  Let $\eta(G)$ be the number of connected induced subgraphs in a graph $G$,
and $\overline{G}$ the complement of $G$. We prove that
$\eta(G)+\eta(\overline{G})$ is minimum, among all $n$-vertex graphs, if and
only if $G$ has no induced path on four vertices. Since the $n$-vertex star
$S_n$ with maximum degree $n-1$ is the unique tree of diameter $2$,
$\eta(S_n)+\eta(\overline{S_n})$ is minimum among all $n$-vertex trees, while
the maximum is shown to be achieved only by the tree whose degree sequence is
$(\lceil n/2\rceil,\lfloor n/2\rfloor,1,\dots,1)$. Furthermore, we prove that
every graph $G$ of order $n\geq 5$ and with maximum
$\eta(G)+\eta(\overline{G})$ must have diameter at most $3$, no cut vertex and
the property that $\overline{G}$ is also connected. In both cases of trees and
graphs that have the same order, we find that if $\eta(G)$ is maximum then
$\eta(G)+\eta(\overline{G})$ is minimum.
  As corollaries to our results, we characterise the unique connected graph $G$
of given order and number of vertices of degree $1$, and the unique unicyclic
(connected and has only one cycle) graphs $G$ of a given order that minimises
$\eta(G)+\eta(\overline{G})$.
