5. let $n>1$ and $m$ be natural numbers. A parliament consists of $m n$ members who have formed $2 n$ commissions, so that

(i) Each commission shall consist of $m$ deputies.

(ii) Each Member of Parliament is a member of exactly 2 commissions.

(iii) No two commissions shall have more than one joint member.

Determine the largest possible value of $m$ as a function of $n$ so that this is possible.
## 1st solution

Let $K$ be an arbitrary committee. Each of the $m$ members of $K$ is in another committee according to (ii), and these are all different according to (iii). In total, there are at least $m+1$ commissions, so $m \leq 2 n-1$ applies.

We now show that $m=2 n-1$ is possible by recursively constructing an example. For $n=2, m=3$ and six parliamentarians 1,2,3,4,5,6 the four commissions fulfill

$$
K_{1}=\{1,2,3\}, K_{2}=\{3,4,5\}, K_{3}=\{5,6,1\}, K_{4}=\{2,4,6\}
$$

all conditions of the task. We now assume that we have already constructed suitable committees $K_{1}, \ldots, K_{2 n}$ for $n, m=2 n-1$ and consider the case $n+1$, $m=2 n+1$. So $4 n+1$ parliamentarians $P_{1}, \ldots, P_{4 n+1}$ are added. Construct
now construct the new committees $K_{1}^{\prime}, \ldots, K_{2 n+2}^{\prime}$ as follows:

$$
\begin{aligned}
K_{1}^{\prime} & =K_{1} \cup\left\{P_{1}, P_{2}\right\} \\
K_{2}^{\prime} & =K_{2} \cup\left\{P_{3}, P_{4}\right\} \\
& \vdots \\
K_{2 n}^{\prime} & =K_{2 n} \cup\left\{P_{4 n-1}, P_{4 n}\right\} \\
K_{2 n+1}^{\prime} & =\left\{P_{1}, P_{3}, \ldots, P_{4 n-1}, P_{4 n+1}\right\} \\
K_{2 n+2}^{\prime} & =\left\{P_{2}, P_{4}, \ldots, P_{4 n}, P_{4 n+1}\right\}
\end{aligned}
$$