Difference between revisions of "2014 AMC 12A Problems/Problem 13"
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Revision as of 13:32, 8 February 2014
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
