Say that there are $n$ trains running on a railway track $T_{1}, T_{2}, \dots, T_{n} $. Say that their respective speeds are $v_{1},v_{2}, \dots, v_{n}$ such that $v_{1}< v_{2}< \dots < v_{n}$.
Define $\mathbf{a_{n}}$ as the total number of trains after all the link ups have occurred. So, the question asks us to find $\frac{a_{n}}{n!}$. Obviously, $a_{1}=1$.
Refer to the diagram below. $(T_{n},T_{n-1}, \dots, T_{1}), \dots, (T_{k},T_{l}, T_{p}, \dots, T_{r})$ are all the $(n!)$ permutations of these $n$ trains. Say that a new train $T_{n+1}$ comes in with a speed $v_{n+1}$ such that $v_{n+1}>v_{n}> \dots > v_{1}$. Notice that in all $n!$ permutations it can embed itself in one of those $(n+1)$ spots, denoted by the arrows. Since the speed of $T_{n+1}$ is the highest, therefore, if it embeds itself in the spot indicated by the top arrow, then it would add one more train to the total number of existing trains, and there are $n!$ such arrangements. If it embeds itself in any of the other (n) slots, it will link up with the train moving in front of it and start moving with the speed of that train, thereby the total number of trains in these cases will remain the same.
This gives the recurrence:
$$
\begin{aligned}
a_{n+1} &= \underbrace{a_n + n!}_{\text{when } T_{n+1} \text{ is in the top spot}} + \underbrace{n a_n}_{\text{when } T_{n+1} \text{ is in other spots}} \cr
&\implies a_{r+1} = a_r + r! + r a_r \cr
&\implies a_{r+1} = (r+1)a_r + r! \cr
&\implies \frac{a_{r+1}}{(r+1)!} = \frac{a_r}{r!} + \frac{1}{r+1} \cr
&\implies \sum_{r=1}^{n-1} \left( \frac{a_{r+1}}{(r+1)!} - \frac{a_r}{r!} \right) = \sum_{r=1}^{n-1} \frac{1}{r+1} \cr
&\implies \frac{a_n}{n!} - \frac{a_1}{1!} = \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} \cr
&\implies \frac{a_n}{n!} = \frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{n} \quad \square
\end{aligned}
$$
The answer is (C).
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