WakeUp wrote: A set of $8$ problems was prepared for an examination. Each student was given $3$ of them. No two students received more than one common problem. What is the largest possible number of students? The maximum number is 8 students: 123 345 567 781 247 683 461 825 Let the problems be vertices of $K_8$. Then each student corresponds to a triangle, and we want these triangles edge-distinct. Each problem can only be given to at most 3 students, as there are only 7 edges for each vertex. So 8 triangles is the maximum.

Maximum no. of students= 8 F G H F G H G H C D E D E F D E A A A B B B C C * * * * * * * * The stars are the problems and the alphabets are the students. We start with A putting it over 3 stars and then B and then C on two stars and the remaining one C on the same star as A. We put one D with A, one with B and one with C so that no two alphabets share more than one star. We carry this on in the similar fashion. Now as we get to I , we observe we cant put it over more than 2 stars since two alphabets will share more than one star and thus it doesn't satisfies the condition that each student gets three problems. So the maximum no. of alphabets we can get to is till H which is equal to 8 students.

Generalization: If a set of $n$ problems was prepared for an examination and each student was given $m$ of them, such that no two students received more than one common problem, the largest possible number of students is $\Bigg \lfloor \frac{ n* \Bigg \lfloor \frac{n-1}{m-1} \Bigg \rfloor } {m} \Bigg \rfloor$ Proof: Suppose that some problem was given to $x$ students. Each of these $x$ students would receive $m-1$ different additional problems, thus implying that we need atleast a total of $x(m-1) + 1$ problems. Thus $n \geq x(m-1) + 1$, or $x \leq \Bigg \lfloor \frac{n-1}{m-1} \Bigg \rfloor$. Thus each problem was assigned to atmost $\Bigg \lfloor \frac{n-1}{m-1} \Bigg \rfloor$ students and so there were at most $n* \Bigg \lfloor \frac{n-1}{m-1} \Bigg \rfloor$ assignments of problems. As each student was assigned $m$ problems, there were at most $\Bigg \lfloor \frac{ n* \Bigg \lfloor \frac{n-1}{m-1} \Bigg \rfloor } {m} \Bigg \rfloor$ students. Note: Converse problem (harder than the original in my opinion) is to find the minimum number of problems needed for an examination of $n$ students where each student was given $m$ problems, such that no two students received more than one common problem.