One class needs to choose $n$ students to join a competition. The teacher hold $n$ exams in order to choose $n$ students. In each exam, there are no two students who have the same score. The teacher will hold one exam and pick the student whose score is the highest in that exam. Then, the teacher will hold another exam and pick the student whose score is the highest in this exam from the remaining students. In this way, the teacher will choose $n$ students. Obviously, the order that the teacher hold the exams will affect the students will be chosen. So, how many students have the chance to be chosen at most?
For example, when $n=2$ and in exam A, Alice is the first, Bob is the second, Tom is the third. In exam B, Alice is the first, Tom is the second, Bob is the third. If the teacher choose exam A at first and then choose exam B, Alice and Tom will be chosen. If the teacher choose exam B at first and then choose exam A, Alice and Bob will be chosen. So Alice, Bob and Tom have the chance to be chosen.
Suppose that $f(n)$ students have the chance to be chosen at most. For small $n$ I have $f(1)=1,f(2)=3,f(3)=5,f(4)=8,f(5)=10,f(6)=13$. I don't know how to get the general formula.
We use number $i$ to represent the $i_{th}$ student and we use $[~]$ to represent the exams. The order in $[~]$ means the rank in the exams. Then we have: $$ f(4)=8: [1,2,3,4], [1,2,3,5], [1,2,6,7], [1,2,6,8] $$
$$ f(5)=10: [1,2,3,4,5], [1,2,3,4,6], [1,2,3,7,8], [1,2,3,7,9], [1,2,3,10] $$
$$ f(6)=13: [1,2,3,4,5,6], [1,2,3,4,5,7], [1,2,3,4,8], [1,2,3,9,10,11], [1,2,3,9,10,12], [1,2,3,9,13] $$
In this way we can show that $f(2n) \geqslant 2f(n)+n$ and $f(2n+1) \geqslant f(n)+f(n+1)+n$. I think these two inequalities are actually two equations, but I don't know how to prove it.
Thank you very much!