组合数学 (Fall 2015)/Problem Set 2: Difference between revisions
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:<math>N_m=\sum_{k=m}^r(-1)^{k-m}{k\choose m}s_k</math>, | :<math>N_m=\sum_{k=m}^r(-1)^{k-m}{k\choose m}s_k</math>, | ||
:and please explicitly give the <math>s_k</math>. | :and please explicitly give the <math>s_k</math>. | ||
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Revision as of 09:49, 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].
Problem 3
Let [math]\displaystyle{ N_m }[/math] denote the number of objects from a collection of [math]\displaystyle{ N }[/math] objects that possess exactly [math]\displaystyle{ m }[/math] of the properties [math]\displaystyle{ a_1,a_2,\ldots,a_r }[/math]. Generalize the principle of inclusion-exclusion by computing [math]\displaystyle{ N_m }[/math] as the following form
- [math]\displaystyle{ N_m=\sum_{k=m}^r(-1)^{k-m}{k\choose m}s_k }[/math],
- and please explicitly give the [math]\displaystyle{ s_k }[/math].
Problem 4
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