The number of functions from $[5]$ to $[5]$ such that $f(f(x)) \neq x $ for $ x=1, \ldots 5$ is $44$

Question

True or false: The number of functions from $[5]$ to $[5]$ such that $f(f(i)) \neq i $ for $ i=1, \ldots 5$ is $44.$

(Here $[5] = \{1,2,3,4,5\}.)$

The number of derangements on $5$ elements is $44$ but the functions in question are clearly not just derangements, since the existence of $x, y$ such that $f(x)=y$ and $f(y)=x$ is also prohibited. Further, there is no restriction on $f$ to be bijective.

My attempt: for each function $f,$ define $g= f^2$ and count functions $g: [5] \to [5]$ such that $g(i) \neq i$ by inclusion-exclusion as follows: $5^5 -\binom{5}{1}(4^4) + \binom{5}{2}(3^3)- \ldots $ and so on. The problem though is that the set of $g's$ aren't in bijective correspondence with the set of $f's$ and I'm not convinced that the expression I have above is right because the resultant answer is too large.

Any help would be appreciated.

what is the meaning of [5]? – miracle173 Sep 24 '17 at 13:19 — miracle173, Sep 24 '17 at 13:19
Are the functions supposed to be bijective? – drhab Sep 24 '17 at 13:21 — drhab, Sep 24 '17 at 13:21
No, that isn't given to us. – Aryaman Jal Sep 24 '17 at 13:23 — Aryaman Jal, Sep 24 '17 at 13:23

score 6 · Accepted Answer · edited Sep 24 '17 at 14:18

6

According to OEIS's A134362, for $[5]$ it should be $444$. A general formula for the number of functions $f:[n]\to[n]$ such that for every $x\in[n]$, $f(f(x)) \not =x$ is $$n!\sum_{k=0}^{\lfloor n/2\rfloor}(-1)^k\cdot\frac{(n-1)^{n-2k}}{2^k k!(n-2k)!}.$$

P.S. Note that such $f$ are not necessarily injective: $$[1,2,3,4,5]\stackrel{f}{\to} [2, 3, 1, 1, 1]\stackrel{f}{\to} [3, 1, 2, 2, 2]$$

A proof for this formula by Marco Riedl is alredy published at math.stackoverflow.

edited Sep 24 '17 at 14:18

miracle173

11,049

answered Sep 24 '17 at 13:31

Robert Z

145,942

I confirm that among the $5^5=3125$ functions there are 444 with such property. – Robert Z Sep 24 '17 at 13:40
I added a link to the proof of this formula that was already published here. I hop this is ok – miracle173 Sep 24 '17 at 14:20
@miracle173 It's wonderful.Thanks – Robert Z Sep 24 '17 at 14:22
@Aryaman Jal Any further doubt? – Robert Z Sep 24 '17 at 17:55

Donald Splutterwit · Answer 2 · 2017-09-24T14:18:58.020

4

The condition $f(f(i)) \neq i$ will imply bijectivity. There are indeed $44$ derrangements of $[5]$, there are $24$ of the "flavour" $(abcde)$ and $20$ of the "flavour" $(ab)(cde)$. The latter "flavour" will not satisfy the condtion $f(f(i)) \neq i$. So there are $\color{blue}{24}$ functions on $[5]$ whose second iterated map also has no fixed points.

Edit : Double functions with no fixed points.

On a three element set there are $2$ functions of the form $ a \rightarrow b \rightarrow c \rightarrow a $. ($a \neq b \neq c \neq a$)

On $[4]$, there are $6$ of the form $ a \rightarrow b \rightarrow c \rightarrow d \rightarrow a$ & there are $24$ of the form $ a \rightarrow b \rightarrow c \rightarrow d \rightarrow b$. So $30$ in toto.

On $[5]$,

There are $24$ of the form $ a \rightarrow b \rightarrow c \rightarrow d \rightarrow e \rightarrow a$

There are $120$ of the form $ a \rightarrow b \rightarrow c \rightarrow d \rightarrow e \rightarrow b$

There are $120$ of the form $ a \rightarrow b \rightarrow c \rightarrow d \rightarrow e \rightarrow c$

There are $60$ of the form $ a \text{ and } b \rightarrow c \rightarrow d \rightarrow e \rightarrow c$

There are $120$ of the form $ a \rightarrow c , b \text{ and } c \rightarrow d \rightarrow e \rightarrow c$

So there are $444$ in toto.

edited Sep 24 '17 at 14:18

answered Sep 24 '17 at 13:30

Donald Splutterwit

36,613
2
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How does $f(f(i)) \ne i$ imply bijectivity? $f(i) = i + 1$ for $i = 1, : 2, : 3, : 4$ and $f(5) = 3$ has the above property, yet it's not a permutation. In fact, if I haven't miscounted, there should be way more than 44 such functions. – cip999 Sep 24 '17 at 13:39
@cip999 Robert Z's answer is actually right ... My opening statement about bijectivity is nonsense ... give me 5 or 10 mins to edit ? – Donald Splutterwit Sep 24 '17 at 13:46
@Donald Splutterwit Nice counting argument (+1) – Robert Z Sep 24 '17 at 14:05
Where is (a and b and c) => d =>e => d ? – miracle173 Sep 24 '17 at 14:08
@RobertZ We got there in the end. I feel into the trap of assuming the function must be a permutation. Your solution was right all along. – Donald Splutterwit Sep 24 '17 at 14:08
@miracle173 that map will have $f(f(d))=d$ (contrary to hypothesis) – Donald Splutterwit Sep 24 '17 at 14:09
thank's, I see now. – miracle173 Sep 24 '17 at 14:10

The number of functions from $[5]$ to $[5]$ such that $f(f(x)) \neq x $ for $ x=1, \ldots 5$ is $44$

2 Answers2