3 It is known that there are 12 theater troupes participating in a 7-day performance, and it is required that each troupe can see the performances of all other troupes, with only those not performing on a given day being able to watch from the audience. How many performances are needed at a minimum? (1994 China National Training Team Practice Question)
3. Let $A=\{1,2, \cdots, 7\}$ be the set of performance dates, and $A_{i}$ be the set of performance dates for the $i$-th troupe $(i=1,2, \cdots, 12)$. It is clear that $A_{1}, A_{2}, \cdots, A_{12}$ do not contain each other. Otherwise, for $i \neq j$, if $A_{i}$ is contained in $A_{j}$, then the $j$-th troupe cannot see the performance of the $i$-th troupe. Consider the total number of performances $t=\sum_{i=1}^{12}\left|A_{i}\right|$. If there is some $\left|A_{i}\right|=0$, then the other troupes cannot see the performance of the $i$-th troupe, which is a contradiction. Therefore, for any $i$, $\left|A_{i}\right| \geqslant 1$. Introduce a parameter: Suppose there are exactly $k$ sets that are singletons, and the other sets have at least 2 elements. Thus, $t=\sum_{i=1}^{12}\left|A_{i}\right| \geqslant \underbrace{1+1+\cdots+1}_{k \uparrow 1}+\underbrace{2+2+\cdots+2}_{12-k \uparrow 2} = 24-k$. Without loss of generality, let $\left\{a_{1}\right\}=A_{1},\left\{a_{2}\right\}=A_{2}, \cdots,\left\{a_{k}\right\}=A_{k}$. Since $A_{1}, A_{2}, \cdots, A_{12}$ do not contain each other, $a_{1}, a_{2}, \cdots, a_{k}$ are not in $A_{k+1}, A_{k+2}, \cdots, A_{12}$, so $A_{k+1}$, $A_{k+2}, \cdots, A_{7}$ are subsets of $\left\{a_{k+1}, a_{k+2}, \cdots, a_{7}\right\}$, i.e., $\left\{A_{k+1}, A_{k+2}, \cdots, A_{12}\right\}$ is a family of mutually non-contained subsets of $A \backslash\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}$, so $\mathrm{C}_{7-k}^{\left[\frac{\gamma-k}{2}\right]} \geqslant 12-k$. This inequality does not hold for $7-k=1,2,3,4$, so $7-k \geqslant 5$, i.e., $k \leqslant 2$. Therefore, $t \geqslant 24-k \geqslant 22$. Finally, when $t=22$, let $A_{1}=\{1\}, A_{2}=\{2\}$, and $A_{3}, A_{4}, \cdots, A_{12}$ be 10 distinct two-element subsets of $\{3,4,5,6,7\}$. For example, $A_{3}=\{3,4\}, A_{4}=\{3,5\}, A_{5}=\{3,6\}, A_{6}=\{3,7\}, A_{7}=\{4,5\}, A_{8}=\{4,6\}, A_{9}=\{4,7\}, A_{10}=\{5,6\}, A_{11}=\{5,7\}, A_{12}=\{6,7\}$. Therefore, the minimum number of performances is 22.