The sample space is
\begin{align}
\Omega = \{(N_1, N_2, \ldots, N_n): N_i \in \{1, \ldots, m\}, 1 \leq i \leq n\}.
\end{align}
Clearly, $|\Omega| = m^n$.
To determine the probability of interest $P(A) = |A|/|\Omega|$, it suffices to count $|A|$. Suppose $m \geq n$. Note that if $k \in \{1, \ldots, m\}$ is the "exactly one number" that is assigned to more than one person, then $k$ can be assigned to $2, 3, \ldots, n$ people. In view of this, $|A|$ can be then determined by applying the rule of product and the rule of sum.
In detail, there are $\binom{m}{1}$ ways of choosing $k$ that will be assigned to $j$ ($2 \leq j \leq n$) people. Once $k$ is chosen, there are $\binom{n}{j}$ ways to assign it to $j$ people picked randomly from a cohort of $n$ people, and $\binom{m - 1}{n - j} \times (n - j)!$ ways of assigning the remaining $n - j$ people each a distinct number. It then follows by the rule of product that the number of ways of assigning exactly one number to $j$ people ($2 \leq j \leq n$) is:
\begin{align}
\binom{m}{1} \times \binom{n}{j} \times \binom{m - 1}{n - j} \times (n - j)!.
\end{align}
By the rule of sum:
\begin{align}
|A| = \binom{m}{1}\sum_{j = 2}^n \binom{n}{j}\binom{m - 1}{n - j}(n - j)!.
\end{align}
I am not sure though, if $|A|$ can be further simplified.
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