Is there a way to solve this problem using combinations? I've been getting to the given correct answer using trial and error but I want to know if there's a way using something related to combinations or permutations? Here's the problem:
A teacher is providing her students with 4 special lessons. (a) Each lesson has exactly 3 participants, (b) Any two students must attend at least one special lesson together. According to these two conditions, what is the maximum number of students who can join these special lessons?
Thank you
Suppose there are $n$ students attending the lessons. This means there are $\binom{n}2$ pairs which need to attend a lesson together. Since every lesson takes care of at most $\binom32=3$ pairs of students, and there are four lessons, we must have $$ \binom{n}2\le 3\cdot 4=12 $$ This immediately implies $n\le 5$, because $\binom{6}2=15>12$.
This does not prove that $5$ students is sufficient, but it gives you a place to start looking. Namely, try to see if $5$ students is possible, and if you suspect it is not, then come up with a proof of this. In general, solving "combinatorial design" problems like these are very hard, so you cannot expect it to be entirely solved using elementary techniques like combinations.