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Yes, there is a simple reason why this happens, but instead of giving the "magic" bijection, I will describe how to construct it. I will need to express this in the language of partitions, where I'm more familiar, although it shouldn't be hard to switch back to the dominoes interpretation. Below, I assume the weight of a partition of $n$ is $q^n$, and the weight of a pair of partitions is the product of their weights.

Proposition A: There is a weight preserving bijection between tilings of an $n\times 1$ rectangle with $k$ dominoes and $a$ black squares and pairs of partitions $(P,Q)$ where $P$ has $k+a$ parts each of size $\le n-2k-a$, and $Q$ has $k+a$ parts $q_1\geq q_2\geq \cdots \geq q_{k+a}=1$ and $q_{i}-q_{i+1}\in \lbrace 1,2\rbrace$, where these differences are $=2$ exactly $k$ or $k-1$ times.

Proof: Let $p_{k+a}$ denote the number of white squares before the first black square or domino, $p_{k+a-1}$ denote the number of white squares before the second black square or domino etc. We will let $P$ be the partition $p_1,p_{2},\dots$. We let $q_{i+1}-q_i=2$ q_{k+a-i}-q_{k+a-i+1}=2$ if the $i$'th black square or domino is a domiono, and $q_{i+1}-q_i=1$ q_{k+a-i}-q_{k+a-i+1}=1$ otherwise. It should be clear that this is a bijection. Proving that it is weight preserving is also easy and is left as an exercise.

Proposition B: The partitions $Q$ from the previous theorem are in a weight preserving bijection (up to a power of $q$ which depends only on $k$ and $a$) with partitions which have $k$ parts which are $\le a$.

Proof: Take the partition $Q$ and subtract the partition $(k+a,k+a-1,\dots,1)$, so we obtain $Q'=(q_1-k-a,q_2-k-a+1,\dots,q_{k+a}-1)$. Now let $\hat{Q}$ be the transpose partition of $Q'$. Check that $Q'$ has $k$ distinct parts which are $\in [0,k+a-1]$. Therefore we make a fourth partition $R=\hat{Q}-(k-1,\dots,1)$. Check that $R$ has $k$ parts which are $\le a$.

Now let's denote by $B(x,y)$ the set of partitions contained in an $x\times y$ rectangle . (so partitions that have at most $x$ parts and each part is at most $y$). It is well known that the generating function for these is ${x+y\brack x}$.

Proposition C There is a weight preserving bijection between pairs of partitions $B(k+a,n-2k-a)\times B(k,a)$ and pairs of partitions $B(n-2k,k)\times B(n-2k-a,a)$.

Proof: Look up your favourite bijective proof of $${n-k \brack k+a}{k+a\brack a}={n-k\brack k}{n-2k\brack a}.$$ (Note that this is the part which messes things up a bit, so that the resulting final bijection won't be "obvious".)

So propositions A,B give a bijection between tilings with $k$ dominos and $a$ black squares and $B(k+a,n-2k-a)\times B(k,a)$. Applying proposition C, we get a bijection with $B(n-2k,k)\times B(n-2k-a,a)$. Finally since $B(n-2k,k)$ is in a weight preserving bijection (up to an inconsequential power of $q$) with domino tilings (no black squares) and $B(n-2k-a,a)$ is up to a power of $q$ the coefficient of $z^a$ in $(1+zq^{k+1})\cdots(1+zq^{n-k})$, we get the desired result. $\square$
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Yes, there is a simple reason why this happens, but instead of giving the "magic" bijection, I will describe how to construct it. I will need to express this in the language of partitions, where I'm more familiar, although it shouldn't be hard to switch back to the dominoes interpretation. Below, I assume the weight of a partition of $n$ is $q^n$, and the weight of a pair of partitions is the product of their weights.

Proposition A: There is a weight preserving bijection between tilings of an $n\times 1$ rectangle with $k$ dominoes and $a$ black squares and pairs of partitions $(P,Q)$ where $P$ has $k+a$ parts each of size $\le n-2k-a$, and $Q$ has $k+a$ parts $q_1\geq q_2\geq \cdots q_{k+a}=1$ and $q_{i}-q_{i+1}\in \lbrace 1,2\rbrace$, where these differences are $=2$ exactly $k$ or $k-1$ times.

Proof: Let $p_{k+a}$ denote the number of white squares before the first black square or domino, $p_{k+a-1}$ denote the number of white squares before the second black square or domino etc. We will let $P$ be the partition $p_1,p_{2},\dots$. We let $q_{i+1}-q_i=2$ if the $i$'th black square or domino is a domiono, and $q_{i+1}-q_i=1$ otherwise. It should be clear that this is a bijection. Proving that it is weight preserving is also easy and is left as an exercise.

Proposition B: The partitions $Q$ from the previous theorem are in a weight preserving bijection (up to a power of $q$ which depends only on $k$ and $a$) with partitions which have $k$ parts which are $\le a$.

