B3. Six scouts go on a scavenger hunt. On Saturday, they go to the forest and on Sunday, they go into the mountains. On both days, they have to find their way in pairs. Their leader wants to divide them into pairs for each trip, so that no one has the same partner on the second day as on the first day. In how many ways can he do this?
B3. 120 Let the oldest scout be $A$. There are 5 possibilities to find a partner $B$ for him for the first day. Then there are 4 possible partners $C$ for $B$ on the second day, as he cannot pair with $A$ again. For $C$, there are still 3 possible partners $D$ for the first day, as he cannot pair with $B$ twice, and $A$ is already occupied. For $D$, there are then 2 possible partners $E$ for the second day, as $B$ and $C$ are already occupied and $A$ cannot be his partner because that would leave two scouts who form a pair on both days. Finally, there is 1 scout left who has no choice and goes with $E$ on the first day and with $A$ on the second day. In total, there are thus $5 \times 4 \times 3 \times 2=120$ possibilities.