In a circus, there are $n$ clowns who dress and paint themselves up using a selection of 12 distinct colours. Each clown is required to use at least five different colours. One day, the ringmaster of the circus orders that no two clowns have exactly the same set
of colours and no more than 20 clowns may use any one particular colour. Find the largest number $n$ of clowns so as to make the ringmaster's order possible.
Let $C$ be the set of $n$ clowns. Label the colours $1,2,3, \ldots, 12$. For each $i=1,2, \ldots, 12$, let $E_{i}$ denote the set of clowns who use colour $i$. For each subset $S$ of $\{1,2, \ldots, 12\}$, let $E_{S}$ be the set of clowns who use exactly those colours in $S$. Since $S \neq S^{\prime}$ implies $E_{S} \cap E_{S^{\prime}}=\emptyset$, we have

$$
\sum_{S}\left|E_{S}\right|=|C|=n
$$

where $S$ runs over all subsets of $\{1,2, \ldots, 12\}$. Now for each $i$,

$$
E_{S} \subseteq E_{i} \quad \text { if and only if } \quad i \in S
$$

and hence

$$
\left|E_{i}\right|=\sum_{i \in S}\left|E_{S}\right| .
$$

By assumption, we know that $\left|E_{i}\right| \leq 20$ and that if $E_{S} \neq \emptyset$, then $|S| \geq 5$. From this we obtain

$$
20 \times 12 \geq \sum_{i=1}^{12}\left|E_{i}\right|=\sum_{i=1}^{12}\left(\sum_{i \in S}\left|E_{S}\right|\right) \geq 5 \sum_{S}\left|E_{S}\right|=5 n
$$

| $1234 \mid 5678$ | $9101112 \mid$ |  |
| :---: | :---: | :---: |
| $23 \mid 8567$ | 1291011 \| |  |
| 7856 | $1112910 \mid$ |  |
| 6785 | $1011129 \mid$ | 234 |

Therefore $n \leq 48$.

Now, define a sequence $\left\{c_{i}\right\}_{i=1}^{52}$ of colours in the following way:

The first row lists $c_{1}, \ldots, c_{12}$ in order, the second row lists $c_{13}, \ldots, c_{24}$ in order, the third row lists $c_{25}, \ldots, c_{36}$ in order, and finally the last row lists $c_{37}, \ldots, c_{52}$ in order. For each $j, 1 \leq j \leq 48$, 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 $n \leq 48$ can be obtained in a much simpler observation that

$$
5 n \leq 12 \times 20=240 \text {. }
$$

There are many other ways of constructing 48 distinct sets consisting of 5 colours. For example, consider the sets

$$
\begin{array}{ccccc}
\{1,2,3,4,5,6\}, & \{3,4,5,6,7,8\}, & \{5,6,7,8,9,10\}, & \{7,8,9,10,11,12\}, \\
\{9,10,11,12,1,2\}, & \{11,12,1,2,3,4\}, & \{1,2,5,6,9,10\}, & \{3,4,7,8,11,12\} .
\end{array}
$$

Each of the above 8 sets has 6 distinct subsets consisting of exactly 5 colours. It is easy to check that the 48 subsets obtained in this manner are all distinct.