Sphere packing inequality for error-correcting codes
For a codeword $x$, let $B_k(x)$, the ball of radius $k$ around $x$, consist of all words at distance at most $k$ from $x$. Notice that
$$|B_k(x)| = \sum_{i=0}^k \binom{m+r}{i}. $$
If the code supports correcting up to $k$ errors, then the sets $\{ B_k(x) : x \in C \}$ must be disjoint. The code contains $2^m$ many different codewords, hence we get
$$
2^m \sum_{i=0}^k \binom{m+r}{i} \leq 2^{m+r},
$$
which is the same as your inequality. Here $m+r$ is the length of each codeword, and $2^{m+r}$ is the number of possible vectors.

If your code is $q$-ary rather than binary, then the same calculation shows that
$$
\sum_{i=0}^k (q-1)^i \binom{m+r}{i} \leq q^r.
$$