Draw a simple graph where its vertices can be divided into 2 sets where every edge joins a vertex in one set to a vertex in the other

Question

The question is:

Draw a simple graph with the following properties or explain why they do not exist. The graph has 6 vertices and 8 edges and its vertices can be divided into two sets in such a way that every edge joins a vertex in one set to a vertex in the other.

I don't understand this question but I tried doing this. Since every vertex needs to join every other vertex (for example, in the scenario with two sets where 1 set has 1 vertex and the other has 5 vertices), does this mean that for this graph to exist, there has to be n(n-1)/2 edges?

Answer 1

Suppose our two vertex sets have $m$ and $n$ vertices respectively; if every edge links one vertex type to another, there can be at most $mn$ edges, when every pair of different-type vertices is connected by an edge.

In this question, $m+n=6$, so $(m,n)$ can be $(1,5),(2,4),(3,3)$, up to reversal of variables. These vertex sets allow for 5, 8 and 9 edges respectively. We reject $(m,n)=(1,5)$ as we need 8 edges. The other two choices lead to the two distinct solutions to the problem (the two vertex sets are marked with white and black circles):

Answer 2

Without planarity requirement, the problem is only asking to have a bipartite graph : you need to be able to split the vertices in two "stable sets", i.e. two sets of vertices with no edges between vertices of the same set. As per wikipedia

a bipartite graph (or bigraph) is a graph whose vertices can be divided into two disjoint and independent sets $U$ and $V$ such that every edge connects a vertex in $U$ to one in $V$

For example :

The easiest exemple fitting the question would be $K_{2,4}$, the bipartite complete graph on 6 vertices, split in two groups. $U$ has size 4, $V$ has size $2$, and all elements of $U$ are linked to all elements of $V$