Prove using combinatorics $\sum\limits_{k=0}^n (k-1)^2 D_n(k)=n!$.

Question

$D_n(k)$ is the number of permutations of $n$ numbers that exactly $k$ numbers are in their place.

With some calculations I saw that it is not true for $n=1$ so the condition $n \ge 2$ should also be added.I got an algebric proof here.But I need one using combinatorics.For expressing $(k-1)^2$ maybe it is useful to extend it because we cant use $(k-1)^2$.So we got:

$\sum\limits_{k=0}^n k^2*D_n(k)-2*\sum\limits_{k=0}^n k*D_n(k)+2*\sum\limits_{k=0}^n D_n(k)$

It is easy to prove that the second one is $-2n!$ and the third is ,$n!$ so we have to prove:

$\sum\limits_{k=0}^n k^2*D_n(k)=2*n!$

But how should I do that?

$k-1$ is just $k-1$, the biggest problem is to notice that the LHS is telescopic. — Jack D'Aurizio, Jun 06 '17 at 15:49
@JackD'Aurizio But we need a combinatoral proof so we can't simplify and then show that. — Taha Akbari, Jun 06 '17 at 16:15
By mimicking the usual approach for finding $D_n(0)=n!\sum_{k=0}^{n}\frac{(-1)^k}{k!}$. — Jack D'Aurizio, Jun 06 '17 at 16:33
Have a look at here, for starters: https://math.stackexchange.com/questions/1056610/combinatorial-argument-for-recursive-formula — Jack D'Aurizio, Jun 06 '17 at 16:35
[link] (https://oeis.org/wiki/Rencontres_numbers) it is true here but without $(k-1)^2$ — serg_1, Jun 06 '17 at 17:20

score 0 · Accepted Answer · answered Jun 11 '17 at 10:29

$\sum\limits_{k=0}^n k^2*D_n(k)=\sum\limits_{k=0}^nk*D_n(k)+\sum\limits_{k=0}^nk(k-1)*D_n(k)=n!+\sum\limits_{k=0}^nk(k-1)*D_n(k)$

For finding the amount of $\sum\limits_{k=0}^nk(k-1)*D_n(k)$ notice that it is equal to $\sum\limits_{k=2}^nk(k-1)*D_n(k)$.And that is putting $k$ people in their seats and choose one boss and one manger.For that we can choose the boss and manager first($n(n-1)$) ways and put the other randomly so we have:

$$\sum\limits_{k=0}^n k^2*D_n(k)=2*n!$$

Prove using combinatorics $\sum\limits_{k=0}^n (k-1)^2 D_n(k)=n!$.

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