Thanks in advance!
Let $\displaystyle A$ be a balanced subset of $\displaystyle \{1,2,...,2n\}$. Then the number of elements of $\displaystyle A$ must be even. Furthermore, $\displaystyle |A|$ can be either $\displaystyle 2,4,6,...,2n$. Assume we wish to find all balanced subsets with $\displaystyle 2k$ elements, $\displaystyle 1\leq k\leq n$. To do this we need to choose $\displaystyle k$ even integers, and and $\displaystyle k$ odd integers. There are $\displaystyle {n\choose k}$ ways to choose an even integer and $\displaystyle {n\choose k}$ to choose odd integers. In total there are $\displaystyle {n\choose k}^2$ ways to create a balanced subset with $\displaystyle 2k$ integers. This sum is taken over $\displaystyle k=1,...,n$. And thus, there are a total of $\displaystyle \sum_{k=1}^n {n\choose k}^2$ balanced subsets.
Thank you,
Now a continuation of this question, I have came up with the formula
$\displaystyle bn$ = $\displaystyle {2n\choose n}$
Prove combinatorially that this is true for all integers n>=1.
I have this so far:
1. Choose $\displaystyle n$ integers from the $\displaystyle 2n$
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.