What is the largest possible first place score in olympic sport climbing?

Question

Consider a competition with $n$ competitors, and $k$ contests. Each competitor receives a combined score, which is the product of their placements in each of the contests (assume no ties in the contests). The competitor with the lowest combined score wins.

For example, olympic sport climbing has $n=20$ competitors and $k=3$ for speed, boulder, and lead. Brooke Raboutou got 12th place in speed, 2nd place in boulder, and 8th place in lead for a combined score of 192. Janja Garnbret's combined score was $14\times 1\times 4=56$, which was the first-place score.

What is the largest possible score of the winner(s) of such a contest?

By brute force (i.e., computing all possible outcomes), here is the answer for some values of $n$ and $k$.

n\k	1	2	3	4	5
1	1	1	1	1	1
2	1	2	2	4	4
3	1	3	6	9	18
4	1	4	8	24	48
5	1	5	15	40	-
6	1	6	24	-	-
7	1	7	35	-	-

Answer 1

Some observations:

By taking the geometric mean we get $f(n,k) \leq \lfloor (n!)^{k/n} \rfloor$, this result is achievable when $k$ is a multiple of $n$, by making each contest a rotation by one place of the previous one.

The following table shows $\lfloor (n!)^{k/n} \rfloor - f(n,k) $ for $1\leq,n,k \leq 5$, maybe it helps.

0 0 0 0 0
0 0 0 0 1
0 0 0 1 1
1 0 2 0 5
1 1 2 6 0

Answer 2

A competitor finishing in first place in one contest can do no worse than 20th in each of the other two contests, for a score of $1\times20\times20=400$. If this were to happen, a competitor finishing first in one of the other contest can do no worse than $1\times19\times20=380$. This provides a strict upper bound on the largest possible first place score.

A most unlikely result showing that this is at least mathematically possible: $$\begin{matrix} 1&19&20&=&380\\ 19&20&1&=&380\\ 20&1&19&=&380\\ 7&7&8&=&392\\ 8&8&7&=&448\\ 6&9&10&=&540\\ 9&10&6&=&540\\ 10&6&9&=&540\\ 2&17&18&=&612\\ 17&18&2&=&612\\ 18&2&17&=&612\\ 5&11&12&=&660\\ 11&12&5&=&660\\ 12&5&11&=&660\\ 3&15&16&=&720\\ 15&16&3&=&720\\ 16&3&15&=&720\\ 4&13&14&=&728\\ 13&14&4&=&728\\ 14&4&13&=&728 \end{matrix}$$

Answer 3

For fixed $n$ and $k$, you can solve the problem via integer linear programming as follows. Let $P=\{1,\dots,n\}$ be the set of players, let $E=\{1,\dots,k\}$ be the set of events, and let $R=\{1,\dots,n\}$ be the set of ranks. Let binary decision variable $x_{p,e,r}$ indicate whether player $p$ in event $e$ has rank $r$. The problem is to maximize $$\min_{p\in P} \prod_{e\in E} \sum_{r\in R} r x_{p,e,r} \tag1$$ subject to \begin{align} \sum_{r\in R} x_{p,e,r} &= 1 &&\text{for $p\in P$ and $e\in E$} \tag2 \\ \sum_{p\in P} x_{p,e,r} &= 1 &&\text{for $r\in R$ and $e\in E$} \tag3 \\ \end{align} The objective $(1)$ maximizes the lowest score among all players. Constraint $(2)$ assigns one rank for each player and event. Constraint $(3)$ assigns one player for each rank and event. The objective $(1)$ can be linearized by introducing variable $z$, maximizing $z$, and imposing constraints \begin{align} z &\le \sum_{e\in E} \sum_{r\in R} \log(r) x_{p,e,r} \tag4 \\ \end{align}