how many ways to fill the room How many ways are there to fill a $3\times10$ room with $1\times2$ tiles?
I tried to solve this problem $\binom{30}{2}\times\binom{28}{2}\times\cdots$ but then I noticed that they may not be next to each other. What to do?
 A: Here's a more general approach, in which we'll find the number of ways $1 \times 2$ dominos can tile a $3 \times n$ room. Let $a_n$ be the number of ways we can tile a $3 \times n$ room. Similarly let's define $b_n$ as the number of ways we can tile a $3 \times n$ room, without tiling one corner.
Take a look at the right bottom corner. If the tile covering it is vertical, then it implies that the tile covering the right upper corner is horizontal. Therefore there are $b_{n-1}$ ways to tile the rest of the room.
If the tile covering the right bottom corner is horizontal we can consider 2 cases. If the tile above it horizontal, then we must have horizontal tile above it and there are $a_{n-2}$ ways to tile the rest of the room. If the tile above it is veritcal, then we have $b_{n-1}$ ways to tile the rest of the room. Hence:
$$a_n = a_{n-2} + 2b_{n-1}$$
Now let's consider a $3 \times n$ room without one corner. Then if we cover the two hanging fields, we have $a_{n-1}$ ways to cover the rest of the room. If it's horizontal, then we must have the 2 tiles above it horizontally too, so there are $b_{n-2}$ ways to cover the rest of the room. Hence:
$$b_n = a_{n-1} + b_{n-2}$$
Now $a_n - a_{n-2} = a_{n-2} - a_{n-4} + 2(b_{n-1} - b_{n-3}) = a_{n-2} - a_{n-4} + 2a_{n-2} \implies a_n = 4a_{n-2} - a_{n-4}$
From now here you can solve the recurrence relation and get the general formula for $a_n$. Or if you want you can recurrsively calculate $a_{10}$ starting with $a_0 = 1$ and $a_2 = 3$. 
Obviously when $n$ is odd, there are no solutions. So solving the equation we get:
$$a_{2n} = \frac{\sqrt{3} - 1}{2\sqrt{3}}(2-\sqrt{3})^n + \left(\frac 12 + \frac 1{2\sqrt{3}}\right)(2+\sqrt{3})^n$$
