2

Give a combinatorial argument to prove that the number of derangements satisfies the recursive formula $d_n = (n − 1)(d_{n−1} + d_{n−2})$ for $n ≥ 2$. (Hint: For a derangement $σ$, consider the integer $k$ with $σ(k) = 1$. Argue based on the number of choices for $k$ and then whether $σ(1) = k$ or not.)

I tried working through this but I am having trouble using the hint. I also tried to use inclusion-exclusion excluding all cases where $σ(k)=k$ but it went no where. Any help would be appreciated. To be honest, I really have no idea where to go with this.

Meecolm
  • 683

2 Answers2

5

SKETCH: Let $\sigma$ be a derangement of $[n]=\{1,\ldots,n\}$, and suppose that $\sigma(k)=1$. There are two possibilities.

  • $\sigma(1)\ne k$: Define a function $$\tau:[n]\setminus\{1\}\to[n]\setminus\{1\}:\ell\mapsto\begin{cases}\sigma(\ell),&\text{if }\ell\ne k\\\sigma(1),&\text{if }\ell=k\;.\end{cases}$$ Show that $\tau$ is a derangement of the $(n-1)$-element set $[n]\setminus\{1\}$.

  • $\sigma(1)=k$: Then the restriction of $\sigma$ to $[n]\setminus\{1,k\}$ is a derangement of the $(n-2)$-element set $[n]\setminus\{1,k\}$.

Show that every derangement of $[n]$ arises uniquely in one of these two ways, and see how many there are of each kind.

Brian M. Scott
  • 616,228
3

We seperate into two cases.

Case 1: Let $i$ be the image of $1$, case one is when $1$ is the image of $i$, There are $n-1$ possible values for $i$. And once $i$ has been selected there are $d_{n-2}$ ways to biject the remaining numbers.

Case 2: The image of $i$ is not $1$. Then every element has exactly one forbidden element in the co-domain. And the forbidden elements are different for everyone, so there are $(n-1)(d_{n-1})$ of this type.

adding up and grouping we get $d_n=(n-1)(d_{n-1}+d_{n-2})$

Asinomás
  • 105,651