# Combination's problem (AMC 12A 2014)

A fancy bed and breakfast inn has 5 rooms, each with a distinctive color-coded decor. One day 5 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 2 friends per room. In how many can the innkeeper assign the guests to the rooms?

The possible answers are: (A) 2100 (B) 2220 (C) 3000 (D) 3120 (E) 3125

• For a contest of this type, cases: (i) all singles; (ii) one couple; (iii) two couples. It is easy to miscount the two couple case. – André Nicolas Feb 6 '14 at 9:55

The innkeeper can organize the guests into rooms with the following unordered frequencies: $$1,1,1,1,1, \qquad 1,1,1,2 \qquad \text{or} \qquad 1,2,2.$$
• Case $\{1,1,1,1,1\}$: there are $5!$ ways to assign the guests.
• Case $\{1,1,1,2\}$: there's $5 \times 4$ ways to assign the frequencies to rooms, and $\binom{5}{2} 3!$ ways to assign the guests to rooms (once the frequencies have been assigned).
• Case $\{1,2,2\}$: there's $5 \times \binom{4}{2}$ ways to assign the frequencies to rooms, and $\binom{5}{2}\binom{3}{2}$ ways to assign the guests to rooms (once the frequencies have been assigned).
In total this gives: $$5!+5 \times 4 \times \binom{5}{2}3!+5 \times \binom{4}{2} \times \binom{5}{2}\binom{3}{2}=2220.$$
• Can you please explain the $\dbinom{5}{2}3!$ in case {$1 1 1 2$[.I counted it in the following process:5 possibilities for the first,4 for the second,3 for the third and only 1for the last 2.This yields the same answer as yours,but I am interested in knowing how you counted it. – rah4927 Feb 6 '14 at 10:26
• The way I see it, we place 2 of the 5 people for the 2-person room: $\binom{5}{2}$, and the remaining 3 people can be put in the non-empty rooms in $3!$ ways. (There's probably several equivalent ways of interpreting these numbers.) – Rebecca J. Stones Feb 6 '14 at 10:30