Students in a class form groups each of which contains exactly three members such that any two distinct groups have at most one member in common. Prove that, when the class size is $ 46$, there is a set of
- a) $ 7$;
- b) $10$ students in which no group is properly contained.
For a) Say we have $n$ sets. Since each pair of students is in at most one of these sets and each set contains 3 pairs, we have $${46\choose 2}\geq 3n \implies n\leq 345$$ Let's take at random and independetly a student with a probability $p={1\over 5}$. Let $X$ be a number of choosen students and $Y$ be a number of "bad" subsets i.e. number of subsets among those $n$ that are entirely in the set of choosen students. Then we have $$E(X-Y) = 46p-np^3\geq {46\over 5} -{345\over 125}>6$$ and thus we have a set of $7$ as demanded.
Now, my question is why this does not work for $b)$?
And is there a way to overcome this situation?
Note: I know how to solve b) without probabilistic method, so don't need nonprobabilistic solution for it.