2018 AIME II 第 11 题

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11.

1,2,3,4,5,61, 2, 3, 4, 5, 6 的排列个数,使得对每个满足 1k51 \le k \le 5kk,排列的前 kk 项中至少有一项大于 kk

Find the number of permutations of 1,2,3,4,5,61, 2, 3, 4, 5, 6 such that for each kk with 1k5,1 \le k \le 5, at least one of the first kk terms of the permutation is greater than k.k.

答案:461
知识点:排列补集计数递推计数
难度评级:3060
解答:

条件失败恰好当某个 k5.k \le 5. 使前 kk 项是 {1,,k}\{1, \ldots, k\} 的一个排列。对 1,,n,1, \ldots, n, 的一个排列,令 kk 为前缀正好是 {1,,k}\{1, \ldots, k\} 的最小长度(全长 nn 总是可行),并令 cnc_n 为这个最小的 kkn.n. 的排列数。我们要求的是 c6.c_6.

每个 1,,n1, \ldots, n 的排列都唯一分解为长度 kk 的最小前缀(有 ckc_k 种选择),后接剩余 nkn - k 个数的任意排列,所以 k=1nck(nk)!=n!.\sum_{k=1}^{n} c_k \,(n-k)! = n!.c1=1c_1 = 1 开始,得到 c2=1c_2 = 1c3=3c_3 = 3c4=13c_4 = 13c5=71c_5 = 71,并且 c6=720(1201+241+63+213+171)=720259=461. \begin{aligned} c_6 &= 720 \\ &\quad {}- \tiny(120 \cdot 1 + 24 \cdot 1 + 6 \cdot 3 + 2 \cdot 13 + 1 \cdot 71) \\ &= 720 - 259 = 461. \end{aligned}

The condition fails exactly when the first kk terms are a permutation of {1,,k}\{1, \ldots, k\} for some k5.k \le 5. For a permutation of 1,,n,1, \ldots, n, let kk be the smallest length for which the prefix is {1,,k}\{1, \ldots, k\} (the full length nn always works), and let cnc_n be the number of permutations whose smallest such kk is n.n. We want c6.c_6.

Every permutation of 1,,n1, \ldots, n decomposes uniquely as a minimal prefix of length kk (ckc_k choices) followed by any arrangement of the remaining nkn - k values, so k=1nck(nk)!=n!.\sum_{k=1}^{n} c_k \,(n-k)! = n!. Starting from c1=1,c_1 = 1, this gives c2=1,c_2 = 1, c3=3,c_3 = 3, c4=13,c_4 = 13, c5=71,c_5 = 71, and c6=720(1201+241+63+213+171)=720259=461. \begin{aligned} c_6 &= 720 \\ &\quad {}- \tiny(120 \cdot 1 + 24 \cdot 1 + 6 \cdot 3 + 2 \cdot 13 + 1 \cdot 71) \\ &= 720 - 259 = 461. \end{aligned}

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