组合数学 (Fall 2015)/Problem Set 2

Problem 4

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].