Encouraged by David's proof for my earlier MO question, let's consider a similar problem.

I can prove the below equality by **computing** each of the two sides, directly. That means, there is an algebraic proof. Observe that one of the matrices is symmetric (in fact, Hankel), while the other carries a different structure.

QUESTION.Is there a combinatorial or conceptual reason for this equality? $$\det\left[\binom{x}{i+j+a}\right]_{i,j=1}^n =\det\left[\binom{x+n-i}{n+j+a}\right]_{i,j=1}^n.$$ Here, $a\in\mathbb{Z}^{\geq0}$ and $x$ is an indeterminate.

**POSTSCRIPT.**

Let me add a $q$-analogue. Denote $(q)_n=(1-q)\cdots(1-q^n)$ and $\binom{n}k_q=\frac{(q)_n}{(q)_k(q)_{n-k}}$. Then, $$\det\left[q^{ij}\binom{x}{i+j+a}_q\right]_{i,j=1}^n =\det\left[q^{ij}\binom{x+n-i}{n+j+a}_q\right]_{i,j=1}^n.$$