Combinations of bijective derangement functions

Question

Let $S = \{1,2,3,4,5\}$ How many bijective functions $f:S \to S$ do we have such that $f(x) \neq x$ for all $x\in S$?

I was trying to solve by taking the total number of such functions and subtract all the cases where $f(x) = x$.

$5!$ is all cases

$4!$ is where one $f(x)=x$ and rest are not

$3!$ is where two $f(x)=x$ and rest are not

$2!$ is where three $f(x)=x$ and rest are not

$1!$ is where all five $f(x)=x$

so $5!-(4!-3!-2!-1)$.

But then I realized that, for example, $4!$ contains also the case where all $f(x)=x$ which mean I am subtracting too much.

Btw, the things you're looking for are called derangements: permutations which do not fix any points. — The Count, Feb 20 '17 at 18:48
these types of permutations are called derangements. You can try to read more about them if you directly want the answer or what i would suggest is you learn about inclusion - exclusion. Then try to find all permutations for which f(x)=x and subtract those from all permutations . Good luck — asddf, Feb 20 '17 at 18:48
I learned about inclusion exclusion, but i'm still stuck. I know i need to subtract the intersection between each group( 4! and 3!, 4! and 2!, 4! and 1!, 3! and 2!, 2! and 1!) and then add the intersection between all of them. @asddf — Gigalala, Feb 20 '17 at 20:22
Ill write it out as an answer becuase there is no place here. — asddf, Feb 20 '17 at 20:49

score 1 · Accepted Answer · answered Feb 20 '17 at 20:59

so suppose $A_1$ is the set of all permutations where atleast one $f(x)=x$ (ie one point is fixed).$A_2$ is the set of all permutations where atleast two $f(x)=x$. $A_3$... and $A_n$ is the set where all $f(x)=x$. So you want to find $n!-|A_1 \cup A_2 \cup A_3...\cup A_n|$ right ? (ie the number of total permutations minus the number of permutations you dont want: that will give us the number of permutations you want).

Well what we see here is a the cardinality of a union of sets that arent disjoint, so we will use inclusion exclusion principle.

How many ways are there to fix k points ? Well thats easy there are $\binom{n}{k}$ possiblities right

How many possible permutations are left for the other (n-k) that we didnt fix?

Well thats easy too right: (n-k)! so what we want to calculate is now using inclusion exclusion :

$n!-$$\sum_{k=1}^n$$(-1)^{k-1}\binom{n}{k}(n-k)!$

This can be simplified i will let you do the rest :)))

Combinations of bijective derangement functions

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