My hunch was wrong, it has to do with the exclusion-inclusion principle. Consider a set of $n+m$ elements (labeled $1,2,... n+m$) Define the set $S$ as the set of all subsets of $S$ with $j$ elements (ignoring the ordering), such set has cardinality $|S| = \binom{n+m}{j}$, an element of $S$ looks something like $(1,3,4,9,\cdots, \cdot)$ with $j$ numbers inside. Now, let $A_k$ be the subset of $S$ given by all elements of $S$ which contain the number $k$, we let this number take the values $k=1,2, \cdots, n$, since repetitions are not allowed, and all possible collections are counted, the cardinality of such set is easily computed by fixing one element to be $k$, leaving $j-1$ slots to fill, from $n+m-1$ numbers, hence:
$|A_k| = \binom{n+m-1}{j-1}$
There are $n$ such sets
It is easy also to show that (using the same argumentation, now two numbers are fixed $k,l$ with $k\neq l$ so there remain $j-2$ slots to be filled with $n+m-2$ numbers)
$|A_k \cap A_l| = \binom{n+m-2}{j-2} $
In general now we have
$$|\bigcap_{a=1}^i A_{l_a} | =\binom{n+m-i}{j-i} $$
and there are $\binom{n}{i}$ such intersections (all sets carry $i$ distinct subindices). The exclusion inclusion principle is now applied to the set $S$ and the subsets $A_i$ this is just substitution:
$$
|S - \bigcup_{i=1}^n A_i| = \sum_{i=0}^n (-1)^i \binom{n}{i} \binom{n+m-i}{j-i}
$$
Compute now the cardinality of the set directly: it is the subset of elements of $S$ which do not contain any of the numbers $1,2, \cdots, n$, so these are all the combinations of just $m$ numbers in $j$ slots. This number is $\binom{m}{j}$, so we conclude that:
$$
\sum_{i=0}^n (-1)^i \binom{n}{i} \binom{n+m-i}{j-i} = \binom{m}{j}
$$