Hamming distance for binary codes can be calculated using the binomial coefficient.

Let $x$ be a binary codeword of length $n$ , i.e. $x \in Z_{2}^{n}$ .

There are exactly $(\binom{n}{k})$ elements of $Z_{2}^{n}$ that are exactly Hamming distance $k$ away from $x$ .

Proof

Let $x \in Z_{2}^{n}$ .
Then $x$ has $n$ positions which each can be either element of $Z_{2} = {0, 1}$ ,
and there is exactly one way each position can differ from its current state;

Since the binomial coefficient is isomorphic to a combination, i.e. there are $(\binom{n}{k})$ ways to "choose" an unordered set of $k$ elements from a fixed set of $n$ elements;
Since the Hamming distance $d_{H} (x, y)$ is defined as the number of positions where $x$ and $y$ disagree,
and to be distance $k$ away from a codeword is to choose (order independent) $k$ positions to flip,

there are exactly $(\binom{n}{k})$ possible ways to change $k$ positions.

QED