= Solution
First choose the $l$ absent labels, which will be the <leaves of a graph>, in $\binom nl$ ways. Every other label must occur exactly $r-1$ times in the <Prüfer sequence>. There are $n-l$ such labels and their total multiplicity is $n-2$ by (a). The number of orderings of this <multiset> is $(n-2)!/((r-1)!)^{n-l}$. The <bijection> in (i) turns each ordering into one distinct <labelled tree> and exhausts all possibilities. Therefore
$$
\boxed{\binom nl\frac{(n-2)!}{((r-1)!)^{n-l}}.}
$$
The formula is used with the feasible integer parameters from (a), and gives one for the two-<vertex> <edge> with no internal labels.
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