What is the number of one-to-one functions f from the set {1, 2, . . . , n} to the set {1, 2, . . . , 2n − 1} so that f(x) $\neq$ 2x − 1 for all x?

Alright so I did see this question, but it really didn't cover it for me so I'm asking myself.

Check my math - Number of one-to-one functions $f$ from $\{1, \ldots, n\}$ to $\{1, \ldots, 2n-1\}$ such that $f(x) \neq 2x - 1$ for all $x$

So i'm new to this, and need to solve this using inclusion and exclusion. Here's what I got, but I"m not sure how to progress.

Going off of my notes, here's what I got ..

If we take $A_{i}$ to be a set of one-to one functions so f(x) = 2x − 1

$A_{1} \cup A_{2} \cup A_{3} .... \cup A_{n}$; is the set of one to functions so f(x)= 2x-1 for some x (complement of our original)

Now the next step according to my notes is to find intersections, but I'm a little confused on how to progress.

I think

|$A_{i}$| has I think (2n-1)! intersections

|$A_{i} \cap A_{j} $| has (2n-2)! intersections

|$A_{i} \cap A_{j} \cap A_{k}$| has (2n-3)! intersections

This work might be completely wrong

What we want is the total # of one to one functions minus |$A_{1} \cup A_{2} \cup A_{3} .... \cup A_{n}$|

I think, but I'm not sure how to get there, and I"m a beginner so if you could help I'd appreciate it.


The question is similar to finding a number of permutations having no fixed elements. Such permutations are known as derangements.

The proof from wikipedia can be adapted for your case as follows:

For $1\leq k \leq n$ we define $S_k$ to be the set of permutations of $n$ objects that map $k\mapsto 2k-1$. Any intersection of a collection of $i$ of these sets fixes a particular set of $i$ objects and therefore contains $\frac{(2n-1-i)!}{(n-1)!}$ functions. There are ${n\choose i}$ such collections, so the inclusion–exclusion principle yields

$$\begin{align}|S_1\cup\cdots\cup S_n| &=\sum_{i=1}^n (-1)^{i-1}{n\choose i}\frac {(2n-1-i)!}{(n-1)!} \end{align}$$

and since none of the $n$ objects fixed, we get

$$\frac {(2n-1)!}{(n-1)!} - |S_1\cup\cdots\cup S_n|=\sum_{i=0}^n (-1)^i{n \choose i} \frac {(2n-1-i)!}{(n-1)!}.$$

  • $\begingroup$ This is a bit out of the scope, of what I was looking at. Is there a solution using inclusion and exclusion principles? $\endgroup$ – Brownie Apr 20 at 21:05
  • $\begingroup$ Yes the expression can be derived by inclusion/exclusion principle. I would expect that you find the derivation in the given reference. $\endgroup$ – user Apr 20 at 21:09
  • $\begingroup$ Thanks for the help, but I think this a little out side of my skill level, and I'm not exactly sure how to progress with you proposed method. $\endgroup$ – Brownie Apr 20 at 21:20
  • $\begingroup$ So derangement's are something I haven't even heard of, but the way I was looking to solve it was like so.... the following is from my notes. |$A_{1} \cup A_{2} \cup A_{3} .... \cup A_{n}$| = $n \choose 1$ (n-1)! - $n \choose 2$ (n-2)! + ... + $(-1)^{n+1} $ $((n \choose n)) ^{(n-n)!} $. Then you would subtract that from the total number of functions n! in this case. This particular solution is for a different problem but I'm trying to apply it to this situation. I'm having trouble progressing from finding number of intersections for |$A_{i}$|, because I'm still trying to understand. $\endgroup$ – Brownie Apr 20 at 21:48

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