How many binary strings of length 2n + 1 have more 1's than 0's? Use bijection to prove. NOTE: This is a homework question, so I only ask for hints and suggestions to nudge me in the correct direction.
Question:
Let $n \in \mathbb{N}$. How many binary strings of length 2n + 1 have more 1's than 0's? Use bijection to prove.
Progress so far:
We need a set of known size, B, along with a bijective function $f: A \rightarrow B$, where A is the set with binary strings of length $2n + 1$ and where each element in A has more 1's than 0's.
I tried to find a pattern by setting n = 1:
$$ 
\begin{align*}
000 \rightarrow &\lbrace empty set \rbrace \\
001 \rightarrow &\lbrace 1 \rbrace \\
010 \rightarrow &\lbrace 2 \rbrace \\
011 \rightarrow &\lbrace 1,2 \rbrace \\
100 \rightarrow &\lbrace 3 \rbrace \\
101 \rightarrow &\lbrace 1,3 \rbrace \\
110 \rightarrow &\lbrace 1,2 \rbrace \\
111 \rightarrow &\lbrace 1,2,3 \rbrace \\
\end{align*}
$$
I realized that binary strings with more 1's than 0's map to elements in B where the elements are of size $\geq 2n$. 
Therefore the size of set $S$ where $S = \lbrace x : |x| \geq 2n\rbrace$ is the number of binary strings of length $2n+1$ with more 1's than 0's.
This is where I am stuck. I don't know how to fit bijection into this solution. Also I think my mapping from set A to set B is incorrect since set A it should only contain the binary strings that have more 1's than 0's.
Any help would be greatly appreciated.

Hint: If we toss a fair coin $2015$ times, the probability there are more heads than tails is, by symmetry, equal to the probability there are more tails than heads. We cannot have equality, so the probability is $\frac{1}{2}$.  
