How many permutations of cycle-shape $(3,2^2,1)$ are there in $S_8$? I am not familiar with this kind of counting problem, so I googled the key words. From what I found it looks like Stirling Numbers of First Kind do the job(?). These numbers are denoted [$\frac nk$ ] where $n$ is the length of permutations and $k$ is the number of cycles with $0 \le k \le n$.
In the problem, the length of permutations is $8$ and the number of cycles is $4$. So, there are [$\frac 84$ ] permutations. Does that work?
 A: There are
$
\frac{{8!}}
{{\left( 3 \right)\left( {2^2  \times 2!} \right)\left( {8 - 7} \right)!}} = 1,680
$ 1 3-cycle, 2 2-cycles in 8 objects.  See this page for more details.  (Also note that he uses the term "cycle class", but the common term is "cycle type".)
EDIT:
Okay, I will explain this a little more.  Let me abbreviate "1 3-cycle and two 2-cycles" as {3,2,2}.
I will be complete concrete and show you all 24 redundancies for the cycle type {3,2,2} in 8 objects which you have asked about.
Firstly, I will rewrite my calculation as the following, and I will refer to each factor in the denominator from left to right in the explanation that follows.
$\frac{{8!}}
{{\left( 3 \right)\left( 2 \right)\left( 2 \right)\left( {2!} \right)\left( {8 - (3+2+2)} \right)!}}$
Now, let (1$\to$2$\to$3)(4$\leftrightarrow$5)(6$\leftrightarrow$7)(8) represent the cycle type {3,2,2}.
Then the 24 redundancies we divide by are the following:
For the 3-cycle (rotate the 3-cycle clockwise each time): $=(3)$
Set A
{$(1\to2\to3)$, $(3\to1\to2)$, $(2\to3\to1)$}
For the 2 2-cycle (Rotate one of the 2-cycles each time): $=(2)(2)$
Set $B_1$
{$(5\leftrightarrow4)(7\leftrightarrow6)$,
$(4\leftrightarrow5)(7\leftrightarrow6)$,
$(5\leftrightarrow4)(6\leftrightarrow7)$,
$(4\leftrightarrow5)(6\leftrightarrow7)$}
and... switch the two 2-cycles in the above work: $=(2)(2)$
Set $B_2$
{$(7\leftrightarrow6)(5\leftrightarrow4)$,
$(7\leftrightarrow6)(4\leftrightarrow5)$,
$(6\leftrightarrow7)(5\leftrightarrow4)$,
$(6\leftrightarrow7)(4\leftrightarrow5)$}
$B_1$ and $B_2$ are a result of $(2!)$.  That is, when we switch/permute these two 2-cycles, we are saying that there are only two ways we can order these two same length cycles.
For the 1-cycle $=(8 - (3+2+2))$
Set $C$
{$(8)$}
We now take the Kronecker product $\{$A$\}\times\{B_1\cup B_2\}\times\{C\}$ to get $(3)(4+4)(1) = 24$ equivalent permutations.  Therefore, we divide by 24 to only use one of them.
Key Point:  Because the product of disjoint cycles is commutative, and since we can represent an $n$-cycle in $n$ different ways (by rotating all the numbers inside of it---and keep their cyclic order the same as we do), 23/24 of these permutation representations are redundant.  Therefore, we divide by 24 to only use one of them.
A: No, $\left[ 8 \over 4 \right]$ is the number of ways to partition $8$ objects into $4$ cycles, where the cycles could be of any lengths as long as they add up to $8$.  So the $6769$ cycles counted by $\left[ 8 \over 4 \right]$ would include, for example, partitioning into four $2$-cycles.
Your problem is a bit easier than finding $\left[ 8 \over 4 \right]$ would be (if you didn't have a handy table of Stirling numbers) because there is only one breakup to deal with. First, observe that for a 1-cycle and for the two 2-cycles, once you choose the elements there is only one choice as to what the cycle is.  But for the 3-cycle, there are two choices (cycles isomorphic to $(e_1e_2e_3)$ aand cycles isomorphic to  $(e_3e_2e_1)$ ).
So pick the element in the 1-cycle: This can be done in $8$ ways.
Then pick the four elements forming the two $2$-cycles: This can be done in $\binom74 = 35$ ways.  But for each of those ways there are $3$ distinct ways to break those four into two pairs. Finally, make a choice of "direction" among the $2$ possibilities for the 3-cycle.
The product is then
$$
8\cdot 35\cdot 3\cdot 2 = 1680
$$
ways to partition the eight elements as required.
A: I post the species equation for future reference. The species equation for permutations of cycle shape $a_1 a_2^2 a_3$ is
$$\mathfrak{C}_{=1}(\mathcal{Z})
\times \mathfrak{P}_{=2}( \mathfrak{C}_{=2}(\mathcal{Z}) )
\times \mathfrak{C}_{=3}(\mathcal{Z}).$$
This yields the EGF
$$f(z) = \frac{z}{1}\times \frac{1}{2} \left(\frac{z^2}{2}\right)^2 
\times \left(\frac{z^3}{3}\right)
= \frac{z^8}{24}.$$
Extracting coefficients from this EGF we get
$$8! [z^8] \frac{z^8}{24} = \frac{8!}{24} = 1680.$$
Wikipedia has an entry for the notation.
