2014 AMC 12A Problems/Problem 13
A fancy bed and breakfast inn has rooms, each with a distinctive color-coded decor. One day friends arrive to spend the night. There are no other guests that night. The friends can room in any combination they wish, but with no more than friends per room. In how many ways can the innkeeper assign the guests to the rooms?
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 .
(Solution by AwesomeToad16)
We can work in reverse by first determining the number of combinations in which there are more than friends in at least one room. There are three cases:
Case 1: Three friends are in one room. Since there are possible rooms in which this can occur, we are choosing three friends from the five, and the other two friends can each be in any of the four remaining rooms, there are possibilities.
Case 2: Four friends are in one room. Again, there are possible rooms, we are choosing four of the five friends, and the other one can be in any of the other four rooms, so there are possibilities.
Case 3: Five friends are in one room. There are possible rooms in which this can occur, so there are possibilities.
Since there are possible combinations of the friends, the number fitting the given criteria is .
Video Solution by OmegaLearn
|2014 AMC 12A (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25|
|All AMC 12 Problems and Solutions|