There's a formula in Counting: The art of enumerative combinatorics by George E. Martin which I don't quite understand.
Let $\Pi(r,n)$ be the number of partitions of $r$ into $n$ parts. If we want to know the total number of partitions of the $r$ number, $\Pi(r)$, we have to sum such $\Pi(r,n)$ (agreed!)
$$ \Pi(r) = \sum_{j=1}^r \Pi(r,j) \stackrel{?}{=} \Pi(r+r,r), $$
So, why is the sum $\sum_{j=1}^r \Pi(r,j)$ equal to $\Pi(r+r,r)$? I know that by looking the Pascal's triangle this is always true, but I can't imagine why in principle both problems are the same.
This is exactly the problem of $r$ indistinguishable balls into indistinguishable boxes with no box empty, so I can't find an argument in favor of this equality.
Assuming that $\Pi(r,j)$ means the count of partitions of $r$ with exactly $j$ parts, and $\Pi(r)$ means the count of all partitions of $r$, the equality $\Pi(r) = \Pi(2r,r)$ can be understood with a simple thought experiment.
Consider a partition of $2r$ with exactly $r$ parts. Now subtract one from each part, which gives a partition of $r$ with at most $r$ parts (since some former parts may have been one and disappeared on subtraction). But any partition of $r$ will have at most $r$ parts, so this gives a mapping of partitions of $2r$ with exactly $r$ parts to partitions of $r$.
It is easily argued that the mapping is surjective, by giving an inverse mapping. That is, given a partition of $r$ (with at most $r$ parts), create a partition of $2r$ with exactly $r$ parts by padding the summation with zero summands to get exactly $r$ summands and then adding one to each summand. Note that this is precisely the reversal of the first mapping.
Finally we want to demonstrate that the first mapping is injective. Suppose for contradiction that we have initially two partitions of $2r$, each with exactly $r$ parts:
$$ x_1 \le x_2 \le \ldots \le x_r \;,\; \sum x_i = 2r $$ $$ y_1 \le y_2 \le \ldots \le y_r \;,\; \sum y_i = 2r $$
such that the mapping of subtracting one from each part produces the same partition of $r$. That is, we get the same number of zero summands and the same number of nonzero summands, as indeed $x_i - 1 = y_i - 1$ for all $i$, whether these reduced parts are zeros or not. Then $x_i = y_i$ all along, and the original partitions of $2r$ were identical. Ergo the mapping is injective.
Therefore $\Pi(r) = \Pi(2r,r)$.