1. ## Combinatorial Proof

2. Let $A$ be a balanced subset of $\{1,2,...,2n\}$. Then the number of elements of $A$ must be even. Furthermore, $|A|$ can be either $2,4,6,...,2n$. Assume we wish to find all balanced subsets with $2k$ elements, $1\leq k\leq n$. To do this we need to choose $k$ even integers, and and $k$ odd integers. There are ${n\choose k}$ ways to choose an even integer and ${n\choose k}$ to choose odd integers. In total there are ${n\choose k}^2$ ways to create a balanced subset with $2k$ integers. This sum is taken over $k=1,...,n$. And thus, there are a total of $\sum_{k=1}^n {n\choose k}^2$ balanced subsets.

3. Thank you,
Now a continuation of this question, I have came up with the formula
$bn$ = ${2n\choose n}$
Prove combinatorially that this is true for all integers n>=1.

I have this so far:
1. Choose $n$ integers from the $2n$
2. Choose all the even integers in that set
3. Choose the odd that are not in that set

How can I turn this into a more formal proof?

Thanks again.