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How would I go about proving this mathematically? Having looked at a proof for a similar question I think it requires proof by induction.

It seems obvious that it would be even by thinking about the first few cases. As for $n=0$ there will be no horizontal dominoes which is even, and for $n=1$ there can only be one vertical domino so there are $0$ horizontal dominoes, which is again even. Then for $n=2$ you can have either two horizontal or two vertical dominoes which again gives $0$ or $2$ horizontal dominoes which is again an even number. And so on for n greater than $2$.

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?

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    Do you mean $2\times1$ dominoes laid vertically on a space which is two squares wide? – 2017-02-05
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    ok I think you mean a space 2 squares high and n wide. – 2017-02-05
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    Okay, what am I not understanding? Let $n = 4$ and the strip be 2x4. You can have i) 0 even horizontal and 4 vertical, ii) 1 odd horizontal and 2 vertical, or iii) 2 even horiontal and 0 vertical. The answer is obviously false. Or did you mean and even number of *vertical* dominoes. In that case the answer is false but you do have that the parity of vertical must be the same parity as n which is pretty obvious as for the number to be less then n the potential vertical dominoes are being replace be horizontal dominos which must take up an even number of spaces. – 2017-02-06

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Mathematically you can design $t_n$ to be the number of horizontal dominoes and then solve the recursion $ t_n = 2 + t_{n-2}$ for with inital values $ t_1 = 0 , t_2 = 2$.

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If it's only $2$ high then any domino laid horizontally, and offset from the one above or below, leaves a single space above and a single below, which cannot be filled without leaving another single space.

Therefore any horizontal domino must be one of a pair laid one directly above the other.

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Ah, you mean the number of horizontal dominos in BOTH rows combined is even!

Consider each square in the grid. Either it is occupied by a vertical or horizontal domino. If it is occupied by a vertical domino, than so is the square in the grid directly below or directly above. If it is occupied by a horizontal domino then it is not occupied by a vertical domino and neither is the square directly above or below.

So each of the two rows have the exact same composition. So the number of horizontal dominos in the first row are the same number (and int the exact same positions) as the number of horizontal dominoes in the second row.

Thus there is an even number of them.

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    I'm not sure you've understood the question as OP intends it. The rectangle is a horizontal strip with height $2$ and width $n$. – 2017-02-06
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    Um... so? Lay out any $k \le n/2$ horizontal dominoes and leave the rest vertical. $k$ can be either even or odd if $n \ge 2$. (Otherwise it must be $0$ if $n \le 1$ but that obviously stops being the case if $n \ge 2$. – 2017-02-06
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    Consider a rectangle of height $2$ and width $3$. Cover the rectangle using an odd number of horizontal dominoes. I don't see how you manage that. – 2017-02-06
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    I suppose it should be stipulated that you're not allowed to chop any of the dominoes in half. – 2017-02-06
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    Oh, in *BOTH* rows. well the number of horizontal dominoes in each row must be equal as any veritcal domino cuts through both rows. – 2017-02-06
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    Well, yes. Just a further word to the wise that before one judges a question bizarre, one might make sure that there's a common interpretation. Of course, sometimes the question really just is bizarre... :-P – 2017-02-06
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We assume, given the description of the problem in the OP, that the rectangle is a horizontal strip of height $2$ and width $n$.

Proof by Contradiction. Suppose we had a covering of the $2 \times n$ rectangle containing an odd number of horizontal dominoes. Then either the upper row or the lower row contains more dominoes than the other; without loss of generality, assume the upper row contains more. Then the upper-row horizontal dominoes cover more squares than the lower-row ones do. But vertical dominoes must cover an equal number of upper and lower dominoes. So more upper-row squares are covered than lower-row squares, violating the definition of the rectangle. Ergo, no such covering is possible.

Proof by Induction. As you point out the number of horizontal dominoes is clearly zero (and therefore even) for $n = 0$ or $1$. For $n \geq 2$, consider the last two squares on the right. There are two cases: In one case, they are covered by a single vertical domino, in which case the number of horizontal dominoes is equal to the number in the first $2 \times n-1$. In the other case, they are covered by two horizontal dominoes, in which case the number of horizontal dominoes is two more than the number in the first $2 \times n-2$. In either case, the parity of the number of horizontal dominoes is unaffected, so the result obtains.

You've added a second question: how to show that the number of ways to cover the $2$-by-$n$ rectangle with $2j$ horizontal dominoes (and therefore $n-j$ vertical ones).

The easiest way I can think of to proceed is to observe that each row must look identical. Vertical dominoes perforce look identical in each row, and if the horizontal ones don't line up one on top of the other, then there will be an extra space left over somewhere. (Actually, two extra spaces, but at any rate—extra spaces.)

Therefore, we can just count the number of ways that we can cover a single strip of $n$ squares with pieces of either one square (those represent half a vertical domino) or two squares (those represent a horizontal domino with a matching domino in the other row). If there are $j$ two-square pieces altogether in the strip, then there must be $n-j$ pieces in the row, because the two-square pieces collectively cover $2j$ squares, leaving the remaining $n-2j$ squares to be covered by $n-2j$ one-square pieces: $n-2j+j = n-j$ pieces in all.

Each arrangement corresponds to choosing $j$ of those $n-j$ pieces to be the two-square pieces, so the number of ways of doing that is

$$ \binom{n-j}{j} $$

I'm assuming you can handle the rest of the deduction.

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    Is there a way to prove the second statement? @BrianTung – 2017-02-08
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    @DavinaGoodman: Which statement do you mean by "the second statement"? – 2017-02-08
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    Sorry I meant the second part of the 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 $\sum_j {n-j\choose j}$ and deduce that $U_n$.... and so on – 2017-02-09
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    @DavinaGoodman: Added. – 2017-02-09
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Coloring Argument

Let's say that the $n$-by-$2$ board is colored black on one $n$-by-$1$ row and white on the other row. Every vertical domino takes a square of each color, whilst a horizontal domino can only take two squares of the same color. Since the number of black squares is the same as the number of white squares, we conclude that the number of black horizontal dominoes is the same as the number of white horizontal dominoes, rendering the total number of horizontal dominoes even.

Algebraic Argument

Suppose that, for some $A(x,y),B(x,y)\in\mathbb{Z}[x,y]$, $$\left(1+x+x^2+\ldots +x^{n-1}\right)\,(1+y)=A(x,y)\cdot (1+x)+B(x,y)\cdot (1+y)\,.$$ (In other words, if $x^{i-1}y^{j-1}$ is a term in $A(x,y)$, then there is a horizontal domino with the left square on the $i$-th row and $j$-th column. Likewise, if $x^{i-1}y^{j-1}$ is a term in $B(x,y)$, then there is a vertical domino with the top square on the $i$-th row and $j$-th column.) Plugging in $x=1$ and $y=-1$ to get $$A(1,-1)=0\,.$$ However, $A(1,1)$ is the number of horizontal dominoes and $$A(1,1)\equiv A(1,-1)=0\pmod{2}\,.$$ Thus, for any such arrangement, there are always an even number of horizontal dominoes.