TITLE: How to count number of ways to tile a board of the size $2 \times 10$?
QUESTION [6 upvotes]: In how many ways can we tile the board of the size $2 \times 10$ using the pieces of the format $2×1$, $1×2$, and $2×2$?

The picture above shows six different tilings of the board $2×10$ using the described pieces.
I have an idea of how to do it the long way by first counting the number of ways to tile with just using the $2×2$ and $1×2$, then I could do the same thing for the other two combinations then eventually counting the number of ways when using all three sizes, but I want to learn how to solve it the quicker way and using math.  
I have a textbook in applied combinatorics, what do I need to learn in order to solve a problem like this? I think this has to do with recursion but I am not completely sure.

REPLY [2 votes]: Let the number of tilings of a $2\times n$ board with the given tiles be $z_n$. We have $z_1=1$ and $z_2=3$.
Given a $2×n$ board whose leftmost $n-1$ columns are filled, I can add a single vertical domino to complete the tiling. If the leftmost $n-2$ columns are filled, I can add either a $2×2$ block or two horizontal dominoes to complete the tiling without forming a tiling of the leftmost $n-1$ columns (two vertical dominoes do this, so are rejected). This establishes the recurrence relation
$$z_n=z_{n-1}+2z_{n-2}$$
and $z_{10}$ can be found by continuing up the indices:
$$z_1=1,z_2=3$$
$$\{z_1,z_2,z_3,\dots,z_{10}\}=\\
\{1,3,5,11,21,43,85,171,341,683\}$$
Hence there are 683 ways to tile a $2×10$ board with the given conditions. This is the Jacobsthal sequence, indexed as A001045 in the OEIS.