Final answer:
There are 30 different ways to accommodate five guests in four rooms with no room left empty, by selecting the two guests that will share a room in 5 ways, and then distributing the remaining three across the other rooms in 6 ways.
Step-by-step explanation:
The question asks about ways to accommodate five guests in four rooms in such a way that no rooms remain vacant. This can be approached as a problem in combinatorial mathematics, specifically a partition problem where we want to partition a set of five guests into four non-empty subsets (rooms).
Since every room must be occupied, we have to distribute the guests in such a way that one room will have two guests and the remaining three rooms will have one guest each. This scenario doesn't care about the order of the rooms but does consider the distribution of guests to be different if guests are swapped. There are 5 ways to choose the two guests that will share a room. Once chosen, there are then 3! ways to distribute the remaining three guests across the three rooms, which multiplies out to 6 ways.
So, in total, there are 5 * 6 = 30 different ways to accommodate five guests in four rooms so that no room is left empty.