Simplifying (and/or bounding) a sum of product of binomial coefficients The question title is quite overused, but I hope I haven't duplicated something.
Can this be simplified?
$$\sum_{k=1}^{p} \binom{q-1}{k-1}\cdot\binom{n-1}{b-1+k-1}\cdot \binom{m-1}{a-1+k-1}$$
Edit: All variables are non-negative integers, and $p \leq \min(m,n,q)$. Also,$m,n \geq q$.
I suspect that it might be possible because I was able to simplify a similar expression with two terms:
$$\sum_{k=1}^p \binom{m-1}{k-1}\cdot\binom{n-1}{a-1+k-1} = \binom{m+n-2}{a+n-2}$$
Edit: One of the reasons for simplifying this was to get a good upper bound on the expression; in the latter, it would just be when $a+p-2 = [\frac{m+n-2}{2}]$. Would it be possible to get an tight upper bound without simplifying it?
 A: You are entering the world of hypergeometric functions and I do not think that this is a simplification since moreover these functions corresponds to infinite summations.
To give you a taste of it, this would write
$$\binom{m-1}{a-1} \binom{n-1}{b-1} \, _3F_2(a-m,b-n,1-q;a,b;-1)-$$ $$\binom{q-1}{p}
   \binom{m-1}{a+p-1} \binom{n-1}{b+p-1} \, _4F_3(1,a-m+p,b-n+p,p-q+1;p+1,a+p,b+p;-1)$$
If $p=q$, the result is
$$\binom{m-1}{a-1} \binom{n-1}{b-1} \, _3F_2(a-m,b-n,1-q;a,b;-1)$$ and now, I am stuck !
A: Gosper's algorithm by hand:
Our term is $$t(k) = \binom{q-1}{k-1}\cdot\binom{n-1}{b-1+k-1}\cdot \binom{m-1}{a-1+k-1}$$ so $$\frac{t(k+1)}{t(k)} = \frac{(q-k)(n-k-b+1)(m-a-k+1)}{k(b+k-1)(a+k-1)}$$
The numerator and denominator are both in linear factors already and none of them differ by a constant integer, so we seek a polynomial $s(k)$ such that $$1 = (q-k)(n-k-b+1)(m-a-k+1)s(k+1) - (k-1)(b+k-2)(a+k-2)s(k)$$
If the leading term of $s(k)$ is $\alpha_i k^i$, the leading term of the RHS is $-2 \alpha_i k^{3+i}$, so the RHS can't possibly be a constant polynomial. Therefore there is no solution, and Gosper's algorithm tells us that the sum does not have a closed form as a hypergeometric term.
