Question

一支足球队有 $22$ 名可用球员。固定的 $11$ 名球员首发，另外 $11$ 名可作为替补。比赛中，教练最多可以换人 $3$ 次，每次用一名替补替换场上的 $11$ 名球员之一。被换下的球员不能再上场，但已经上场的替补之后可以被换下。任意两次换人不能同时发生。参与换人的球员以及换人的顺序都要计入不同方式。设 $n$ 为教练在比赛中换人的方式数（包括不换人的可能）。求 $n$ 除以 $1000$ 的余数。

A soccer team has $22$ available players. A fixed set of $11$ players starts the game, while the other $11$ are available as substitutes. During the game, the coach may make as many as $3$ substitutions, where any one of the $11$ players in the game is replaced by one of the substitutes. No player removed from the game may reenter the game, although a substitute entering the game may be replaced later. No two substitutions can happen at the same time. The players involved and the order of the substitutions matter. Let $n$ be the number of ways the coach can make substitutions during the game (including the possibility of making no substitutions). Find the remainder when $n$ is divided by $1000.$

答案

答案：122

知识点：乘法原理分类讨论

难度评级：2350

解答：

任一时刻场上都有 $11$ 名球员，其中任意一人都可以被换下，而替补席每次换人后少一人。因此第一次换人有 $11 \cdot 11$ 种方式，第二次有 $11 \cdot 10$ 种方式，第三次有 $11 \cdot 9$ 种方式。

对换 $0, 1, 2$ 或 $3$ 次求和，除以 $1000$ 的余数为 $122$ 。 $\begin{aligned} n &= 1 + 11 \cdot 11 \\ &\quad {}+ 11^2 \cdot 11 \cdot 10 \\ &\quad {}+ 11^3 \cdot 11 \cdot 10 \cdot 9 \\ &= 1 + 121 + 13310 + 1317690 \\ &= 1331122. \end{aligned}$

At every moment there are $11$ players in the game, any of whom may be removed, while the bench shrinks by one with each substitution. So the first substitution can be made in $11 \cdot 11$ ways, the second in $11 \cdot 10$ ways, and the third in $11 \cdot 9$ ways.

Summing over $0, 1, 2,$ or $3$ substitutions, $\begin{aligned} n &= 1 + 11 \cdot 11 \\ &\quad {}+ 11^2 \cdot 11 \cdot 10 \\ &\quad {}+ 11^3 \cdot 11 \cdot 10 \cdot 9 \\ &= 1 + 121 + 13310 + 1317690 \\ &= 1331122. \end{aligned}$ The remainder upon division by $1000$ is $122.$

2019 AIME I 第 4 题

答案：122

难度评级：2350

解答：

其他年份的第 4 题