Proving some identities in Derangements.

Question

Let $D_n$ denote the number of derangements of {1,2,3,...,n}.

We know that for $n\geq1$, we have:

\begin{equation*} D_n=n!(1-\frac{1}{1!}+\frac{1}{2!}+...+(-1)^n\frac{1}{n!}). \end{equation*}

Given $D_n=nD_{n-1}+(-1)^n$ for $n=2,3,...$

My question is: How can I prove the formula for $D_n$ above using the formula given. I am planning to use induction but I cant proceed. Thanks for your help.

related question: I have a problem understanding the proof of Rencontres numbers (Derangements) — robjohn, May 08 '15 at 10:50

Olivier Oloa · Answer 1 · 2015-05-08T12:03:59.607

2

Here is a way to obtain the formula. Assume that $$ D_n=nD_{n-1}+(-1)^n,\quad n\geq1. \tag1 $$ Dividing $(1)$ by $n!$ gives $$ \frac{D_n}{n!}=\frac{nD_{n-1}}{n!}+\frac{(-1)^n}{n!},\quad n\geq1, $$ or $$ \frac{D_n}{n!}-\frac{D_{n-1}}{(n-1)!}=\frac{(-1)^n}{n!},\quad n\geq1, \tag2 $$ then summing from $n=1$ to $n=N$ we get, by telescoping on the left hand side: $$ \frac{D_N}{N!}-\frac{D_{0}}{0!}=\sum_{n=1}^N\frac{(-1)^n}{n!}, $$ or

$$ D_N=N!\sum_{n=0}^N\frac{(-1)^n}{n!}, \quad N\geq0, $$

as wanted.

edited May 08 '15 at 12:03

answered May 08 '15 at 09:47

Olivier Oloa

120,989

Thanks for your help sir Olivier Oloa – Jr Antalan May 08 '15 at 20:42

score 1 · Answer 2 · answered May 08 '15 at 09:45

You can use induction by considering: $ X_n=\frac{D_n}{n!}$ so that the given formula can be written as: $$X_{n}=X_{n-1}+\frac{(-1)^n}{n!} \tag 1$$

Basis step :$X_0=1$
Induction step assume that $X_{n-1}=1-\frac{1}{1!}-\frac{1}{2!}+\cdots +\frac{(-1)^{n-1}}{(n-1)!}$ hence : $$X_n=X_{n-1}+\frac{(-1)^n}{n!}=1-\frac{1}{1!}-\frac1{2!}+\cdots +\frac{(-1)^{n-1}}{(n-1)!}+\frac{(-1)^n}{n!}\tag 2 $$

which terminates the proof.

score 1 · Accepted Answer · edited Apr 13 '17 at 12:20

This is essentially shown at the end of this answer. I will extract the relevant portion here:

Derivation of the Closed Form from the Recursion:

Given $\mathcal{D}(0)=1$ and $\mathcal{D}(1)=0$, and the recursion $$ \mathcal{D}(n)-n\mathcal{D}(n-1)=(-1)^n\tag{1} $$ Dividing both sides of $(1)$ by $n!$ yields $$ \frac{\mathcal{D}(n)}{n!}-\frac{\mathcal{D}(n-1)}{(n-1)!}=\frac{(-1)^n}{n!}\tag{2} $$ Equation $(2)$ is very simple to solve for $\frac{\mathcal{D}(n)}{n!}$: $$ \frac{\mathcal{D}(n)}{n!}=\sum_{k=0}^n\frac{(-1)^k}{k!}+C\tag{3} $$ Plugging $n=0$ into equation $(3)$ yields that $C=0$. Therefore, $$ \mathcal{D}(n)=n!\sum_{k=0}^n\frac{(-1)^k}{k!}\tag{4} $$

Proving some identities in Derangements.

3 Answers3