We choose one of the $4$ z-fields to contract with the single x-field. We then choose one of the remaining $3$ z-fields to contract with one of the $4$ w-fields. The remaining two z-fields just contract with themselves. Now choose one of the remaining $3$ w-fields to contract with the single y-field.
(Here is where we have to be careful). There are $2$ choices for the w-field contraction with one of the $4$ u-fields, and then $3$ choices for the other w-u contraction. In computing this last combination we have over counted by a factor of $2$.
To see this more clearly, consider one of the contractions,
The subscript denote which fields are contracted with which other fields (I'm not sure how to express contractions in Latex).
There are two ways to get this particular contraction: we could either choose the first w-field to be contracted with the first u-field, and THEN choose the second w-field to be contracted with the second u-field; OR we could choose the second w-field to be contracted with the second u-field, and THEN choose the first w-field to be contracted with the first u-field.
Clearly both of these are equivalent. However, in the combinatorics we have counted both of them, and so we must divide by a factor of $2$. So the total number of different contractions giving the same expression as $(4.45)$ is
$3! \, \times \, 4 \cdot 3 \, \times \, 4 \cdot 3 \cdot 2 \, \times \, 4 \cdot 3 \, \times \, 1/2$
Where the $3!$ comes from the interchange of vertices.
EDIT: If that isn't clear, think about the following scenario. There are two boxes, in the first there are two objects, $A$, and $B$, and in the second there are two more, $C$, and $D$. How many different ways are there to pair off the objects so that each object in the first box has a partner in the second box? Clearly the answer is two: $A,C$ and $B,D$; and $A,D$ and $B,C$.
One might think the answer is $2\cdot 2$, but we can see that this produces duplicates
First Pair & Remaining Pair \\
A,C & B,D\\
A,D & B,C\\
B,C & A,D\\
B,D & A,C\\
So we must multiply by a factor of $1/2$ to fix the overcounting.
Hope that helps.