I'm a bit surprised that this exercise appears so early in a combinatorics book without any hints.
Regarding your first question, yes, I understand the question the same way. What I don't understand is how you originally expected us to help you in interpreting the question without linking to your book or explaining what the "first row" was. I'd suggest to always directly link to any book you're referring to if it's available online, and also not to change the spelling if you quote from a book – I searched for the problem before you provided the link and would have found it if you hadn't changed the spelling of "labeled" to "labelled" :-)
Regarding your second question: What you want to count isn't the ways of labelling the vertices of a given graph; you want to count all such graphs with all labellings for all structures; so I don't think this approach will get you very far.
The problem itself doesn't define what exactly a rooted trivalent tree is. In particular, it's not immediately clear whether the root itself can have degree $1$ or $3$ or only one of the two. Unfortunately Google Books doesn't show (at least to me) the page with Figure 14.3 that the problem refers to, which would probably resolve this. However, this page seems to indicate that the root may be required to have degree $1$, which would fit with how this book seems to be use the term, and also with this solution, which apparently makes that assumption.
The given condition is equivalent to the condition that the vertices are labelled in decreasing order as viewed from the root. Thus we're counting the number of labelled full/proper/strictly binary trees with vertex labels increasing from the root.
The recurrence
$$a_n=\frac12\sum_{k=1}^{n-1}\binom{2n-2}{2k-1}a_na_{n-k}$$
derived in the solution linked to yields the initial terms $1,1,4,34,496,\dotsc$, which lead to OEIS sequence A002105.
Here's a way to find the number. It looks like it might not be the best (Sloane's A055315 has other formulas), but it works.
Let $n$ be the number of vertices and $e=n-1$ be the number of edges.
Claim: There is exactly one vertex of degree $3$, all others have degree $2$ or $1$. (Assume otherwise, and we violate the Handshaking Lemma, which stipulates the sum of degrees is $2e=2(n-1)$.)
So, the trees look like this:
If we delete the degree-$3$ vertex, we obtain three components. These three components only have vertices of degree $1$ and $2$, and so must be paths. Suppose the paths have $a$, $b$ and $c$ vertices. Then $a+b+c=n-1$ and $a,b,c$ are all positive integers. Conversely, given three paths of lengths $a \geq 1$, $b \geq 1$ and $c \geq 1$ and $a+b+c=n-1$, we can join their endpoints to a new vertex to create an $n$-vertex tree with exactly three $3$ leaves.
Let $a(n)$ be the number of partitions of $n-1$ into $3$ non-zero parts; this is equivalent to Sloane's A001399 which includes the formula $$a(n)=1+\left\lfloor\frac{(n-1)^2+6(n-1)}{12}\right\rfloor$$ among others.
Given an unlabeled tree, we can label the vertices in $n!$ (not necessarily distinct) ways. In this way, we generate $a(n)\ n!$ labeled trees.
In some cases (not many), there are symmetries which need to be accounted for. Because of symmetry, in this case
we count every labeling twice. This happens whenever $\{a,b,c\}$ have two parts of the same size.
And in this case
we count every labeling six times. This happens when $\{a,b,c\}$ has all three parts of the same size.
In these cases, we'll need to add a correction.
In the first case: we subtract $$\sum_{a \geq 1:n-1-a \text{ is even and } \geq 2} \tfrac{1}{2}n!=\frac{\lfloor (n-2)/2 \rfloor\ n!}{2}$$ since we have double-counted the $\frac{1}{2} n!$ graphs with $b=c=\tfrac{n-1-a}{2}$.
In the second case: if $n-1 \equiv 0 \pmod 3$, we first "uncorrect" the first correction, so we add back in $\tfrac{1}{2}n!$ and then subtract $\frac{5}{6}n!$ since we originally counted six times the $\frac{1}{6} n!$ graphs with $a=b=c$.
This gives $$\begin{cases} a(n)\ n!-\tfrac{\lfloor (n-2)/2 \rfloor\ n!}{2} & \text{if } n-1 \not\equiv 0 \pmod 3 \\ a(n)\ n!-\tfrac{\lfloor (n-2)/2 \rfloor\ n!}{2}+\tfrac{1}{2}n!-\tfrac{1}{6} n! & \text{if } n-1 \equiv 0 \pmod 3. \end{cases}$$ in agreement with Sloane's A055315.
Best Answer
You have the right Prufer codes, but you have made some mistakes in counting. It's easy to check your formula doesn't work: for $n=4$ there are $4$ labelled trees with this property, because the tree must be a star and the only choice you have is which label goes in the middle.
First, to choose $n-3$ from $n$ elements in order is $\binom n3\times (n-3)!=\frac{n!}{6}$.
Secondly, when you choose an element to duplicate, you not only need to choose which element is duplicated ($n-3$ options) but also where to put the duplicate ($n-2$ possible places). However, this actually counts each code twice, because it distinguishes the "original" and "duplicate" version of the label that appears twice, meaning you need to divide by $2$.
So overall there are $\frac{n!}{6}\times\frac{(n-3)(n-2)}{2}=\frac{n!(n-2)(n-3)}{12}$ such trees.