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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, $$\det\left[q^{ij}\binom{x}{i+j+a}_q\right]_{i,j=1}^n =\det\left[q^{ij}\binom{x+n-i}{n+j+a}_q\right]_{i,j=1}^n.$$

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  • $\begingroup$ What if we write weights on edges for which the weighted paths are enumerated by q-binomials times certain powers of q? There is such a construction, coming from q-Pascal identity. $\endgroup$ – Fedor Petrov Mar 3 at 18:57
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Since everything is polynomial in $x$, we may consider large positive integer values of $x$.

The first is Lindström–Gessel–Viennot determinant for paths from the points $(-i,i)$ (collection $A$) to $(a+j,x-j-a)$ (collection $B$), always $i$ and $j$ go from 1 to $n$.

The second is LGV determinant for the paths from $(-j,j)$ (aha, again collection $A$) to $(n+a,x-i-a)$ (collection $C$). It is clear that the disjoint paths from $A$ to $C$ must go through $B$ and after $B$ they have unique continuation to $C$.clear

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Let $L_i$ and $R_i$ be the $i$-th rows of the left and right matrices. Then $$\begin{align} R_n &= L_n \\ R_{n-1} &= L_n + L_{n-1} \\ R_{n-2} &= L_n + 2L_{n-1} + L_{n-2} \\ R_{n-3} &= L_n + 3L_{n-1} + 3L_{n-2} + L_{n-3} \\ R_{n-4} &= L_n + 4L_{n-1} + 6L_{n-2} + 4L_{n-3} + L_{n-4}, \end{align}$$ and so forth, with the coefficients being binomial coefficients. Each sum is just a Vandermonde convolution.

So the second matrix can be derived from the first by elementary row operations, which we know preserve the determinant. (The simplest way is to work from the first row downwards, adding the required multiples of the rows below.)

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  • $\begingroup$ This is nice, thanks. Yet, it seems algebraic in nature. I hope for some combinatorial argument, such as David's answer to my earlier question. $\endgroup$ – T. Amdeberhan Mar 2 at 18:20

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