Proving that the number of ways to divide a 2-high-by-n-wide rectangle by 2-wide-by-1-high dominoes is equal to $\sum_j {n-j\choose j}$

Question

I would like to prove that the number of way of dividing a 2-high-by-n-wide rectangle into dominoes so that $2j$ dominoes are horizontal is ${n-j\choose j}$ and deduce that $U_n$ (where $U_n$ is the number of ways to divide a 2-high-by-n-wide rectangle into 2-wide-by-1-high dominoes) =$$\sum_j {n-j\choose j}$$ where this sum is over all the integers $j$ with $0\le j\le \frac{n}{2}$.

I understand that trivially for a 2-high-by-n-wide rectangle you can divide it by exactly $2j=n$ horizontal dominoes or by $n$ vertical dominoes or some combination of vertical and horizontal dominoes, but how can I use this knowledge to construct the proof?

Answer 1

Since within each pair of horizontal dominos the two pieces must necessarily be aligned (otherwise there is no way to complete the large rectangle), each pair create a $2\times 2$ square. If we have $2j $ pairs of horizontal dominos, i.e. $j $ squares, they cover an area corresponding to a $2 \times 2j \,\,$ rectangle. So, to cover the remaining area corresponding to a $2\times (n-2j) \,\,$ rectangle with vertical dominos, we need $n-2j \,\,$ vertical pieces. In total, therefore we have $(n-2j)+j=n-j \,\,\,\,$ pieces to place.

Now note that we can place the $n-2j \,$ vertical dominos and the $j $ squares in any order to create the sequence of $n-j \, $ elements. Also note that, when we have decided the positions of the $j $ squares, the positions of the vertical dominos are automatically defined (they fill all remaining blank places in the sequence). Therefore, for a given $j $, the number of ways in which we can place the pieces is equal to the combinations of $j$ items chosen among $n-j \,$ items. Summing this result over all values of $j $ you obtain the result.