Find closed formula for $S(n,n-4)$. Find closed formula for $S(n,n-4)$.
These are stirling numbers of 2nd kind.
My attempt: I uses the recurrence relation $$S(n,k)=S(n-1,k-1)+kS(n-1,k)$$
$S(n,n-4)=S(n-1,n-5)+(n-4)S(n-1,n-4)=S(n-2,n-6)+(n-5)S(n-2,n-5)+(n-4)\{S(n-2,n-5)+(n-4)S(n-2,n-4)\}$
But it seems not helpful for me.How I can find closed formula for $S(n,n-4)$?
 A: Working with the recursion looks cumbersome; to use the combinatoric interpretation seems easier: $S(n,k)$ gives the ways of placing $n$ distinguishable balls in $k$ undistinguishable urns with no empty urn. Then, for $k=n-4$, we have these possible distributions:
A. {5, 1 ,1 ...} : ${n \choose 5}$
B  {4, 2 ,1 ...} :  ${n \choose 4}{n-4 \choose 2}$
C. {3, 3 ,1 ...} :  ${n \choose 3}{n-3 \choose 3} \frac{1}{2}$
D. {3, 2 , 2, 1 ...} : ${n \choose 3}{n-3 \choose 2} {n-5 \choose 2}\frac{1}{2}$
E. {2, 2 , 2, 2, 1 ...} : ${n \choose 2}{n-2 \choose 2} {n-4 \choose 2}{n-6 \choose 2}\frac{1}{4!}$
You just need to sum all those terms to get $S(n,n-4)$, excepting (for small $n$) the terms that are impossible (or simply using the convention ${a \choose b}=0$ for $a<b$) - and perhaps try to simplify a little all that...
For example, for the first "full" case, we get:
$S(8,4)=56+ 420 +280+  840 + 105 =1701$
A: Here is  a solution by  generating functions for  verification purpose
until something simpler appears.

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

The species of set partitions with sets of length at most five is
is
$$\mathfrak{P}
(\mathcal{A}_1 \mathfrak{P}_{=1}(\mathcal{Z})
+ \mathcal{A}_2 \mathfrak{P}_{=2}(\mathcal{Z})
+ \cdots
+ \mathcal{A}_5 \mathfrak{P}_{=5}(\mathcal{Z})).$$
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-set 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 sets
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-set,  a two-set 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-sets 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-set,  two two-sets  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-sets  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
$$\frac{z^n}{(n-5)!} \times
\left(\frac{1}{120}
+ \frac{n-5}{48}
+ \frac{n-5}{72}
+ \frac{(n-5)(n-6)}{48}
+ \frac{(n-5)(n-6)(n-7)}{384} \right)
\\ = \frac{z^n}{(n-5)!} \times
\left(\frac{1}{384} n^3 - \frac{5}{192} n^2
+ \frac{97}{1152} n - \frac{251}{2880}\right).$$
Performing coefficient extraction on this we obtain the answer
$$n! [z^n]
\frac{z^n}{(n-5)!} \times
\left(\frac{1}{384} n^3 - \frac{5}{192} n^2
+ \frac{97}{1152} n - \frac{251}{2880}\right)
\\ = \frac{n!}{(n-5)!}
\left(\frac{1}{384} n^3 - \frac{5}{192} n^2
+ \frac{97}{1152} n - \frac{251}{2880}\right).$$
This is very similar to the computation at this 
MSE link.
A: Stirling Numbers of the Second Kind have an explicit formula:
$$S(n, k) = \frac{1}{k!} \sum_{j=0}^{k} (-1)^{k-j} \binom{k}{j} j^{n}$$
Citation: http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind
A: Derived from Leonbloy’s answer : the same reasoning provides a closed polynomial formulation for $S(n,n-k)$ in general
Let $K=\{(a_1,...,a_k)\}  $ the set of all k-uples made of $k$ non-negative integers $a_j$ such that $$\sum_{j=1}^{j=k}j a_j=k$$
The elements $a\in K$ define the partitions of the integer $k$. 
$$\sum_{j=1}^{j=k} a_j=s(a)$$ $s(a)$ is the number of summands in the partition $a$. Then (if I am not wrong)
$$S(n,n-k)=\sum_{a\in K} \frac{\prod_{j=1}^{j=k+s(a)}(n-j+1)}{\prod_{j=1}^{j=k}(j+1)!^{a_j}{a_j!}}$$
Since it comes from enumeration, each term in the sum is itself an integer-valued polynomial, with degree ranging from $k+1$ to $2k$
EDIT a similar formula has already been published: Eq.(52) in this paper.
