2014 AMC 12A 第 13 题

先试着解答 2014 AMC 12A 第 13 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2014 AMC 12A 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

13.

一家精致的住宿加早餐旅馆有 55 个房间,每个房间都有独特的颜色主题装饰。某天 55 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 22 人。店主有多少种方式把客人分配到房间?
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?
21002100
22202220
30003000
31203120
31253125
答案:B
知识点:分类讨论组合乘法原理
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 55 位朋友分到 55 个房间有 5!=1205!=120 种。
一对同住:选这对有 (52)=10\binom52=10 种,再把 44 个组放进房间有 5⋅4⋅3⋅2=1205\cdot4\cdot3\cdot2=120 种,共 10⋅120=120010\cdot120=1200 种。
两对同住:选出单独住的人有 55 种,剩余四人分成两对有 33 种,因此共有 1515 种分组方式。再把这 33 组安排进房间,有 5⋅4⋅3=605\cdot4\cdot3=60 种方式,共 15⋅60=90015\cdot60=900 种。
总数为 120+1200+900=2220120+1200+900=2220。
所以正确答案是 B。
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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