组合数学 (Fall 2015)/Problem Set 2: Difference between revisions
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* 在<math>n\to\infty</math>时，随机重排并发回作业后，满足上述要求的概率是多少。 | * 在<math>n\to\infty</math>时，随机重排并发回作业后，满足上述要求的概率是多少。 | ||
==Problem | ==Problem 2== | ||
Let <math>\pi</math> be a permutation of <math>[n]</math>. | Let <math>\pi</math> be a permutation of <math>[n]</math>. | ||
Recall that a cycle of permutation <math>\pi</math> of length <math>k</math> is a tuple <math>(a_1,a_2,\ldots,a_k)</math> such that <math>a_2=\pi(a_1), a_3=\pi(a_2),\ldots,a_k=\pi(a_{k-1})</math> and <math>a_1=\pi(a_k)\,</math>. Thus a fixed point of <math>\pi</math> is just a cycle of length 1. | Recall that a cycle of permutation <math>\pi</math> of length <math>k</math> is a tuple <math>(a_1,a_2,\ldots,a_k)</math> such that <math>a_2=\pi(a_1), a_3=\pi(a_2),\ldots,a_k=\pi(a_{k-1})</math> and <math>a_1=\pi(a_k)\,</math>. Thus a fixed point of <math>\pi</math> is just a cycle of length 1. | ||
* Fix <math>k\ge 1</math>. Let <math>f_k(n)</math> be the number of permutations of <math>[n]</math> having no cycle of length <math>k</math>. Compute this <math>f_k(n)</math> and the limit <math>\lim_{n\rightarrow\infty}\frac{f_k(n)}{n!}</math>. | * Fix <math>k\ge 1</math>. Let <math>f_k(n)</math> be the number of permutations of <math>[n]</math> having no cycle of length <math>k</math>. Compute this <math>f_k(n)</math> and the limit <math>\lim_{n\rightarrow\infty}\frac{f_k(n)}{n!}</math>. | ||
Revision as of 09:29, 2 November 2015
Problem 1
假设我们班上有n+2个人，其中两个人是DNA完全相同的双胞胎。我们收上n+2份作业后，将这些作业打乱后发回给全班同学，每人一份。要求每个人不可以收到自己那一份作业或者与自己DNA相同的人的作业。令[math]\displaystyle{ T_n }[/math]表示满足这个要求的发回作业的方式，问：
- 计算[math]\displaystyle{ T_n }[/math]是多少；
- 在[math]\displaystyle{ n\to\infty }[/math]时，随机重排并发回作业后，满足上述要求的概率是多少。
Problem 2
Let [math]\displaystyle{ \pi }[/math] be a permutation of [math]\displaystyle{ [n] }[/math]. Recall that a cycle of permutation [math]\displaystyle{ \pi }[/math] of length [math]\displaystyle{ k }[/math] is a tuple [math]\displaystyle{ (a_1,a_2,\ldots,a_k) }[/math] such that [math]\displaystyle{ a_2=\pi(a_1), a_3=\pi(a_2),\ldots,a_k=\pi(a_{k-1}) }[/math] and [math]\displaystyle{ a_1=\pi(a_k)\, }[/math]. Thus a fixed point of [math]\displaystyle{ \pi }[/math] is just a cycle of length 1.
- Fix [math]\displaystyle{ k\ge 1 }[/math]. Let [math]\displaystyle{ f_k(n) }[/math] be the number of permutations of [math]\displaystyle{ [n] }[/math] having no cycle of length [math]\displaystyle{ k }[/math]. Compute this [math]\displaystyle{ f_k(n) }[/math] and the limit [math]\displaystyle{ \lim_{n\rightarrow\infty}\frac{f_k(n)}{n!} }[/math].