C2) Let $n$ be a positive integer. Roger has a $(2 n+1) \times(2 n+1)$ square garden. He puts down fences to divide his garden into rectangular plots. He wants to end up with exactly two horizontal $k \times 1$ plots and exactly two vertical $1 \times k$ plots for each even integer $k$ between 1 and $2 n+1$, as well as a single $1 \times 1$ square plot. How many different ways are there for Roger to do this?
Solution: Consider the 4 largest plots Roger will fence off. We will prove they will comprise the border of the garden.

Consider a vertical $1 \times 2 k$ piece. Clearly, one of its short (horizontal) edges must touch the border, because otherwise there would be a narrow margin of width smaller than 1 on either side of the piece which cannot belong to any of the other rectangles.

We will show that a long (vertical) edge also touches the border. If this is not the case, then the horizontal space on either side of the rectangle is strictly less than $2 n$, meaning that both horizontal plots would need to be situated in the remaining $(2 n+1) \times 1$ strip above or below our rectangle. This is clearly impossible since $2 n+2 n>2 n+1$.

Therefore it is clear that a $1 \times 2 k$ piece must touch both a vertical and horizontal border, and therefore a corner; the same is true by symmetry for a $2 k \times 1$ piece. We therefore have one such piece for every corner, and it is simple to see there are only two configurations possible:
![](https://cdn.mathpix.com/cropped/2024_04_28_5d50c375e30aaac9179ag-21.jpg?height=412&width=1008&top_left_y=1161&top_left_x=570)

After removing these 4 pieces, we are now left with a $(2 n-1) \times(2 n-1)$ square in the middle, which has to be subdivided exactly like in the initial problem statement (for $n-1$ instead of $n$ ). Iterating the same argument should give us the answer $2^{n}$. We prove this more formally with induction.

Induction hypothesis: There are exactly $2^{k}$ possibilities to cover the $(2 k+1) \times(2 k+1)$-square. Base case: For $n=1$ we have the $2=2^{1}$ possibilities described above for the border. The remaining area is exactly the $1 \times 1$ square, which means we don't get more possibilities.

Induction step: By the reasoning above, We first choose one of two possibilities for the border of the $(2 n+1) \times(2 n+1)$-square and end up with a $(2 n-1) \times(2 n-1)$-square in the middle, which can be covered in $2^{n-1}$ different ways by the induction hypothesis for $k=n-1$. Since we can combine both possibilities for the border with all possibilities of the interior, we obtain $2 \cdot 2^{n-1}=2^{n}$ possibilities in total.

This proves that $2^{n}$ is indeed the answer.