In $K_{23},$ the degree of each vertex is exactly $22.$ We color the edge satisfying the following condition: If two edge $e_1,e_2$ comes from a common vertex, we color them with different color. And there is a well-known result: Coloring $K_{n}$ satisfying such condition, the number of color we should use is at least $n-1$(if $n$ is even), or $n$(if $n$ is odd). Edit: oops! Thanks Eary for pointing out, my above statement only give us there are at least $23$ different movies.

There might be an example for $23$ different collections, but the problem asks for the minimal number of different collections which an example can be gicen. You're wrong, as the answer is not always $n$ for odd integers $n$. For instance, it's not hard to give an example for $4$ different collections when there are $5$ students total.

All the students may have the same movie collection and it gives the answer 1. Am I wrong somewhere?

Can you explain your construction?

For $n$ students from $1$ to $n$ each pair of students $(i,j)$ watch together movie $i-j \mod n$, is it right?

mszew wrote: For $n$ students from $1$ to $n$ each pair of students $(i,j)$ watch together movie $i-j \mod n$, is it right? For a pair $x,y$ of students, how do we decide if it is $x-y$ or $y-x$, modulo $n$?

Something is missing i guess... If everyone has seen the same movie, the answer is 1, so is it that every pair of student watched a different movie?

godfjock wrote: Something is missing i guess... If everyone has seen the same movie, the answer is 1, so is it that every pair of student watched a different movie? As stated, every $\binom{23}{2}$ pairs have watched a movie, and one student has watched $22$ different movies