Determinantal symmetry: proof requested: Part I 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*}
 A: Use Lindstrom-Gessel-Viennot lemma for points $(-a, - j) $ (the first collection of points) and $(i, b) $ (the second collection), both $i, j$ vary from 0 to $c-1$. We see that the determinant counts the families of disjoint lattice paths starting in the points of the first collection and finishing in the second (allowed path directions are North and East).
Of course it may be possible only if we join respective points $(-a, - j) $ and $(j, b) $. For $j=0$ this path is a Young diagram $Y_1$ in $a\times b$ rectangle. For $j=1$ the first and the last steps must be East and North respectively, and intermediate steps form a Young diagram $Y_2$ in the $a\times b$ rectangle, and the assumption that the paths are disjoint is equivalent to the assumption that $Y_2$ is contained in $Y_1$ if we naturally identify the ground rectangles (shifting by the vector $(-1,1)$). Proceeding this way we see that the whole thing is the number of 3d Young diagrams in the $a\times b\times c$ box. Thus the symmetry. 
A: Just for the record:
In view of Fedor's nice reply, the interpretation gives away
$$\det\left[\binom{i+j+a+b}{i+a}\right]_{i,j=0}^{c-1}
=\prod_{i=0}^{a-1}\prod_{j=0}^{b-1}\prod_{k=0}^{c-1}
\frac{i+j+k+2}{i+j+k+1}.\tag1$$
An apparent symmetry!
As a follow-up to Mark Wildon's comment, let me add the below $q$-construction.
Let $(q)_k=(1-q)(1-q^2)\cdots(1-q^k)$ with $(q)_0:=1$. Also, define the $q$-binomial by
$$\binom{n}k_q=\frac{(q)_n}{(q)_k(q)_{n-k}}.$$ 
Then, we may generalize the above identity (1) as
$$\det\left[q^{-i^2-ai-bj}\binom{i+j+a+b}{i+a}_q\right]_{i,j=0}^{c-1}
=\prod_{i=0}^{a-1}\prod_{j=0}^{b-1}\prod_{k=0}^{c-1}\frac{1-q^{i+j+k+2}}{1-q^{i+j+k+1}}.$$