Proof: Take the partition $Q$ and subtract the partition $(k+a,k+a-1,\dots,1)$, so we obtain $Q'=(q_1-k-a,q_2-k-a+1,\dots,q_{k+a}-1)$. Now let $\hat{Q}$ be the transpose partition of $Q'$. Check that $Q'$ has $k$ distinct parts which are $\in [0,k+a-1]$. Therefore we make a fourth partition $R=\hat{Q}-(k-1,\dots,1)$. Check that $R$ has $k$ parts which are $\le a$.

Now let's denote by $B(x,y)$ the set of partitions contained in an $x\times y$ rectangle. It is well known that the generating function for these is $\binom{x+y}{x}_q$.{x+y\brack x}$.

Proposition C There is a weight preserving bijection between pairs of partitions $B(k+a,n-2k-a)\times B(k,a)$ and pairs of partitions $B(n-2k,k)\times B(n-2k-a,a)$.

Proof: Look up your favourite bijective proof of $$\binom{n-k}{k+a}_q\binom{k+a}{a}_q=\binom{n-k}{k}_q\binom{n-2k}{a}_q.$$ ${n-k \brack k+a}{k+a\brack a}={n-k\brack k}{n-2k\brack a}.$$ (Note that this is the part which messes things up a bit, so that the resulting final bijection won't be "obvious".)

So propositions A,B give a bijection between tilings with $k$ dominos and $a$ black squares and $B(k+a,n-2k-a)\times B(k,a)$. Applying proposition C, we get a bijection with $B(n-2k,k)\times B(n-2k-a,a)$. Finally since $B(n-2k,k)$ is in a weight preserving bijection (up to an inconsequential power of $q$) with domino tilings (no black squares) and $B(n-2k-a,a)$ is up to a power of $q$ the coefficient of $z^a$ in $(1+zq^{k+1})\cdots(1+zq^{n-k})$, we get the desired result. $\square$
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I will provide

Yes, there is a sketch simple reason why this happens, but instead of giving the "magic" bijection, I will describe how to get such a bijective proofconstruct it. I will need to express this in the language of partitions, where I'm more familiar, although it shouldn't be hard to switch back to the dominoes interpretation. Below, I assume the weight of a partition of $n$ is $q^n$, and the weight of a pair of partitions is the product of their weights.

Proposition A: There is a weight preserving bijection between tilings of an $n\times 1$ rectangle with $k$ dominoes and $a$ black squares and pairs of partitions $(P,Q)$ where $P$ has $k+a$ parts each of size $\le n-2k-a$, and $Q$ has $k+a$ parts $q_1\geq q_2\geq \cdots q_{k+a}=1$ and $q_{i+1}-q_i\in q_{i}-q_{i+1}\in \lbrace 1,2\rbrace$, where these differences are $=2$ exactly $k$ or $k-1$ times.

Proof: Let $p_{k+a}$ denote the number of white squares before the first black square or domino, $p_{k+a-1}$ denote the number of white squares before the second black square or domino etc. We will let $P$ be the partition $p_1,p_{2},\dots$. We let $q_{i+1}-q_i=2$ if the $i$'th black square or domino is a domiono, and $q_{i+1}-q_i=1$ otherwise. It should be clear that this is a bijection. Proving that it is weight preserving is also easy and is left as an exercise.

Proposition B: The partitions $Q$ from the previous theorem are in a weight preserving bijection (up to a power of $q$ which depends only on $k$ and $a$) with partitions which have $k$ parts which are $\le a$.

Proof: Take the partition $Q$ and subtract the partition $(k+a,k+a-1,\dots,1)$. Then add a row of (k+a,k+a-1,\dots,1)$, so we obtain $k$ squares and then take Q'=(q_1-k-a,q_2-k-a+1,\dots,q_{k+a}-1)$. Now let $\hat{Q}$ be the transpose partition . Now subtract the of $Q'$. Check that $Q'$ has $k$ distinct parts which are $\in [0,k+a-1]$. Therefore we make a fourth partition $(a,a-1,\dots,1)$.R=\hat{Q}-(k-1,\dots,1)$. Check that $R$ has $k$ parts which are $\le a$.

Now let's denote by $B(x,y)$ the set of partitions contained in an $x\times y$ rectangle. It is well known that the generating function for these is $\binom{x+y}{x}_q$.

Proposition C There is a weight preserving bijection between pairs of partitions $B(k+a,n-2k-a)\times B(k,a)$ and pairs of partitions $B(n-2k,k)\times B(n-2k-a,a)$.

Proof: Look up your favourite bijective proof of $$\binom{n-k}{k+a}_q\binom{k+a}{a}_q=\binom{n-k}{k}_q\binom{n-2k}{a}_q.$$ (Note that this is the part which messes things up a bit, so that the resulting final bijection won't be "very pretty".)obvious".)

So propositions A,B give a bijection between tilings with $k$ dominos and $a$ black squares and $B(k+a,n-2k-a)\times B(k,a)$. Applying proposition C, we get a bijection with $B(n-2k,k)\times B(n-2k-a,a)$. Finally since $B(n-2k,k)$ is in a weight preserving bijection (up to an inconsequential power of $q$) with domino tilings (no black squares) and $B(n-2k-a,a)$ is up to a power of $q$ the coefficient of $z^a$ in $(1+zq^{k+1})\cdots(1+zq^{n-k})$, we get the desired result. $\square$
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