor plan, i.e. equal to $(\sqrt{2})^{n}=2^{\frac{n}{2}}$.

We have thus shown that $b_{n}=2^{\frac{n}{2}} a_{n}$ always applies regardless of the shape of the floor plan. The condition $a_{n} \neq 0$ is not required for this.

## 2nd solution

We show again that every colorless cover can be colored in $2^{\frac{n}{2}}$ ways. Assume that the colorless cover consists of $k$ dominoes and $l$ tetrominoes. Then $2 k+4 l=n$, and the number of colorings is equal according to the product rule

$$
2^{k} \cdot 4^{l}=2^{k+2 l}=(\sqrt{2})^{n}
$$