Question

一家精致的住宿加早餐旅馆有 $5$ 个房间，每个房间都有独特的颜色主题装饰。某天 $5$ 位朋友来过夜，当晚没有其他客人。这些朋友可以按任何组合住房，但每个房间最多住 $2$ 人。店主有多少种方式把客人分配到房间？

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?

$2100$

$2220$

$3000$

$3120$

$3125$

答案：B

知识点：分类讨论组合乘法原理

难度评级：1660

小提示：

按房间入住人数分类：全是单人、一对同住、或两对同住。

Split into cases by the room occupancies: all singles, one pair, or two pairs

大提示：

每种情况按“分组方式数”乘以“把这些组放进不同房间的方式数”来数。

Count each case as (ways to form the groups) times (ways to place the groups into distinct rooms)

解答：

全是单人：把 $5$ 位朋友分到 $5$ 个房间有 $5!=120$ 种。

一对同住：选这对有 $\binom52=10$ 种，再把 $4$ 个组放进房间有 $5\cdot4\cdot3\cdot2=120$ 种，共 $10\cdot120=1200$ 种。

两对同住：选出单独住的人有 $5$ 种，剩余四人分成两对有 $3$ 种，因此共有 $15$ 种分组方式。再把这 $3$ 组安排进房间，有 $5\cdot4\cdot3=60$ 种方式，共 $15\cdot60=900$ 种。

总数为 $120+1200+900=2220$ 。

所以正确答案是 B。

All singles: assign $5$ friends to $5$ rooms in $5!=120$ ways.

One pair: choose the pair in $\binom52=10$ ways, then place the $4$ groups into rooms in $5\cdot4\cdot3\cdot2=120$ ways, giving $10\cdot120=1200.$

Two pairs: choose the solo friend in $5$ ways and split the rest into two pairs in $3$ ways ( $15$ groupings), then place the $3$ groups into rooms in $5\cdot4\cdot3=60$ ways, giving $15\cdot60=900.$

The total is $120+1200+900=2220.$

Thus, the correct answer is B.

2014 AMC 12A 第 13 题

答案：B

难度评级：1660

小提示：

大提示：

解答：

其他年份的第 13 题