Three trucks need to transport $k$ full containers, $k$ half-full containers, and $k$ empty containers in such a way that each truck is loaded identically, and each has the same number of containers. How many ways can this be done for $k=7$, if the trucks and the containers of the same fill level are not considered distinguishable?
We have a total of $3k$ barrels, so we need to load $k$ barrels onto each truck. This ensures from the outset that the load derived from the weight of the barrels is the same, so we only need to balance the weight of the wine in the barrels. Let's denote the weight of a full barrel of wine as $H$, then we need to load $H / 2$ weight of wine onto each truck. Let's place, for example, $A, B, C$ full barrels on each truck, then on the one hand,

$$
A+B+C=k
$$

on the other hand, these weights cannot be greater than $H / 2$, so

$$
A \leqq k / 2, \quad B \leqq k / 2, \quad C \leqq k / 2
$$

If we want to have $H / 2$ weight of wine on each truck, we need to place $(k-2A), (k-2B), (k-2C)$ half-full barrels on each truck, respectively. Due to (2), these numbers are non-negative, and due to (1), we have thus distributed the half-full barrels. Finally, let's place $A, B, C$ empty barrels on each truck, and we are done (since we need to load a total of $k$ barrels onto each truck, we have no other option). The number of possibilities is therefore equal to the number of triples of non-negative integers that satisfy (1) and (2), where triples that differ only in the order of the numbers are not considered different. If $k=7$, there are two decompositions: $3+3+1$ and $3+2+2$.