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2014 AMC 12A Problem 13

Problem 13 of 25IntermediateCounting & Probability

A fancy bed and breakfast inn has 55 rooms, each with a distinctive color-coded decor. One day 55 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 22 friends per room. In how many ways can the innkeeper assign the guests to the rooms?

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Solution

All singles: assign 55 friends to 55 rooms in 5!=1205!=120 ways. One pair: choose the pair in (52)=10\binom52=10 ways, then place the 44 groups into rooms in 5⋅4⋅3⋅2=1205\cdot4\cdot3\cdot2=120 ways, giving 10⋅120=1200.10\cdot120=1200. Two pairs: choose the solo friend in 55 ways and split the rest into two pairs in 33 ways (1515 groupings), then place the 33 groups into rooms in 5⋅4⋅3=605\cdot4\cdot3=60 ways, giving 15⋅60=900.15\cdot60=900. The total is 120+1200+900=2220.120+1200+900=2220. Thus, the correct answer is B.

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Concepts: casework · combinations · multiplication principle

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.