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Let $p_{n}(k)$ denote the number of permutations of length $n$ with exactly $k$ fixed points (a fixed point of $\pi$ is an element $t$ such that $\pi(t)=t$ ). Show that the generating function $$ f_{n}(x)=\sum_{i=0}^{n} x^{i} p_{n}(i) $$ satisfies $f_{n}(1)=f_{n}^{\prime}(1)=n !$

EDIT: Question context: Math graduate course - Probabilistic methods (2020)
My work: since every permutation of length n has a number of fixed points between 0 and n included it implies that $f_{n}(1)=n !$

Mattiatore
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    Have you got any thoughts of your own on this? Hint: $p_n(0)+p_n(1)$ is the number of permutations in $S_n$ with zero or with one fixed point. What can you say about $p_n(0)+p_n(1)+p_n(2)+\dots+p_n(n)$? How many fixed points are possible for permutations in $S_n$? – JMoravitz Aug 03 '21 at 21:26
  • Well I understand that the hint solves the first question, because every permutation of length n has a number of fixed points between 0 and n included. For the second part about the derivative I am not sure. After calculating I get $f_{n}^{\prime}(1)= \sum_{i=0}^{n} i p_{n}(i) $ but nothing after this. – Mattiatore Aug 03 '21 at 21:37
  • Note that the number of permutations of length $n$ with exactly $k$ fixed points is $\binom{n}{k}\mathcal{D}(n-k)$ because there are $\binom{n}{k}$ ways to choose the fixed points and $\mathcal{D}(n-k)$ to arrange the others, where $\mathcal{D}(k)$ is the number of Derangements of $k$ items. – robjohn Aug 04 '21 at 07:08
  • Context added following the guidelines. – Mattiatore Aug 04 '21 at 19:01

2 Answers2

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Note that the number of permutations of length $n$ with exactly $k$ fixed points is $\binom{n}{k}\mathcal{D}(n-k)$ because there are $\binom{n}{k}$ ways to choose the fixed points and $\mathcal{D}(n-k)$ to arrange the others, where $\mathcal{D}(k)$ is the number of Derangements of $k$ items.

Therefore, the generating function is $$ \begin{align} \overbrace{\sum_{k=0}^n\binom{n}{k}\mathcal{D}(n-k)x^k}^{f_n(x)} &=\sum_{k=0}^n\overbrace{\frac{n!}{k!(n-k)!}\vphantom{\sum_0^k}}^{\binom{n}{k}}\overbrace{\sum_{j=0}^{n-k}(-1)^j\frac{(n-k)!}{j!}}^{\mathcal{D}(n-k)}x^k\tag{1a}\\[6pt] &=n!\!\!\sum_{\substack{j,k\ge0\\j+k\le n}}\!\!\frac{(-1)^j}{j!\,k!}x^k\tag{1b}\\ &=n!\sum_{m=0}^n\sum_{j=0}^m\frac{(-1)^j}{j!\,(m-j)!}x^{m-j}\tag{1c}\\[3pt] &=n!\sum_{m=0}^n\frac1{m!}\sum_{j=0}^m\binom{m}{j}x^{m-j}(-1)^j\tag{1d}\\[3pt] &=\sum_{m=0}^n\frac{n!}{m!}(x-1)^m\tag{1e} \end{align} $$ Explanation:
$\text{(1a)}$: expand $\mathcal{D}(n-k)$ (from $(5)$ in this answer) and $\binom{n}{k}$
$\text{(1b)}$: cancel terms and reindex
$\text{(1c)}$: substitute $m=j+k$
$\text{(1d)}$: multiply and divide by $m!$
$\text{(1e)}$: apply the Binomial Theorem

Thus, for $0\le m\le n$, $$ f^{(m)}(1)=n!\tag2 $$

robjohn
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Consider permutations of $[n]$: Each permuatution will have $i$ fixed points (where $i $ ranges from $0$ to $n$) & for a given value $i$, $p_n(i)$ of them will have $i$ fixed points, so \begin{eqnarray*} p_n(0)+p_n(1) +\cdots+p(_n(n)=n! \end{eqnarray*} and thus $f_{n}(1)=n !$.

To show the second part we need to notice that $ip_n(i)=np_{n-1}(i-1)$. This can be shown by the following combinatorial arguement: Given a permuatation of $[n]$ with $i$ fixed points, choose one of these fixed points. Now delete this element and reorder the other larger elements by decreasing them by $1$ to give a permutation of $n-1$ with $i-1$ fixed points & each of these permutations can be obtained in $n$ possible ways.

Donald Splutterwit
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