Number of words within Hamming distance $\delta$

This is a problem I'm having reading Arora & Barak's book, page 378-379. They define:


For two words $x, y \in \{0, 1\}^m$, the fractional Hamming distance of $x$ and
$y$ is
equal to the fraction of bits on which they differ, i.e. $\frac{1}{m} \cdot |\{i : x_i \neq y_i\}|$, where the $x_i$ refers to the $i$-th bit of $x$ (same
for $y_i$).

Now, in the proof of Lemma 18.9, they use the fact that


for every string $y_i$ of $m$ bits, the number of strings
that are of distance at most $\delta$ to it is ${m \choose \lceil \delta m \rceil}$

meaning that there are ${m \choose \lceil \delta m \rceil}$ words that disagree on not more than a $\delta$ fraction of bits.

But, I can't convince myself of that.
Suppose for simplicity that $y_i = 0000 \ldots 000$. Here are the words that are of distance at most $\delta m$: all the words that have exactly one $1$, or two, or three, up to $\lfloor \delta m \rfloor$ of bits set to $1$. Thus I rather think that this number is

$$
\sum_{k = 1}^{\lfloor \delta m \rfloor} {m \choose k}
$$

Can anyone explain?

That's an inaccuracy. If $\delta 0$ and large enough $m$ (depending on $\epsilon,\delta$).

StackExchange Computer Science Q#51071, answer score: 3