It is double-counting, I think it can be solved with $card(S)$ $S:((p, q)$$| p$ is pair of students, $q$ is performance) but not sure

The answer is $n=4$.This can be achieved as follows:-Label the students $S_1,S_2,\cdots S_{12}$.Then Construction

works Count the triples $T=(P_i,P_j,X)$ where stuent $X$ participated in both performance $P_i$ and $P_j$.Let student $X$ be in $d_X$ performances.Since $\binom{\cdot}{2}$ is a convex function whence have \[2\binom{n}{2}\geq |T|=\sum \binom{d_X}{2}\geq 12\binom{\tfrac{6n}{12}}{2}\]Computing we have $n\leq 4$.This completes the proof$\blacksquare$