Determine the number of graphs with no vertices of degree 0 on a given $n$-element vertex set V.
The total number of simple graphs with $n$ vertices is $2^{\binom{n}{2}}$. We want to find the number of graphs with degree 0 and subtract it from the total.
Let $v$ be an arbitrary vertex in V. Let's denote the set $$A_{v} = \{E \in 2^{\binom{n}{2}}: deg(v) = 0\}$$ It is the set of edge sets (which correspond to graphs) for which the degree of $v$ is 0. Let's calculate the size of $A_{v}$: We discard all edge sets (graphs) in which there are edges that are incident to $v$. For the rest of the $n-1$ vertices we do not care if there any incident or not, thus the total number of such graphs is $$|A_{v}| = 2^{\binom{n-1}{2}}$$ Now consider $|A_{u} \cap A_{v}|$, such that $u \neq v$. We discard edges sets (graphs) in which there are edges adjacent to $u$ or to $v$ or both (if there is an edge $u,v$). $$|A_{v} \cap A_{u}| = 2^{\binom{n-2}{2}}$$ More generally, if there are $|U|=k$ such vertices with degree zero, the total number of graphs is $$|\bigcap_{u \in U}| = 2^{\binom{n-k}{2}}$$ Now, we can apply the inclusion-exclusion formula $$|\bigcup \limits_{i=1}^{n}A_{i}| = \sum \limits_{\emptyset \neq U \subseteq \{1, 2, \ldots , n\}} (-1)^{|U| - 1} |\bigcap \limits_{u \in U} A_{u}|$$
$$|\bigcup \limits_{i=1}^{n}A_{i}| = \sum \limits_{\emptyset \neq U \subseteq \{1, 2, \ldots , n\}} (-1)^{k- 1} 2^{\binom{n-k}{2}}$$
So, the total number is $2^{\binom{n}{2}}$ minus $|\bigcup \limits_{i=1}^{n}A_{i}|$.
Is this line of reasoning correct?