In answering to this post (which actually parallel this other one about the more common phenomenon in car traffic)
I came across to this interesting Sum-Product, defined for non-negative integers $n,q,g$
$$
N(n,q,g) = \left\{ {\begin{array}{*{20}c}
0 & {\left| {\;q < g} \right.\;} \\
{\left[ {0 = g} \right]} & {\left| {\;0 = q} \right.} \\
{\sum\limits_{\left\{ \begin{subarray}{l}
1\, \leqslant \,k_{\,1} \, < \,k_{\,2} \, < \, \cdots \, < \,k_{\,g} \, \leqslant \,n\, \\
\,j_{\,1} + j_{\,2} \, + \,\, \cdots \, + \,j_{\,g} \, = \,q\;\;\left| {\;1\, \leqslant \,j_{\,k} \,} \right.
\end{subarray} \right.} {k_{\,1} ^{\,j_{\,1} - 1} \;k_{\,2} ^{\,j_{\,2} - 1} \cdots k_{\,g} ^{\,j_{\,g} - 1} } } & {\left| {\;1 \leqslant q} \right.\;} \\
\end{array} } \right.
$$
which gives the number of ways into which a linear flock of $q$ sheeps may finally gather
into $g$ groups, given that each sheep is randomly assigned with one of the $n$ values for speed.
With no sheep we can form $0$ groups (the empty set), for whichever $n$ ( [X] represents the Iverson's bracket).
The probability of having $g$ groups is therefore given by $$ P(n,q,g) = \frac{{N(n,q,g)}} {{n^{\,q} }} = \left\{ {\begin{array}{*{20}c} {\left[ {0 = g} \right]} & {\left| {\;0 = q} \right.\;} \\ {\frac{1} {{n^{\,g} }}\sum\limits_{\left\{ \begin{subarray}{l} 1\, \leqslant \,k_{\,1} \, < \,k_{\,2} \, < \, \cdots \, < \,k_{\,g} \, \leqslant \,n\, \\ \,j_{\,1} + j_{\,2} \, + \,\, \cdots \, + \,j_{\,g} \, = \,q\;\;\left| {\;1\, \leqslant \,j_{\,k} \,} \right. \end{subarray} \right.} {\left( {\frac{{k_{\,1} }} {n}} \right)^{\,j_{\,1} - 1} \;\left( {\frac{{k_{\,2} }} {n}} \right)^{\,j_{\,2} - 1} \cdots \left( {\frac{{k_{\,g} }} {n}} \right)^{\,j_{\,g} - 1} } } & {\left| {\;1 \leqslant q} \right.\;} \\ \end{array} } \right. $$ which when the range of speeds becomes continuous gives $$ P(\infty ,q,g) = \frac{1} {{q!}}\left[ \begin{gathered} q \\ g \\ \end{gathered} \right] $$ The details on the derivation of the above formulas can be read in the answer to the referred post.
A few values of $N(n,q,g)$ are given in the Table.
Many known paths are discernible in the columns, rows and diagonals there.
However I could not yet figure out a more compact expression for that.
Any hint or suggesting a reference (on Stirling p.m.f.) is appreciated.
So, not having received hints, I proceeded in the analysis and could obtain a recurrence relation for $N(n,q,g)$ as $$ \left\{ \begin{gathered} N(n,q,g) = 0\quad \left| {\;n < 0\; \vee \;q < 0\; \vee \;g < 0} \right. \hfill \\ N(n,q,g) - N(n - 1,q,g)\quad = \hfill \\ = n\;\left( {N(n,q - 1,g) - N(n - 1,q - 1,g)} \right) + N(n - 1,q - 1,g - 1) + \left[ {0 = n} \right]\left[ {0 = q} \right]\left[ {0 = g} \right] \hfill \\ \end{gathered} \right. $$
And, applying a binomial tranform to the table obtained, to realize that $$ \bbox[lightyellow] { N(n,q,g) = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,\min (n,q)} \right)} {\left( \begin{gathered} n \\ k \\ \end{gathered} \right)\left\{ \begin{gathered} q \\ k \\ \end{gathered} \right\}\left[ \begin{gathered} k \\ g \\ \end{gathered} \right]} } \tag{1}$$
which in fact satisfies the recurrence above, and gives $$ \begin{gathered} n^{\,q} = \sum\limits_{\left( {0\, \leqslant } \right)\,g\,\left( { \leqslant \,\min (n,q)} \right)} {N(n,q,g)} = \sum\limits_{\left( {0\, \leqslant } \right)\,k,\;j\,\left( { \leqslant \,\min (n,q)} \right)} {\left( \begin{gathered} n \\ k \\ \end{gathered} \right)\left\{ \begin{gathered} q \\ k \\ \end{gathered} \right\}\left[ \begin{gathered} k \\ j \\ \end{gathered} \right]} = \hfill \\ = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,\min (n,q)} \right)} {\left( \begin{gathered} n \\ k \\ \end{gathered} \right)\left\{ \begin{gathered} q \\ k \\ \end{gathered} \right\}k!} = \sum\limits_{\left( {0\, \leqslant } \right)\,k\,\left( { \leqslant \,\min (n,q)} \right)} {\left\{ \begin{gathered} q \\ k \\ \end{gathered} \right\}n^{\,\underline {\,k\,} } } \hfill \\ \end{gathered} $$