Find closed form formula for $c(n,n-4)$. Find closed form formula for $c(n,n-4)$.
Where $c(n,k)$ are signless stirling numbers of first kind.
I need help.This is my last question all others problems of my exercise I have solved this is last one please help me.
Thanks.

Here is  a solution by  generating functions for  verification purpose
until something simpler appears.

Observe that there  cannot be a cycle of length  at least six because
that leaves  $n-6$ items  which can  form at most  $n-6$ cycles  for a
total of  $n-5$ cycles. Similarly for  a cycle of length  seven and so
on.

The species of permutations with  cycles of length at most five is
is
$$\mathfrak{P}
(\mathcal{A}_1 \mathfrak{C}_{=1}(\mathcal{Z})
+ \mathcal{A}_2 \mathfrak{C}_{=2}(\mathcal{Z})
+ \cdots
+ \mathcal{A}_5 \mathfrak{C}_{=5}(\mathcal{Z})).$$
This gives the generating function
$$G(z) = \exp\left(\sum_{q=1}^5 a_q \frac{z^q}{q}\right)
= \prod_{q=1}^5 \exp\left( a_q \frac{z^q}{q} \right).$$

Now there  are several  cases. 
First case.
A five-cycle and  $n-5$ fixed points. 
This gives the generating function
$$\frac{1}{(n-5)!} \left(\frac{z}{1}\right)^{n-5}
\exp\left(a_2 \frac{z^2}{2}\right)
\exp\left(a_3 \frac{z^3}{3}\right)
\exp\left(a_4 \frac{z^4}{4}\right)
\frac{1}{1} \left(\frac{z^5}{5}\right)^1.$$
We can  drop the contributions  in $a_2, a_3,  a_4$ as no  such cycles
appear because there aren't any items left over, for an answer of
$$\frac{1}{(n-5)!} \left(\frac{z}{1}\right)^{n-5}
\left(\frac{z^5}{5}\right)^1.$$
Second case.
A four-cycle,  a two-cycle and  $n-6$ fixed points,  for a
contribution of
$$\frac{1}{(n-6)!} \left(\frac{z}{1}\right)^{n-6}
\left(\frac{z^2}{2}\right)^1
\left(\frac{z^4}{4}\right)^1.$$
Third  case. Two  three-cycles and  $n-6$ fixed  points,  for a
contribution of
$$\frac{1}{(n-6)!}\left(\frac{z}{1}\right)^{n-6}
\frac{1}{2} \left(\frac{z^3}{3}\right)^2.$$
Fourth  case. A  three-cycle,  two two-cycles  and $n-7$  fixed
points for a contribution of
$$\frac{1}{(n-7)!} \left(\frac{z}{1}\right)^{n-7}
\frac{1}{2} \left(\frac{z^2}{2}\right)^2
\left(\frac{z^3}{3}\right)^1.$$
Fifth  case. Four two-cycles  and $n-8$  fixed
points for a contribution of
$$\frac{1}{(n-8)!}\left(\frac{z}{1}\right)^{n-8}
\frac{1}{24} \left(\frac{z^2}{2}\right)^4.$$
Adding the contributions from these generating functions we obtain
$$\frac{z^n}{(n-5)!} \times
\left(\frac{1}{5}
+ \frac{n-5}{8}
+ \frac{n-5}{18}
+ \frac{(n-5)(n-6)}{24}
+ \frac{(n-5)(n-6)(n-7)}{384} \right)
\\ = \frac{z^n}{(n-5)!} \times
\left(\frac{1}{384} n^3 - \frac{1}{192} n^2
+ \frac{1}{1152} n + \frac{1}{2880}\right).$$
Performing coefficient extraction on this we obtain the answer
$$n! [z^n]
\frac{z^n}{(n-5)!} \times
\left(\frac{1}{384} n^3 - \frac{1}{192} n^2
+ \frac{1}{1152} n + \frac{1}{2880}\right)
\\ = \frac{n!}{(n-5)!}
\left(\frac{1}{384} n^3 - \frac{1}{192} n^2
+ \frac{1}{1152} n + \frac{1}{2880}\right).$$
