Select a subset of the set $ \{1, 2, . . . , 200 \}$ in such a way that it contains the same number of even and odd elements. In how many ways can this be done?
My solution:
To ensure that a subset has an equal number of even and odd numbers we need to choose $k$ even numbers (out of $100$) and choose $k$ odd numbers (out of $100$).
The number of ways to choose $k$ even numbers from $100$ is ${{100}\choose{k}}$ and the number of ways to choose $k$ odd numbers from $100$ is ${{100}\choose{k}}$ as well.
So, the correct number seems to be:
$$\sum_{k=0}^{100} {{100}\choose{k}}^2 = {{200}\choose{100}}$$
Is that correct?
Yes, your solution is correct, and the last step uses https://en.wikipedia.org/wiki/Vandermonde%27s_identity
Here's an alternative approach to see that the count is $\binom{200}{100}$. Let $E=\{2,,4,\dots,200\}$, take any $100$-subset $S$ of $\{1,\dots,200\}$, keep the even numbers in $S$, and take the complement of the odd numbers in $S$, yielding a subset with $|S \cap E|$ even numbers and $100-(100-|S \cap E|)=|S \cap E|$ odd numbers. Every good subset arises uniquely this way.