How many hours are needed so that each team has had a meeting? There are 6 teams $E_{1},E_{2},....,E_{6}$ and each team has to have a meeting to discuss about something. How many hours are needed (minimum) so that all the meetings have been done, when:
\begin{align*}
E_{1}&=\{A,B,C\}  \\
E_{2}&=\{A,D,E\} \\
E_{3}&=\{B,C,Z\} \\
E_{4}&=\{Z,H,T\} \\
E_{5}&=\{E,H\} \\
E_{6}&=\{D,E,T\}
\end{align*}
?
 A: The teams are vertices of a graph and join vertices (i.e. meetings) if they share members. The task is to determine the cromatic number $\chi$ of that graph.
We have $\chi\ge 3$ because of odd cycles like
$$E_4\stackrel{H}\longrightarrow
E_5\stackrel{E}\longrightarrow
E_6\stackrel{T}\longrightarrow  E_4 $$
and
$$ E_1\stackrel{A}\longrightarrow E_2\stackrel{D,E}\longrightarrow E_6\stackrel{T}\longrightarrow E_4\stackrel{Z}\longrightarrow E_3\stackrel{B,C}\longrightarrow E_1.$$
On the other hand, three colours (and hence three hours) suffice, e.g.
$$\{E_1, E_5\}\quad\{E_2,E_4\}\quad \{E_3,E_6\}  $$
A: Assuming the problem is that some people are a member in multiple teams and therefore those teams can't meat at the same time, it isn't hard to see that only two teams can meet at any moment. (One way to see this is that any two of $E_1$, $E_2$ and $E_3$ have a common member and the same for the remaining three teams. Only $E_2$ and $E_3$ could meet at the same time, but then, none of $E_4$, $E_5$ and $E_6$ can meet.)
Thus, we need at least $3$ rounds, where during every round, one of the first three and one of the last three teams is meeting. It can be seen that $E_2$ can only meet at the same time as $E_4$ in that case. Then we can pair $E_1$ and $E_3$ with $E_5$ and $E_6$ in some way. Thus, it can be done in three hours.
