Sphere Packing Bound

Suppose that $C$ is a $t$ -error correcting code in $Z_{2}^{n}$ .
Let $| C |$ be the number of codewords in $C$ .
Then

| C | \cdot ((\binom{n}{0}) + \dots + (\binom{n}{t})) \leq 2^{n}

Proof

There are $| C |$ many Hamming balls to consider.

Therefore, there are $| C | \cdot ((\binom{n}{0}) + \dots + (\binom{n}{t}))$ elements contained within Hamming balls.

The cardinality of the Hamming space $Z_{n}^{2}$ is $2^{n}$ .

By An error correcting code has disjoint Hamming balls.,
If $C$ is $t$ -error correcting, then ${B_{t} (x) : x \in C}$ are all disjoint.

Thus, no element of $Z_{n}^{2}$ can be double-counted by the Hamming balls.

Therefore,

| C | \cdot ((\binom{n}{0}) + \dots + (\binom{n}{t})) \leq 2^{n}

QED