Counting the number of subsets of a set of 2n elements satisfying some conditions. Let $X =\{v_1, v_2,\cdots, v_n, v_{n+1},\cdots, v_{2n}\}$ be a set of $2n$ elements. I want to find the number of subsets of $X$ with $n$ elements such that both $v_i$ and $v_{n+i} $ are not together in this subset. 

I tried some small cases with $n=2,3,4$ and think it should be $2^n$ but I would like to prove it rigorously. I'm not sure how to use induction here. Is there a good combinatorial argument that can be used? Is my guess even correct? 

Thank you.
You have set the requirement for subsets that do not contain both $v_i$ and $v_{n+i} $.

However, in order for such a subset to have $n$ elements, it is clear that it must contain one of these.

As such, every subset can be generated by combining all those choices between $v_i$ and $v_{n+i} $. Each of those choices is binary, and there are $n$ choices, giving a total of $2^n$ alternative subsets that meet the conditions.
WLOG, $\{v_1,..v_2n\}=\{0,...,2n-1\}=2n$ (by the Von-Neumann ordinal convention of $k=\{0,..,k-1\}$). 

Our chosen set is determined by its intersection with $n$ which is an arbitrarily chosen element of the powerset $P(n)$ with cardinality $2^n$.