A problem from "solving mathematical problems" by Terence Tao. Find the least no. of concerts This is a problem from "Solving Mathematical Problems" by Terence Tao. I wasn't able to solve it, so I asked y'all. Please help me.
The complete problem-

Six musicians gathered at a music festival. At each concerts some musicians played in the concerts while others listened as part of the audience. What is the least number of concerts needed to be scheduled in order that each musician may listen, as part of the audience, to each other musician.

Hint, also from the book-

Hint: obviously not everyone can listen to every one else in one concert, so more than one concert is needed to exhaust all the listening possibilities think upon these lines and also of the point scoring ideas and you will get a reasonable lower bound for the number of concerts needed find an example satisfying the lower bound and you have solved it

Developments - I made a graph, represented the musicians with the nodes... Graph And it will be a directed graph where $A \rightarrow B$ would mean $A$ has sung and $B$ has listened as the audience... For example if $1$ sang in the first concert then the graph would be...Graph and so on. Now at last we will need a fully complete graph in which there are two edges (opp directions) b/w each pair of nodes so $n(n-1)$ edges in total
Observation- The order of concerts doesn't change the final graph, we can say this because anytime there is a concert and some edges have to be made, if they were already there then there is no change but if they were not there then they will be made so essentially we are adding the set of edges to be made to the set of edges already there to get the new graph's edges and because addition is commutative so therefore the order doesn't matter
i.e. $E_i + E_1 = E_f$ and $E_i +E_1 +E_2 =E_f$ is the same as $E_i+E_2+E_1$
As the hint said, I tried to find a lower bound by calculating the max no. of edges possible with one concert so in the graph if there are n nodes and in the concert $k$ musicians sing then then the max no. of edges increased is $k(n-k)$ And the max value of this is $\frac{n^2}{4}$ And now dividing this with the total no. of edges to be made $n(n-1)$ we get the lower bound to be $\frac{4(n-1)}{n}$ And for the specific case of $6$ musicians is $3.333$ i.e. $4$ but I am not able to make the concerts required for completion. Please help.
 A: 
And for the specific case of 6 musicians is 3.333 i.e. 4 but I am not able to make the concerts required for completion.

Actually, we can make each musician listen (as part of the audience) to each other musician in 4 concerts.
Let the musicians be A, B, C, D, E and F, here is the arrangement:
            1st      2nd      3rd      4th
Player      A,B,C    B,E,F    C,D,E    A,D,F
Audience    D,E,F    A,C,D    A,B,F    B,C,E

A: Solution. Let the musicians be A, B, C, D,
E and F. Suppose there are only three
concerts. Since each of the six must
perform at least once, at least one
concert must feature two or more
musicians. Say both A and B perform in
the first concert. They must still perform
for each other. Say A performs in the
second concert for B and B in the third
for A. Now C, D, E and F must all
perform in the second concert, since it is
the only time B is in the audience.
Similarly, they must all perform in the
third. The first concert alone is not
enough to allow C, D, E and F to
perform for one another. Hence we need
at least four concerts. This is sufficient,
as we may have A, B and C in the first, A,
D and E in the second, B, D and F in the
third and C, E and F in the fourth.
