5. Six musicians gathered at a chamber music festival. . At each scheduled concert some of the musicians played while the others listened as members of the audience. What is the least number of such concerts which would need to be scheduled so that for every two musicians each must play for the other in some concert?
5. Let the musicians be $A, B, C, D, E, F$. 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 3 concerts are not sufficient. Suppose there are only three concerts. Since everyone must perform at least once, there is a conoert where two of the musicians, say $A, B$, played. But they must also play for each other. Thus we have $A$ played and $B$ listened in the second concert and vice versa in the third. Now $C, D, E, F$ 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 .