How to count number of ways to tile a board of the size $2 \times 10$? 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.
 A: Let $A_n$ be the number of tilings of the board $2\times n$.
Note that $A_1=1$ and $A_2=3$. Moreover for $n\geq 3$ the following linear recurrence holds
$$A_n=A_{n-1}+2A_{n-2}.$$
In fact consider the rightmost tile(s) of a tiling of $2\times n$. Either it is a  $2\times 1$, two  $1\times 2$ or one $2\times 2$.
If it is a  $2\times 1$, then removing we have a valid tiling of $2\times (n-1)$.
In the other TWO cases, then after removing them we have a valid tiling of $$2\times (n-2)$.
Finally it is now easy to compute the first ten numbers:  1, 3, 5, 11, 21, 43, 85, 171, 341, 683. 
See here for more properties of this sequence.
A: Let the number of tilings of a $2\times n$ board with the given tiles be $z_n$. We have $z_1=1$ and $z_2=3$.
Given a $2×n$ board whose leftmost $n-1$ columns are filled, I can add a single vertical domino to complete the tiling. If the leftmost $n-2$ columns are filled, I can add either a $2×2$ block or two horizontal dominoes to complete the tiling without forming a tiling of the leftmost $n-1$ columns (two vertical dominoes do this, so are rejected). This establishes the recurrence relation
$$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.
