K2) Let $n$ be a natural number. A volleyball team consisting of $n$ women and $n$ men lines up for a game. Each team member occupies one of the positions $1,2, \ldots, 2 n$, whereby exactly positions 1 and $n+1$ are outside the court. During the game, all team members rotate, switching from position $i$ to position $i+1$ (or from $2 n$ to 1). How many possibilities are there for the starting line-up so that there are always at least $n-1$ women on the pitch, no matter how often the team rotates?

Note: Two starting line-ups are different if at least one team member occupies a different position.
## Solution (Bibin, David):

We visualize the task as follows: The people stand in a circle. The team members in positions $k$ and $n+k$ for $1 \leq k \leq n$ face each other.

Observation: A woman must always stand opposite a man, otherwise there is a rotation in which two women are outside the playing field at the same time.

We can therefore form $n$ pairs of one woman and one man, each facing each other. There are $n$ ! ways of forming these pairs: We place the $n$ women next to each other in a row. Then the men still have $n$ ! possibilities to line up in a second row behind the women. Each of these $n$ ! line-ups correspond to exactly one way of forming the pairs.

Since the pairs are opposite each other, we consider the sets of opposite positions, namely $\{1, n+1\},\{2, n+2\}, \ldots,\{n, 2 n\}$. One of these sets belongs to each pair. Now there are $n$ ! possibilities to assign the pairs to these positions.

In addition, for each pair, the man and woman can be swapped with each other or not. There are 2 possibilities for each of the $n$ pairs, which are independent of each other. The number of possibilities is therefore $2^{n}$.

The starting constellation is already clearly determined by these independent steps. In total, there are therefore $n!\cdot n!\cdot 2^{n}=(n!)^{2} \cdot 2^{n}$ possibilities.