Why is the following not $S(n,3)$ where $S(n,k)$ is a Stirling number of the second kind? (almost solved) In an attempt to relate the number of partitions of integers to that of partitions of distincts objects  I stumbled, in the particular case of $k:=3$, on the following sum
$$\sum_{\genfrac{}{}{0pt}{1}{p\leq q \leq r}{p+q+r =n}} \frac{n!}{p!\, q!\, r!} \times \begin{cases} 1 & \text{if}\ p< q <r\\
 \frac{1}{2} & \text{if exactly two indices are equal}\\
 \frac{1}{3!}& \text{if}\ p=q=r= \frac{n}{3} \end{cases}\enspace =\ \frac{3^n}{3!} \quad \overset{???}{\underset{\textbf{WHY?}}{\neq}} S(n,3)= \frac{3^{n-1} - 2^n +1}{2} $$
where $S(n,k)$ denotes the number of partitions of a set of $n$ distinct objects into $k$ non-empty subsets while the number $p_k(n)$ of summands/indices of summation/ is the number of way of writing $n$ as a sum of $k$ non zero integers.
Remark: one finds the argument leading to the first equality in this Q$\&$A while the last equality is stated in wikipedia and proved for example here.
My reasoning: compute for each given partition $n=p+q+r$ of integers the number of partitions of $\{1,2,\cdots , n\}$ into $3$ subsets of respective cardinal $p,q$ and $r$. That quantity is in fact the cardinal of the orbit $O_{(p,q,r)}$ of e.g. the set partition $\Pi_{(p,q,r)}:=\left\lbrace \vphantom{\frac{a}{a}}\{1, 2 , \cdots , p\}, \{p+1,\cdots , p+q\}, \{p+q+1 , \cdots , n\} \right\rbrace$ under the group $\mathfrak{S}_n$ of permutations, which is given by the formula
$$ \operatorname{Card}\left( O_{(p,q,r)} \right)= \frac{\operatorname{Card}\left( \mathfrak{S}_n\right) }{\operatorname{Card}\left(\operatorname{Stab}\Pi_{(p,q,r)}\right)} = \frac{n!}{p!\, q!\, r!} \times \begin{cases} 1 & \text{if}\ p\neq q \neq r,\; p\neq r\\
 \frac{1}{2} & \text{if exactly two indices are equal}\\
 \frac{1}{3!} & \text{if}\ p=q=r= \frac{n}{3} \end{cases}
$$
where the stabilizer subgroup $\operatorname{Stab}\Pi_{(p,q,r)}$ consists of all the permutations of $\mathfrak{S}_n$ that leave the partition $\Pi_{(p,q,r)}$ unchanged, e.g. permuting only elements withing $\{1, 2 , \cdots , p\}$ or when $p=q$ sending bijectively all elements of $\{1, 2 , \cdots , p\}$ to $\{p+1, p+ 2 , \cdots , p+q\}$ etc.

It's always like this... while writing this question I just noticed that in order for the two members of the first equation to be equal, one must exclude the case where any of $p,q$ or $r$ are $0$: so neither $(0,0,n)$, nor $(0,q,r)$ with $q+r =n$. It seems to work then... but I post the question anyway...
 A: Yeah I think one should not overlook the details... such as the following paradox in the first equality:
$$ \begin{split}
\frac{(1+1+1)^n}{3!} & = \frac{1}{3!} \sum_{p+q+r =n} \frac{n!}{p!\, q!\, r!} \\
&\overset{??!!}{=} \frac{1}{3!} \sum_{\genfrac{}{}{0pt}{1}{p\leq q \leq r}{p+q+r =n}} \frac{n!}{p!\, q!\, r!} \times \begin{cases} 3! & \text{if}\ p< q <r\\
 3 & \text{if exactly two indices are equal}\\
 1 & \text{if}\ p=q=r= \frac{n}{3} \end{cases}
\end{split}
 $$
R.h.s. is supposedly a sum of integer (with the interpretation that we are summing the cardinal of orbits, cf. question)... but l.h.s. isn't (as $3^n$ is not a multiple of $2$...)?!?!

Solution: again it is the case where $p=q=0$ that is problematic. Without the $p\leq q\leq r$ condition, it appears $3$ times ($(p,q,r)= (0,0,n)$ or $(0,n,0)$ or $(n,0,0)$) and hence contributes $\frac{1}{3!}\, \frac{n!}{n!} \times 3 = \frac{1}{2}$ so finally the first equality is valid!

Now we just have to retrieve the number of set-partitions in two or one subsets (reformulation of the cases $(0,q,r)$ with $q+r=n$ or $(p,q,r) = (0,0,n)$) with the caveat that the last partition was counted $\frac{1}{2}$. By analogy, the number of partitions into two or fewer subsets is
$$ \sum_{\genfrac{}{}{0pt}{1}{q\leq r }{q+r=n}} \frac{n!}{q! \, r!} \times \begin{cases} 1 & \text{if}\ q < r  \\
 \frac{1}{2} & \text{if}\ q=r= \frac{n}{2} \end{cases}
= \frac{2^n}{2}
$$
Here the partition $(q,r)=(0,n)$ really contributes $1$ so we'll have to artificially add a $\frac{1}{2}$ at the end to obtain
$$ \sum_{\genfrac{}{}{0pt}{1}{{\color{red} 0 < p}\leq q \leq r}{p+q+r =n}} \frac{n!}{p!\, q!\, r!} \times \begin{cases} 1 & \text{if}\ p< q <r\\
 \frac{1}{2} & \text{if exactly two indices are equal}\\
 \frac{1}{3!} & \text{if}\ p=q=r= \frac{n}{3} \end{cases} = \frac{3^{n-1}-2^n+1}{2}$$
