AMC12 2014 A

AMC12 2014 A · Q13

AMC12 2014 A · Q13. It mainly tests Basic counting (rules of product/sum), Combinations.

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?

一家高档床和早餐旅馆有 $5$ 个房间，每个房间都有独特的颜色装饰。有一天，$5$ 个朋友前来过夜。那天晚上没有其他客人。朋友们可以任意组合入住，但每个房间最多 $2$ 个朋友。旅馆老板可以有多少种方式分配客人到房间？

(A) 2100 2100

(B) 2220 2220

(D) 3120 3120

(E) 3125\qquad 3125\qquad

Answer

Correct choice: (B)

正确答案：（B）

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 $\boxed{\textbf{(B)}}$.

我们可以区分三种情况。情况 1：每个房间住一个客人。在这种情况下，第一房间有 $5$ 个客人选择，第二房间 $4$ 个，...，总共 $5!=120$ 种分配。情况 2：三个房间住一个客人；一个房间住两个。我们有 $\binom{5}{3}$ 种方式选择三个单人房间，$\binom{2}{1}$ 种选择剩下的双人房间。前三个房间放置客人有 $5\cdot4\cdot3$ 种方式，最后两个住在双人房间，总共 $\binom{5}{3}\binom{2}{1}\cdot5\cdot4\cdot3=1200$ 种方式。情况 3：两个房间住两个客人；一个住一个。我们有 $\binom{5}{2}$ 种选择两个双人房间，$\binom{3}{1}$ 种选择剩下的单人房间。然后孤独的人有 $5$ 种选择，第一双人房间两人有 $\binom{4}{2}$ 种。最后两人住在另一个双人房间，总共 $\binom{5}{2}\binom{3}{1}\cdot5\cdot\binom{4}{2}=900$ 种方式。总共，$120+1200+900=2220$ 种分配，即 $\boxed{\textbf{(B)}}$ 。

Topics