Asia Pacific Mathematics Olympiad (APMO)

Let $C$ be the set of $n$ clowns. Label the colours $1,2,3,\dots ,12$. For each $i=1,2,\dots ,12$, let $E_i$ denote the set of clowns who use colour $i$. For each subset $S$ of $\{1,2,\dots ,12\}$, let $E_S$ be the set of clowns who use exactly those colours in $S$. Since $S\ne S^{\prime }$ implies $E_S\cap E_{S^{\prime }}=\varnothing$, we have $$\sum _S\left|E_S\right|=|C|=n,$$ where $S$ runs over all subsets of $\{1,2,\dots ,12\}$. Now for each $i$, $$E_S\subseteq E_i\text{ if and only if }i\in S,$$ and hence $$\left|E_i\right|=\sum _{i\in S}\left|E_S\right|.$$ By assumption, we know that $\left|E_i\right|\le 20$ and that if $E_S\ne \varnothing$, then $|S|\ge 5$. From this we obtain $$20\times 12\ge \sum _{i=1}^{12}\left|E_i\right|=\sum _{i=1}^{12}\left(\sum _{i\in S}\left|E_S\right|\right)\ge 5\sum _S\left|E_S\right|=5n.$$ Therefore $n\le 48$.

Now, define a sequence $\{c_i{\}}_{i=1}^{52}$ of colours in the following way: $$\begin{array}{llllllllllll}1& 2& 3& 4& 5& 6& 7& 8& 9& 10& 11& 12\\ 4& 1& 2& 3& 8& 5& 6& 7& 12& 9& 10& 11\\ 3& 4& 1& 2& 7& 8& 5& 6& 11& 12& 9& 10\\ 2& 3& 4& 1& 6& 7& 8& 5& 10& 11& 12& 9\end{array}$$ The first row lists $c_1,\dots ,c_{12}$ in order, the second row lists $c_{13},\dots ,c_{24}$ in order, the third row lists $c_{25},\dots ,c_{36}$ in order, and finally the last row lists $c_{37},\dots ,c_{48}$ in order. For each $j,1\le j\le 48$, assign colours $c_j,c_{j+1},c_{j+2},c_{j+3},c_{j+4}$ to the $j$-th clown. It is easy to check that this assignment satisfies all conditions given above. So, 48 is the largest for $n$.

Remark: The fact that $n\le 48$ can be obtained in a much simpler observation that $$5n\le 12\times 20=240.$$