A $7\times 1$ board is completely covered by $m\times 1$ tiles without overlap; each tile may cover any number of consecutive squares, and each tile lies completely on the board. Each tile is either red, blue, or green. Let $N$ be the number of tilings of the $7\times 1$ board in which all three colors are used at least once. For example, a $1\times 1$ red tile followed by a $2\times 1$ green tile, a $1\times 1$ green tile, a $2\times 1$ blue tile, and a $1\times 1$ green tile is a valid tiling. Note that if the $2\times 1$ blue tile is replaced by two $1\times 1$ blue tiles, this results in a different tiling. Find $N$.
This problem can be solved by first considering how to join seven $1\times 1$ tiles into no less than $3$ pieces. This is equivalent to partitioning $7$ as the sum of no less than three positive integers. Given the involved number is relatively small, this problem can be tackled by counting the number of positive integers solutions to the equation
$$\begin{equation}\label{eq_join} x_1 + x_2 + \cdots + x_k = 7 \end{equation}$$
where $k=3,\ 4,\ 5,\ 6,\ 7$. Each of these cases is a known basic pattern.
Then, assuming these seven $1\times 1$ titles are joined into $k$ piece, then the next step is to count the ways to color these piece so that each of these three colors appear at least once. This is equivalent to putting $k$ distinguishable balls into $3$ distinct boxes, no empty box is permitted.
Therefore the answer is $$15\times 6+ 20\times 36+ 15\times 150+ 6\times 540 + 1\times 1806=\boxed{8106}$$