Martin has already explained the notation, but you might also find the following connection with Stirling numbers of the second kind useful, since those are the ones mentioned in your title.
If you calculate the $c(n,k)$ for $n=0,1,2,3,4,5$, you get the following array of $0$-th diagonals:
n\k: 0 1 2 3 4 5
---------------------------
0 | 1
1 | 0 1
2 | 0 1 2
3 | 0 1 6 6
4 | 0 1 14 36 24
5 | 0 1 30 150 240 120
This triangle can be found as A131689 at OEIS, where you can discover that the entry in row $n$, column $k$ is $$k!\left\{\matrix{n\\k}\right\},$$ where $\displaystyle\left\{\matrix{n\\k}\right\}$ is the Stirling number of the second kind. And sure enough, if we divide each entry in column $k$ by $k!$ we get the following table:
n\k: 0 1 2 3 4 5
---------------------------
0 | 1
1 | 0 1
2 | 0 1 1
3 | 0 1 3 1
4 | 0 1 7 6 1
5 | 0 1 15 25 10 1
This is readily recognizable as the table of Stirling numbers of the second kind.
$\begin{Bmatrix}
n\\
k
\end{Bmatrix}$ is the number of ways to distribute $n$ distinct objects into $k$ identical boxes with no empty boxes allowed/each box containing at least $1$ object. We will first consider distributions into $k$ distinct boxes.
The total number of ways to distribute $n$ distinct objects into $k$ distinct boxes with empty boxes allowed is: $$k^n$$
Let $A_i$ be the number of ways to perform the distribution with the condition that box $i$ must be an empty box. There are no restrictions on the other boxes. (This is equivalent to distributing the $n$ objects into the remaining $k-1$ boxes, empty boxes allowed.)
For example $A_1$ is the number of ways to distribute the objects into the $k$ boxes with box 1 being empty and is equivalent to distributing the objects into the remaining $k-1$ boxes, empty boxes allowed.
$$\implies |A_1| = (k-1)^n $$
Now we notice that $|A_1|=|A_2|=|A_3|=\dots$ since it doesn't matter which box is empty. (We are always distributing into the remaining $k-1$ boxes).
We know there are $\binom{k}{1}$ possible $A_i $
$$\sum_{i=1}^k|A_i| = \binom{k}{1}(k-1)^n$$
Using a similar argument, $|A_i\cap A_j|=(k-2)^n$ because it involves distribution into $k-2$ remaining boxes. Also, using the fact that $|A_1\cap A_2|=|A_1\cap A_3|=|A_1\cap A_4|...$ and that there are $\binom{k}{2}$ such terms,
$$\sum_{1\leq i,j\leq k}|A_i\cap A_j| = \binom{k}{2}(k-2)^n$$
On a similar note, we try extending this to when $l$ boxes must be empty:
$$\sum_{1\leq i,j\dots l\leq k}|A_i\cap A_j\cap \dots\cap A_l| = \binom{k}{l}(k-l)^n$$
$|A_1\cup A_2\cup \dots \cup A_k|$ is the number of ways to distribute $n$ distinct objects into $k$ distinct boxes with at least one empty box.
We are trying to find the distribution with no empty boxes: $$|\overline{A_1\cup A_2\cup \dots \cup A_k}|=k^n -|A_1\cup A_2\cup \dots \cup A_k|$$
By the principle of inclusion and exclusion,
$$=k^n-(|A_1|+|A_2|+|A_3|+...)+(|A_1\cap A_2|+|A_1\cap A_3|+|A_1\cap A_4|+...)-...$$
$$=k^n-\binom{k}{1}(k-1)^n+\binom{k}{2}(k-2)^n-\binom{k}{3}(k-3)^n+\dots +(-1)^l\binom{k}{j}(k-j)^n$$
$$=\sum_{l=0}^{k}(-1)^l\binom{k}{l}(k-l)^n$$
Let $j=k-l$, when $l=0$, $j=k$. When $l=k$, $j=0$.
$$=\sum_{j=k}^{0}(-1)^{k-j}\binom{k}{k-(k-j)}j^n (\text{this is technically wrong notation})$$
This is just the sum backwards. So we have:
$$=\sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j}j^n$$
Now we divide by $k!$ since we are dealing with identical boxes:
$$\begin{Bmatrix}
n\\
k
\end{Bmatrix}
=\frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j}j^n$$
Best Answer
The Stirling number of the second kind, $\left\{ \begin{array}{c} n\\k \end{array} \right\}$, gives the number of ways to partition the set $A=\{1,2,\cdots,n\}$ into $k$ classes.$^1$
So $$\left\{ \begin{array}{c} n\\2 \end{array} \right\}=\frac{2^n-2}{2}=2^{n-1}-1.$$
We can put each item (an element of $A$) into one of two bins. Since we don't want empty bins, we subtract $2.$ Then, since we don't distinguish between bins, we divide by $2.$
Now we look at $\left\{ \begin{array}{c} n\\n-2 \end{array} \right\}$.
Most of the elements will end up in its own bin. We have two cases: ($1$) three in one bin, or ($2$) there are two bins, each with two elements:
$$\left\{ \begin{array}{c} n\\n-2 \end{array} \right\}=\binom{n}{3}+\frac{1}{2}\binom{n}{2}\binom{n-2}{2}=\binom{n}{3}+3\binom{n}{4}.$$
In both combinatorial expressions, $\binom{n}{3}$ is the number of ways of executing part $(1)$.
For part $(2)$, we could either choose two to be in a bin, then choose two of the remaining $n-2$. Because we could pick the two pairs in either order, we divide by two.
Alternatively (as expressed in the last sum) we can choose the four which gives $\binom{n}{4}$, then we take any of the four and choose the other one of the remaining three to go in the same bin: $\binom{3}{1}=3$.
$^1$Wilf, H. S., $\textbf{generatingfunctionology}$, 3rd edition, CRC Press, p. 18.