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.

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