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We say a permutation $\sigma$ of a set $S = \{1,2,\dots,n\}$ has a property $A$ if for every $1\le k < n$, $$\sigma(\{1,\dots,k\})\not=\{1,\dots,k\}$$ Let $c_n$ be the number of such permutations, prove that the following equation holds

$$\sum_{i=1}^nc_i(n-i)!=n!$$

Not sure how to aproach this, tried it for small numbers and i've got that $c_1=1,c_2=1,c_3=3,c_4=13$. Also not sure if i can somehow prove it combinatorialy with $i$ being lenght of a prefix of a permutation that satisfies property $A$.

For example for $S = \{1,2,3\}$ $$ \sigma= \left( \begin{matrix} 1&2&3\\2&3&1 \end{matrix} \right) $$ works and $$\sigma_2= \left( \begin{matrix} 1&2&3\\2&1&3 \end{matrix} \right)$$ doesn't since $\sigma_2(\{1,2\})=\{1,2\}$

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    $\begingroup$ I think it's meant as a mapping of sets there -- the image of $\{1,\dots,k\}$ is not itself. $\endgroup$ – cwindolf Nov 20 '19 at 21:25
  • $\begingroup$ @Paul No, it requires that $\sigma$ applied to any proper initial segment of $n$ permutes at least one element of the initial segment away from the initial segment. $\endgroup$ – Robert Shore Nov 20 '19 at 21:26
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    $\begingroup$ I believe this is A003319 in oeis. $\endgroup$ – lulu Nov 20 '19 at 21:34
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Call a permutation that has the property you mentioned "good".
You can show that there is a bijection between the set of all of the permutations on $[n]$ and the set $$\{(k,f,g) |k\in [n], f \text{ is a good permutation on } [k] \text{ and } g \text{ is a permutation on } \{k+1,...,n\}\}$$
By looking at the first $k$ such that $f[k]=[k]$ for some permutation. this proves the relation.

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  • $\begingroup$ i am sorry is $[n]$ a set with $n$ elements? $\endgroup$ – Nasal Nov 20 '19 at 21:39
  • $\begingroup$ Yes, $[n] := \{1,...,n\}$ $\endgroup$ – Asaf Rosemarin Nov 20 '19 at 21:40

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