2025 AIME I 第 3 题

Question

一支棒球队的

9

名队员赛后去了冰淇淋店。每名队员都买了一个单球蛋筒，口味为巧克力、香草或草莓。每种口味至少有一名队员选择，并且选择巧克力的人数大于选择香草的人数，选择香草的人数又大于选择草莓的人数。令

N

为满足这些条件的不同口味分配方式数。求

N

除以

1000

的余数。

The

9

members of a baseball team went to an ice-cream parlor after their game. Each player had a single scoop cone of chocolate, vanilla, or strawberry ice cream. At least one player chose each flavor, and the number of players who chose chocolate was greater than the number of players who chose vanilla, which was greater than the number of players who chose strawberry. Let

N

be the number of different assignments of flavors to players that meet these conditions. Find the remainder when

N

is divided by

1000.

答案

答案：16

知识点：多重集排列分类讨论

难度评级：2180

小提示：

先找出所有口味人数三元组：三个正整数严格递减且和为

9

First find every triple of flavor counts: strictly decreasing positive integers summing to

9

大提示：

对人数

(c, v, s)

，把口味分给九名不同队员的方式数是

\frac{9!}{c!\,v!\,s!}

For counts

(c, v, s),

the number of ways to assign flavors to the nine distinct players is

\frac{9!}{c!\,v!\,s!}

解答：

令

c \gt v \gt s \ge 1

分别为选择巧克力、香草和草莓的人数，并且

c + v + s = 9

。检查小的

s

值，可得唯一可能为

(6, 2, 1)

、

(5, 3, 1)

，和

(4, 3, 2)

。

因为队员彼此不同，每个人数三元组贡献一个多项式系数：

\begin{aligned} \frac{9!}{6!\,2!\,1!} &= 252, \\ \frac{9!}{5!\,3!\,1!} &= 504, \\ \frac{9!}{4!\,3!\,2!} &= 1260 \end{aligned}\text{。}

因此

N = 252 + 504 + 1260 = 2016

，除以

1000

的余数为

16

。

Let

c \gt v \gt s \ge 1

be the numbers of players choosing chocolate, vanilla, and strawberry, with

c + v + s = 9.

Checking small values of

s

shows the only possibilities are

(6, 2, 1),

(5, 3, 1),

and

(4, 3, 2).

Since the players are distinct, each triple of counts contributes a multinomial coefficient:

\begin{aligned} \frac{9!}{6!\,2!\,1!} &= 252, \\ \frac{9!}{5!\,3!\,1!} &= 504, \\ \frac{9!}{4!\,3!\,2!} &= 1260. \end{aligned}

Thus

N = 252 + 504 + 1260 = 2016,

and the remainder modulo

1000

is

16.

2025 AIME I 第 3 题

答案：16

难度评级：2180

小提示：

大提示：

解答：

其他年份的第 3 题