2014 AMC 12A Problems/Problem 13

Problem

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 ways can the innkeeper assign the guests to the rooms?

$\textbf{(A) }2100\qquad \textbf{(B) }2220\qquad \textbf{(C) }3000\qquad \textbf{(D) }3120\qquad \textbf{(E) }3125\qquad$

Solution

We can discern three cases.

Case 1: Each room houses one guest. In this case, we have $5$ guests to choose for the first room, $4$ for the second, ..., for a total of $5!=120$ assignments.

Case 2: Three rooms house one guest; one houses two. We have $\binom{5}{3}$ ways to choose the three rooms with $1$ guest, and $\binom{2}{1}$ to choose the remaining one with $2$ . There are $5\cdot4\cdot3$ ways to place guests in the first three rooms, with the last two residing in the two-person room, for a total of $\binom{5}{3}\binom{2}{1}\cdot5\cdot4\cdot3=1200$ ways.

Case 3: Two rooms house two guests; one houses one. We have $\binom{5}{2}$ to choose the two rooms with two people, and $\binom{3}{1}$ to choose one remaining room for one person. Then there are $5$ choices for the lonely person, and $\binom{4}{2}$ for the two in the first two-person room. The last two will stay in the other two-room, so there are $\binom{5}{2}\binom{3}{1}\cdot5\cdot\binom{4}{2}=900$ ways.

In total, there are $120+1200+900=2220$ assignments, or $\boxed{\textbf{(B)}}$ .

(Solution by AwesomeToad)

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2014 AMC 12A Problems/Problem 13

Problem

Solution