Question

九个人坐下吃晚餐，有三种餐食可选。三个人点了牛肉餐，三个人点了鸡肉餐，三个人点了鱼肉餐。服务员按随机顺序给这九个人上餐。求有多少种上餐食类型的方法，使得恰好有一个人得到自己所点的餐食类型。

Nine people sit down for dinner where there are three choices of meals. Three people order the beef meal, three order the chicken meal, and three order the fish meal. The waiter serves the nine meals in random order. Find the number of ways in which the waiter could serve the meal types to the nine people so that exactly one person receives the type of meal ordered by that person.

答案

答案：216

知识点：错位排列分类讨论

难度评级：2400

解答：

先选择唯一被正确上餐的人，有 $9$ 种。由对称性，设此人点的是牛肉。剩下的餐为 $2$ 份牛肉、 $3$ 份鸡肉和 $3$ 份鱼肉，要分给剩下 $8$ 个人，其中有 $2$ 个点牛肉、 $3$ 个点鸡肉、 $3$ 个点鱼肉，且无人匹配。只需追踪剩下 $2$ 份牛肉餐送到鸡肉组还是鱼肉组。

若两份牛肉餐送到同一组，例如送给三个鸡肉点餐者中的两人，则连同送到鱼肉组的情况共有 $3 + 3 = 6$ 种。此时第三个鸡肉点餐者必须得到鱼肉，三个鱼肉点餐者必须得到三份鸡肉餐，两个牛肉点餐者得到剩下的鱼肉餐，全部被迫确定。若一份牛肉送给鸡肉点餐者、一份送给鱼肉点餐者，有 $3 \cdot 3 = 9$ 种；另外两个鸡肉点餐者必须得到鱼肉，另外两个鱼肉点餐者必须得到鸡肉，剩下一份鸡肉和一份鱼肉分给两个牛肉点餐者，有 $2$ 种。

总数为 $9\,(6 + 9 \cdot 2) = 216$ 。

Choose the one person served correctly ( $9$ ways); by symmetry say they ordered beef. The remaining meals — $2$ beef, $3$ chicken, and $3$ fish — must go to the other $8$ people ( $2$ beef, $3$ chicken, and $3$ fish orderers) with nobody matched. Track where the $2$ leftover beef meals go: to chicken or fish orderers.

If both go to the same group, say to two of the three chicken orderers ( $3 + 3 = 6$ ways counting both groups), then the third chicken orderer must receive fish, the three fish orderers must take the three chicken meals, and the two beef orderers take the remaining fish: everything is forced. If one goes to a chicken orderer and one to a fish orderer ( $3 \cdot 3 = 9$ ways), the other two chicken orderers must take fish and the other two fish orderers must take chicken, leaving one chicken and one fish meal to split between the two beef orderers ( $2$ ways).

The total is $9\,(6 + 9 \cdot 2) = 216.$

2012 AIME I 第 3 题

答案：216

难度评级：2400

解答：

其他年份的第 3 题