What bound does the Hamming bound give you for the largest possible size of a $t$-error-correcting code of length $2t + 1$? Let $\mathbb{A}$ = $\{0, 1\}$ and suppose $t$ is a positive integer. What bound does the Hamming bound give you for the largest possible size of a $t$-error correcting code of length $2t+1$?
I have equation 
$$
|C|\le \frac{q^n}{{n\choose0}+{(q-1)}{n\choose1}+\cdots+{(q-1)^t}{n\choose{t}}}
$$
with addition to the $2^n=\sum_{r=0}^n{n\choose{r}}$
I get answer $|C| \le 1$.
Is this correct?
 A: In the equation you gave for the Hamming bound, $n$ is the length of the code and $q$ is the size of the alphabet. In your case, $n=2t+1$ and $q=|\mathbb{A}|=2$. Substituting $q=2$ in the bound:
$$
|C|
\le \frac{2^n}{\binom{n}{0} + \binom{n}{1} + \ldots + \binom{n}{t}}
= \frac{2^{n}}{\binom{2t+1}{0} + \binom{2t+1}{1} + \ldots + \binom{2t+1}{t}}.
$$
To evaluate the denominator, you can use the other equation you listed:
$$\begin{align}
2^n &= \sum_{r=0}^n \binom{n}{r}\\
&= \sum_{r=0}^{2t+1} \binom{2t+1}{r}\\
&= \sum_{r=0}^{t} \binom{2t+1}{r} + \sum_{r=t+1}^{2t+1} \binom{2t+1}{r}\\
&= \sum_{r=0}^{t} \binom{2t+1}{r} + \sum_{r=t+1}^{2t+1} \binom{2t+1}{(2t+1)-r}\\
&= \sum_{r=0}^{t} \binom{2t+1}{r} + \sum_{r=0}^{t} \binom{2t+1}{r}\\
&= 2 \sum_{r=0}^{t} \binom{2t+1}{r}\\
&= 2 \left(\binom{2t+1}{0} + \binom{2t+1}{1} + \ldots + \binom{2t+1}{t}\right).
\end{align}$$
The fourth line is the critical step: $\binom{i}{j} = \binom{i}{i-j}$.
Once you've found an upper bound on $|C|$, you should try to find a binary $t$-error-correcting code of length $2t+1$ which meets that bound.
