A Grand Music Festival

Let the musicians be $A$ , $B$ , $C$ , $D$ , $E$ and $F$ . We first show that four concerts are sufficient. The four concerts with the performing musicians: $\{A,B,C\}$ , $\{A,D,E\}$ , $\{B,D,F\}$ and $\{C,E,F\}$ satisfy the requirement. We shall now prove that three concerts are not sufficient.

Suppose there are only three concerts. Since everyone must perform at least once, there is a concert where two of the musicians, say $A$ and $B$ , played. But they must also have played for each other. Thus we have $A$ played and $B$ listened in the second concert and vice versa in the third. Now $C$ , $D$ , $E$ and $F$ must all perform in the second and third concerts since these are the only times when $A$ and $B$ are in the audience. It is not possible for them to perform for each other in the first concert. Thus, the minimum is 4.

Let two given musicians $A$ and $B$ have a connection if $A$ played to $B$ at least once, and have another connection if $B$ also played to $A$ at least once.

Since there are six musicians ( $A$ , $B$ , $C$ , $D$ , $E$ and $F$ ), a total of $2\times (5+4+3+2+1)=30$ connections must be made.

In any given concert, the maximum number of connections that can be made is $3\times 3=9$ , (since $3\times 3>2\times 4>1\times 5$ ).

So, with three concerts, no more than $3\times 9=27$ connections can be made, and since $27<30$ , three concerts will not be enough.

However, four concerts allow every musician to play to every other musician; the following shows the four concerts with the performing musicians: $\{A,B,C\}$ , $\{A,D,E\}$ , $\{B,D,F\}$ and $\{C,E,F\}$ .