2009 AIME I 第 10 题

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10.

年度星际数学考试(AIME)由五名火星人、五名金星人和五名地球人组成的委员会编写。开会时,委员们围坐在圆桌旁,座位按顺时针顺序编号为 111515。委员会规定,火星人必须坐在 11 号椅,地球人必须坐在 1515 号椅。此外,地球人不能紧坐在火星人的左边,火星人不能紧坐在金星人的左边,金星人不能紧坐在地球人的左边。委员会可能的座位安排数为 N(5!)3N \cdot (5!)^3。求 NN

The Annual Interplanetary Mathematics Examination (AIME) is written by a committee of five Martians, five Venusians, and five Earthlings. At meetings, committee members sit at a round table with chairs numbered from 11 to 1515 in clockwise order. Committee rules state that a Martian must occupy chair 11 and an Earthling must occupy chair 15.15. Furthermore, no Earthling can sit immediately to the left of a Martian, no Martian can sit immediately to the left of a Venusian, and no Venusian can sit immediately to the left of an Earthling. The number of possible seating arrangements for the committee is N(5!)3.N \cdot (5!)^3. Find N.N.

答案:346
知识点:有限制的排列分拆与有序分拆隔板法
难度评级:2990
解答:

先选择每把椅子上坐哪个星球的人;之后每个星球的具体成员都可以用 5!5! 种方式分配到本星球的椅子上,所以 NN 数的是星球模式。相邻规则等价于:按顺时针读取时,每个最大的火星人连续块后面必须接一个金星人连续块,再接一个地球人连续块,然后火星人才可以再次出现。因为 11 号椅坐火星人,1515 号椅坐地球人,所以从 111515 号椅由 (火星人块、金星人块、地球人块)这一模式重复 kk 次组成,其中 1k51 \le k \le 5

对于给定的 kk, 每个星球的五名成员被按顺序分配到 kk 个非空块中,而把 55 写成 kk 个正整数的有序和的方法数为 (4k1)\binom{4}{k-1}。三个星球的块大小相互独立,所以 N=k=15(4k1)3=13+43+63+43+13=346. \begin{aligned} N &= \sum_{k=1}^{5} \binom{4}{k-1}^3 \\ &= 1^3 + 4^3 + 6^3 + 4^3 + 1^3 \\ &= 346. \end{aligned}

First choose which planet sits in each chair; the individuals from each planet can then be assigned to their chairs in 5!5! ways apiece, so NN counts the planet patterns. The adjacency rules say exactly that, reading clockwise, each maximal block of Martians must be followed by a block of Venusians and then a block of Earthlings before Martians can appear again. Since chair 11 holds a Martian and chair 1515 holds an Earthling, the chairs from 11 to 1515 consist of the pattern (Martian block, Venusian block, Earthling block) repeated kk times, for some 1k5.1 \le k \le 5.

For a given k,k, each planet's five members are distributed into kk nonempty blocks in order, and the number of ways to write 55 as an ordered sum of kk positive integers is (4k1).\binom{4}{k-1}. The three planets' block sizes are independent, so N=k=15(4k1)3=13+43+63+43+13=346. \begin{aligned} N &= \sum_{k=1}^{5} \binom{4}{k-1}^3 \\ &= 1^3 + 4^3 + 6^3 + 4^3 + 1^3 \\ &= 346. \end{aligned}

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