A hotel has exactly three rooms: one single room (for 1 person), one double room (for 2 persons), and one four-person room (for 4 persons). In how many distinct ways can 7 different persons be housed in these rooms so that every bed is occupied?

Introduction / Context:
This problem is about distributing distinct people into distinct rooms with fixed capacities. It is a typical application of combinations where each room has a fixed number of beds, and all people and rooms are distinguishable. The question asks for the number of ways to house seven persons in rooms of sizes 1, 2, and 4 such that each bed is filled. There is no mention of ordering within a room, so we consider only which persons share each room, not the seating arrangement inside the rooms.

Given Data / Assumptions:
- There are 7 distinct persons in total.
- Rooms available: one single room (capacity 1), one double room (capacity 2), and one four-person room (capacity 4).
- Total capacity: 1 + 2 + 4 = 7 beds, exactly matching the number of persons.
- Rooms are distinct because they differ by type and capacity.
- Order of people within the same room does not matter; only the grouping matters.

Concept / Approach:
The natural approach is to sequentially choose which person goes into the single room, which two go into the double room, and then allocate the remaining four to the four-person room. Each step uses combinations, as we are choosing subsets of people without regard to order inside the room. The multiplication principle applies because each independent choice can be combined with previous choices to form a complete allocation pattern.

Step-by-Step Solution:
Step 1: Choose 1 person out of 7 to occupy the single room.Step 2: The number of ways to choose this person is 7C1 = 7.Step 3: After assigning the single room, 6 persons remain for the other rooms.Step 4: Choose 2 of the remaining 6 persons to occupy the double room.Step 5: The number of ways to do this is 6C2 = 15.Step 6: The remaining 4 persons automatically go into the four-person room; there is only 1 way to choose them once the first two selections are fixed.Step 7: Total number of ways = 7C1 * 6C2 * 1 = 7 * 15 * 1 = 105.

Verification / Alternative check:
We can confirm by checking that no other hidden arrangements are being ignored. Since we do not distinguish positions within a room (for example, who sleeps in which specific bed of the four-person room), we correctly use combinations. If beds within rooms were labelled, we would need permutations inside each room, which would produce a larger result. Here, the room labels alone distinguish the groups, so 105 is the correct and complete count.

Why Other Options Are Wrong:
- 7!/5! equals 7 * 6 = 42 and would correspond to picking only two ordered positions, not full room assignments.
- 420 is 4 times the correct answer and might come from an incorrect permutation treatment of the double room or four-person room.
- 7! x 6! is astronomically too large and reflects a misunderstanding of the structure of the problem, effectively permuting people multiple times unnecessarily.

Common Pitfalls:
Students sometimes mistakenly treat ordering within rooms as significant, multiplying by extra factorials, or forget that rooms themselves are already distinguished by size. Another error is to choose the group of four first and then overcount by considering the same grouping multiple times in different orders. Sticking to a clean sequence—single room, then double room, then the rest—ensures each allocation is counted exactly once.

Final Answer:
The seven persons can be housed in 105 distinct ways in the three rooms.