COMBINATORICS at its PEAK...

There are 2 ways to arrange the dominoes in general, horizontal and vertical. Horizontal fills 2 spaces horizontally using 2 dominoes and vertical fills a space horizontally using a domino. Assuming A is horizontally and B is vertically.

We must count arrangements of:

$AAAAA$ has 1 arrangement.

$BBAAAA$ has 15 arrangements.

$BBBBAAA$ has 35 arrangements.

$BBBBBBAA$ has 28 arrangements.

$BBBBBBBBA$ has 9 arrangements.

$BBBBBBBBBB$ has 1 arrangement.

So, in total, the number of arrangements is $\boxed{89}$

Let $a_n$ be the number of ways to tile a 2 by n rectangle.

We can either start off with a vertical tile, which leaves a $2 \times (n-1)$ rectangle, or a horizontal tile, which leaves a $2 \times (n-2)$ rectangle. There are $a_n-1$ ways to arrange a 2 x (n-1) rectangle and $a_n-2$ ways to arrange a 2 x (n-2) rectangle.

Hence $a_n = a_n-1 + a_n-2$ .

There is one way to tile a $1 \times 1$ rectangle and 2 ways to tile a $2 \times 2$ rectangle, so $a_1 = 1$ and $a_2 = 2$ .

$a_3 = 3$

$a_4 = 5$

$a_5 = 8$

$a_6 = 13$

$a_7 = 21$

$a_8 = 34$

$a_9 = 55$

$a_{10}= \boxed{89}$

Hint. USE THE CONNECTION OF THE PROBLEM WITH THE FIBONACCI SEQUENCE'S TERM I'll try to post the solution in the near future.