Question

给定八枚可区分的戒指，设 $n$ 为把其中五枚戒指戴在一只手的四根手指（不包括拇指）上的可能排列数。同一根手指上戒指的上下顺序有区别，但不要求每根手指都戴戒指。求 $n$ 从左起前三个非零数字。

Given eight distinguishable rings, let $n$ be the number of possible five-ring arrangements on the four fingers (not the thumb) of one hand. The order of rings on each finger is significant, but it is not required that each finger have a ring. Find the leftmost three nonzero digits of $n.$

答案

答案：376

知识点：组合排列隔板法

难度评级：2300

解答：

选择使用哪五枚戒指有 $\binom{8}{5} = 56$ 种，再把它们排成顺序（依次读第一根手指、第二根手指等）有 $5! = 120$ 种。剩下的是把这个有序列表分成四个可为空的连续段，每段对应一根手指；也就是把 $5$ 分成 $4$ 个非负整数部分的方案数，由隔板法为 $\binom{8}{3} = 56$ 。

因此 $n = 56 \cdot 120 \cdot 56 = 376320$ ，从左起前三个非零数字是 $376$ 。

Choose which five rings to use in $\binom{8}{5} = 56$ ways, and order them (reading down the first finger, then the second, and so on) in $5! = 120$ ways. It remains to split the ordered list into four possibly empty consecutive blocks, one per finger: the number of compositions of $5$ into $4$ nonnegative parts, which by stars and bars is $\binom{8}{3} = 56.$

Therefore $n = 56 \cdot 120 \cdot 56 = 376320,$ whose leftmost three nonzero digits are $376.$

2000 AIME II 第 5 题

答案：376

难度评级：2300

解答：

其他年份的第 5 题