Sum of products of binomials Sums of products of binomial coefficients often have simpler expression which do not involve any summation. Examples are the elementary $$\sum_{i=0}^k\binom{a}{i}\binom{b}{k-i}=\binom{a+b}{k}$$ or the more complicated $$\sum_{i=0}^{\text{min(a,b)}}\binom{x+y+i}{i}\binom{y}{a-i}\binom{x}{b-i}=\binom{x+a}{b}\binom{y+b}{a}$$ or many others like Vandermonde-Chu identity etc. 
I am looking for a similar formula for the following sum $$S(a,b):=\sum_{i=0}^k\binom{a}{k-i}\binom{b+i}{i}$$ where here $k\le a$ and everybody is a positive integer. This expression has arisen to me as a coefficient of $(1+q)^a(1-q)^{-1-b}$ and I am interested in computing determinants of such numbers. Therefore I am not interested in a "generating function" answer, but rather in an expression that does not involve a summation, like the ones before. Alternatively, a natural combinatorial interpretation can also be useful, in order to use Viennot's theory of binomial determinants.
Anybody here familiar with combinatorics can give me any help? Thank you in advance.
 A: Lacking reputation, I am unable to comment. I will add, however,  that a closed ("summation-signless") expression is not forthcoming.  I will adjust your notation modestly  and write
$$
S_k(a,b) = \sum_{0\leq i \leq k}{a\choose k-i}{b+i\choose i}
$$  for $a,b,k\in \mathbb{N}$ with $k\leq a$.  
Let us also write $S_k(a,0) = \sum_{0\leq i \leq k}{a\choose k-i} = \sum_{0\leq i \leq k}{a\choose i}$; when $k<a$, it is well-known that $S_k(a,0)$ (the proper prefix sum of the $a$-th row of Pascal's Triangle) does not admit a closed form. 
We begin by writing 
\begin{alignat}{2}
S_k(a,1) &=\sum_{0\leq i \leq k}{a\choose k-i}(i+1)\notag \\
&= \sum_{0\leq i \leq k}{a\choose k-i}i+ S_k(a,0) \notag\\
&= \sum_{1\leq i \leq k}{a\choose k-i}i+ S_k(a,0) \notag\\
&= \sum_{0\leq i \leq k-1}{a\choose k-(i+1)}(i+1)+ S_k(a,0) \notag\\
&= S_{k-1}(a,1)+ S_k(a,0). \notag\\
\end{alignat}
Thus, $S_k(a,0) = S_k(a,1) - S_{k-1}(a,1)$; a closed form for both $S_{k}(a,1)$ and $S_{k-1}(a,1)$ would lead to a closed difference expression for $S_k(a,0)$.
A: Let $G$ be the (infinite) graph with vertex set $\mathbb{Z}^2$, and the following edges. When $x+y < 0$, the vertex $(x,y)$ has outgoing edges to $(x+1,y)$ and to $(x,y+1)$. When $x+y \geq 0$, the vertex $(x,y)$ has outgoing edges to $(x+1-k,y+k)$ for all $k \geq 0$. That is, to $(x+1,y)$, $(x,y+1)$, $(x-1,y+2)$, and so on. These vertices have infinite outdegree (and when $x+y>0$, infinite indegree) but we will only use finite subgraphs, with finite degrees.
Now the number of directed paths in $G$ from $(-a,0)$ to $(b+1-k,k)$ is equal to $S(a,b)$.
Indeed, each edge in $G$ increases the sum of coordinates $x+y$ by $1$. So every path from $(-a,0)$ to $(b+1-k,k)$ has length $a+b+1$. For a given path, label each of the $a+b+1$ steps by their vertical travel ($0$ for a step east, $1$ for a step north, $2$ for a step in direction $(-1,2)$, etc.). The total of the labels is $k$. The first $a$ steps have labels $0$ or $1$. Subsequent steps have labels $\geq 0$.
The pair $(-a,0)$, $(b+1-k,k)$ may be translated by $(m,-m)$ for any $m$.
In your original question you did not say exactly what determinant you are trying to evaluate. But at least some determinants of values $S(a,b)$ can now be interpreted as counting disjoint path systems in the graph $G$. Well, I don't know how easy it will be to count those paths, but anyway I hope it helps.
A: Mathematica answers
Sum[Binomial[a, k - i]*Binomial[b + i, i], {i, 0, k},Assumptions -> a >= k | b > 0]

$$\binom{a}{k} \, _2F_1(b+1,-k;a-k+1;-1) $$
Addition. Maple performs
sum(binomial(a, k-i)*binomial(b+i, i), i = 0 .. k)assuming a>=k,b>0

$${a\choose k}{\mbox{$_2$F$_1$}(-k,b+1;\,a-k+1;\,-1)} $$
I think both answers are equivalent.
