2024 AMC 10B Problem 12

Give each student the set of languages they speak. The condition says no one’s set sits inside another’s. Everyone speaks the same number $k$ of languages, and two distinct $k$-element sets can never contain each other, so all we need is $100$ different $k$-subsets of the $n$ languages, i.e. $\binom{n}{k} \ge 100.$ With $n = 8$ the best we can manage is $\binom{8}{4} = 70,$ short of $100.$ But $\binom{9}{4} = 126 \ge 100.$ So $9$ languages are both enough and necessary. Therefore, the answer is A.