There are no tilings when \(n\) is odd:

• the area of a \(3\times n\) rectangle is odd, but
• the area of each tile is even.

Thus, we can't put together \(1\times 2\) pieces to make a \(3\times n\) rectangle.

 Answer: \(3\)           

If the rectangle can be split, then each of the top half and bottom half is \(3\times 2\), and so can be tiled in \(3\) ways.

This gives \(3 \cdot 3 = 9\) arrangements for the \(3\times 4\) rectangle.

Suppose now that the rectangle can't be split.

At least one tile crosses this imaginary boundary.

Suppose that the tile is in the leftmost column. This determines the entire layout.

We can reflect this tiling from left to right to obtain another such tiling.

If we start with a tile in the centre column, then we must arrive at one of these two tilings.

Therefore, the answer in the \(n=4\) case is \(9+2=11\).

Let \(h(n)\) be the number of tilings.

We know:

• \(h(2)=3\)
• \(h(4)=11\)
• \(h(n)=0\) whenever \(n\) is odd

Let \(n\) be an even integer with \(n \geq 6\).

There are \(h(n-2)\) possible tilings of the top \(3 \times n-2\) section.

There are \(3\) possible tilings of the bottom \(3 \times 2\) section.

In total, there are \(3\cdot h(n-2)\) tilings.

Arrangement 2

In Arrangement 2a, there are \(h(n-4)\) tilings, because we have essentially removed a \(3 \times 4\) rectangle from the bottom of the \(3 \times n\) rectangle.

This means that \(h(n)=3h(n-2)+2h(n-4)\ +\) \(+\) 

We can continue in this way to see that  \(+\)  \(=\ 2h(n-6)\ +\)  \(+\) 

Because the height of the \(3\times n\) rectangle is finite, this series will stop eventually.

How will this stop?

A \(3\times 2\) rectangle on top in the second-last arrangements

This gives

This allows us to conclude, for example

and so on.

Subtracting the second equation from the first, we obtain

Therefore, the answer to Problem 5 is