Question

求将 $66$ 枚相同硬币分成三个非空堆的方法数，使得第一堆的硬币数少于第二堆，第二堆的硬币数少于第三堆。

Find the number of ways $66$ identical coins can be separated into three nonempty piles so that there are fewer coins in the first pile than in the second pile and fewer coins in the second pile than in the third pile.

答案

答案：331

知识点：分拆与有序分拆隔板法对称性

难度评级：2180

解答：

正整数有序三元组 $(a, b, c)$ 满足 $a + b + c = 66$ 的个数为 $\binom{65}{2} = 2080$ 。其中三项全相等的恰有一个，即 $(22, 22, 22)$ 。恰有两个数值相等的三元组来自 $2a + c = 66$ 且 $c \ne a$ ：这里 $a$ 可取 $1$ 到 $32$ ，但不能取 $22$ ，得到 $31$ 个多重集合，每个可排列成 $3$ 个有序三元组，所以共有 $93$ 个。

因此有三个互不相同数值的有序三元组为 $2080 - 1 - 93 = 1986$ 个，而每个无序选择 $a \lt b \lt c$ 被计数 $6$ 次。有效分法数为 $\frac{1986}{6} = 331$ 。

The ordered triples $(a, b, c)$ of positive integers with $a + b + c = 66$ number $\binom{65}{2} = 2080.$ Exactly one of them has all three values equal, namely $(22, 22, 22).$ Triples with exactly two values equal come from $2a + c = 66$ with $c \ne a:$ here $a$ can be $1$ through $32$ except $22,$ giving $31$ multisets, each arrangeable in $3$ ways, so $93$ ordered triples.

Hence $2080 - 1 - 93 = 1986$ ordered triples have three distinct values, and each unordered choice $a \lt b \lt c$ is counted $6$ times. The number of valid separations is $\frac{1986}{6} = 331.$

2021 AIME I 第 4 题

答案：331

难度评级：2180

解答：

其他年份的第 4 题