A sum involving a binomial

To add an enumerative point of view: Recall that $[n]=\{1,2,\cdots ,n\}.$ You can rewrite your sum as $$\sum _{r=0}^q(-1)^{q-r}\binom{q}{r}\binom{m+r}{r}=\binom{m}{q},$$ doing the change of variable in the sum $r=q-r$ (abuse of notation but i like $r$) and isolating the term $r=0$, you get

$$\binom{m+q}{q}-\sum _{r=1}^q(-1)^{r-1}\binom{q}{r}\binom{m+q-r}{q-r}=\binom{m}{q},$$

So, essentially, by an inclusion-exclusion argument, the LHS is the number of ways to take $q$ objects from $[m+q]$ objects in which you do not take any object bigger than $m,$ which is equivalent to choosing $q$ objects out of $[m]$ so $\binom{m}{q}$ as the RHS.

Trying to prove that

$$\sum_{r=0}^q (-1)^{q-r} {q\choose r} {m+r\choose r} = {m\choose q}$$

we write the LHS as

$$\sum_{r=0}^q (-1)^{q-r} {q\choose r} {m+r\choose m} = [z^m] (1+z)^m \sum_{r=0}^q (-1)^{q-r} {q\choose r} (1+z)^r \\ = [z^m] (1+z)^m ((1+z)-1)^q = [z^m] (1+z)^m z^q \\ = [z^{m-q}] (1+z)^m = {m\choose m-q} = {m\choose q}$$

which is the claim.