2014 AMC 12A Problems/Problem 13
Problem
Solution
We can discern three cases.
Case 1: Each room houses one guest. In this case, we have guests to choose for the first room, for the second, ..., for a total of assignments.
Case 2: Three rooms house one guest; one houses two. We have ways to choose the three rooms with guest, and to choose the remaining one with . There are ways to place guests in the first three rooms, with the last two residing in the two-person room, for a total of ways.
Case 3: Two rooms house two guests; one houses one. We have to choose the two rooms with two people, and to choose one remaining room for one person. Then there are choices for the lonely person, and for the two in the first two-person room. The last two will stay in the other two-room, so there are ways.
In total, there are assignments, or .