A recent question asked for a proof of an identity of restricted partition numbers which looks at first glance like an application of inclusion-exclusion, but for which I have only been able to find a proof by algebraic manipulation of generating functions. Similar manipulation gives other curious identities, and perhaps a combinatorial proof of a simple such identity will provide the intuition for a combinatorial proof of the original, more complex one.
Let $P(n|p|q)$ denote the number of partitions of $n$ into exactly $p$ parts, the largest of which is $q$; and let $P(n|p|\le q)$ denote the number of partitions of $n$ into exactly $p$ parts, the largest of which is no greater than $q$. From the generating function
$$\sum_{n} P(n|p|\le q) x^n = x^p \frac{\prod_{j=q}^{q+p-1} 1-x^j}{\prod_{i=1}^p 1-x^i}$$ we can interpret $$x^p \frac{\prod_{j=q}^{q+p} 1-x^j}{\prod_{i=1}^p 1-x^i}$$ in two ways to get the identity
$$P(n|p|\le q) - P(n-p-q|p|\le q) = P(n|p|\le q+1) - P(n-q|p|\le q+1)$$
or
$$P(n|p|q+1) = P(n-q|p|\le q+1) - P(n-p-q|p|\le q) \tag{*}$$
Is there a bijective proof of $(*)$?
Here is a bijective proof of $$P(n|p|q+1)+ P(n-p-q|p|\le q) = P(n-q|p|\le q+1) $$
Given a partition $\lambda$ of $n-q$ into $p$ parts which are all at most $q+1$, there are two cases.
If all of the parts of $\lambda$ are greater than $1$, then subtracting one from each part leaves a partition of $n-p-q$ into $p$ parts of sizes at most $q$.
If one of the parts of lambda is equal to $1$, then deleting that part and adding a part of size $q+1$ leaves a partition of $n$ into $p$ parts whose largest part is $q+1$.