Binomial coefficient in Andrews' partition book

Consider the generating series

$$\sum_{s=0}^{\infty}\left(\sum_{i+j=s}\binom{A-n+j}{j}\binom{n-j}{i}\right)x^{s}.$$ This equals $$\sum_{j=0}^{\infty}\binom{A-n+j}{j}x^{j}\sum_{i=0}^{\infty}\binom{n-j}{i}x^{i}=(1+x)^{n}\sum_{j=0}^{\infty}\binom{A-n+j}{j}\left(\frac{x}{1+x}\right)^{j},$$ where we use the binomial theorem for the last equality. As $\sum_{k=0}^{\infty}\binom{n+k}{k}x^{k}=\frac{1}{(1-x)^{k}}$ our series becomes $$(1+x)^{n}\frac{1}{\left(1-\frac{x}{1+x}\right)^{A-n+1}}=(1+x)^{A+1}$$ which is $$\sum_{s=0}^{\infty}\binom{A+1}{s}x^{s}$$ by the binomial theorem. Your equality follows from comparing coefficients.

This also follows from the Chu-Vandermonde identity ${s+t \choose n}=\sum_{k=0}^n {s \choose k}{t \choose n-k}$ and the upper negation rule for binomial coefficients $\binom{r}{k} = (-1)^k \binom{k-r-1}{k}$.

$$\sum_{\substack{i+j=s \atop i\geq 0, j \geq 0}}\binom{A-n+j}{j}\binom{n-j}{i} = \sum_{j=0}^s \binom{A-n+j}{j}\binom{n-j}{s-j}.$$ Then apply upper negation to get $$\sum_{j=0}^s (-1)^j \binom{-A+n-1}{j} (-1)^{s-j}\binom{s-n-1}{s-j}. $$ Chu-Vandermonde followed by upper negation again yields $$= (-1)^s \binom{-A+s-2}{s} = \binom{A+1}{s}.$$

Consider all subsets of $\{1,2,\dots,A+1\}$ of cardinality $A-s+1$. There are exactly $\binom{A+1}{s}$ subsets. For each such subset $x_1 < x_2 < \dots < x_{A-s+1}$ consider the element $x_{A-n+1}$. The number of subsets with fixed value $x_{A-n+1}=p:=A-n+j+1$ the number of desired subsets equals $\binom{A-n+j}{A-n}\binom{n-j}{n-s}=\binom{A-n+j}{j}\binom{n-j}{i}$. Then just sum up by all possible values of $j=x_{A-n+1}-A+n-1$.