Consider the determinantal function $$F(a,b,c):=\det\left[\binom{i+j+a+b}{i+a}\right]_{i,j=0}^{c-1}.$$ I would like to ask:

QUESTION.Can you provide an argument, combinatorial or otherwise, to prove the symmetry $$F(a,b,c)=F(c,a,b)=F(b,c,a)=F(a,c,b)=F(c,b,a)=F(b,a,c)$$ without actually computing the its values? I know how to justify this based on direct evaluation.

**Example.** The following are all equal: $F(2,3,4)=F(4,2,3)=F(3,4,2)$ or
$$\det\begin{pmatrix} 10&15&21&28\\20&35&56&84\\35&70&126&210\\56&126&252&462
\end{pmatrix}=
\det\begin{pmatrix} 15&35&70\\21&56&126\\28&84&210
\end{pmatrix}=\det\begin{pmatrix} 35&56\\70&126
\end{pmatrix}.$$

**POSTSCRIPT.** To add more examples to David Speyer's "mincing" in his reply to a related MO question, consider
$$G(a,b,c)=\det\left[\binom{a+b}{a+j-i}\right]_{i,j=0}^{c-1}.$$
Then, we have
\begin{align*} F(a,b,c)&=G(a,b,c) \\
G(a,b,c)&=G(c,a,b)=G(b,c,a)=G(a,c,b)=G(c,b,a)=G(b,a,c).
\end{align*}