2014 AMC 12A Problems/Problem 13
Problem
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?
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
.
(Solution by AwesomeToad)
See Also
2014 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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 |
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