K2) Let $n$ be a natural number. Roger has a square garden of size $(2 n+1) \times(2 n+1)$. He erects fences to divide it into rectangular beds. He wants exactly two horizontal $k \times 1$ beds and exactly two vertical $1 \times k$ beds for every even number $k$ between 1 and $2 n+1$, as well as a single square bed of size $1 \times 1$ when he is finished. How many different ways can Roger divide his garden?
Solution Consider the 4 largest rectangles. We want to show that together they form the edge of the garden.

First consider a single vertical $1 \times 2 n$ rectangle. The short edge of this rectangle must touch a border (otherwise there would be an area with a width smaller than 1 above and below the rectangle, which cannot belong to any other rectangle).

We also claim that the long edge of this vertical rectangle must touch an edge of the garden: If this were not the case, then the horizontal distance to both edges would be a number strictly smaller than $2 n$. Consequently, both horizontal $2 n \times 1$ rectangles would have to lie in the free $(2 n+1) \times 1$ strip, which is either above or below the vertical rectangle. However, this is not possible because $2 n+2 n>2 n+1$.

We therefore conclude that every $1 \times 2 k$ rectangle must touch a corner. Similarly, every $2 k \times 1$ rectangle must touch a corner. We can see that there are exactly two possibilities for such an arrangement:
![](https://cdn.mathpix.com/cropped/2024_04_28_5d50c375e30aaac9179ag-05.jpg?height=418&width=1008&top_left_y=1250&top_left_x=570)

If we now remove these four longest rectangles, we are left with a square $(2 n-1) \times(2 n-1)$ garden in the middle. This must be divided into beds in the same way as in the original task (but with $n-1$ instead of $n$ ). So, in principle, we can repeat the same reasoning and should get the solution $2^{n}$. We prove this a little more formally with induction:

Assertion: There are exactly $2^{k}$ ways to divide a $(2 k+1) \times(2 k+1)$ garden as in the task.

Induction anchoring: For $n=1$, the argument above shows that there are only two possibilities for the beds of dimensions $2 \times 1$ and $1 \times 2$. The remaining area in the middle is then exactly the $1 \times 1$ square.

Induction step: According to the argumentation at the beginning, there are two possibilities to divide the edge of the $(2 n+1) \times(2 n+1)$-garden. According to the induction hypothesis (for $k=n-1$ ), there are exactly $2^{n-1}$ different possibilities for the inner $(2 n-1) \times(2 n-1)$ field. Since we can combine all possibilities for the edge with all possibilities for the inner field, there are a total of $2 \cdot 2^{n-1}=2^{n}$ possibilities.

This proves that the answer you are looking for is actually $2^{n}$.