Let there be $m$ boxes and $k$ balls where the probability that a ball is laid in box $i$ ($1\leq i \leq m)$ is $p_{i}$
Define a probability space $(\Omega, \mathcal{F}, P)$:
Idea:
Define $\Omega:=\{(\omega_{1},...,\omega_{m})\in \mathbb N_{0}^{m}:\sum_{i=1}^{m}\omega_{i}=k\}$
Since $m \in \mathbb N$, we're in a discrete case and therefore $\mathcal{F}=\mathcal{P}(\Omega)$
Now onto the distribution:
Let $\omega:=(\omega_{1},...,\omega_{m})$
$P(\{w\})=\prod_{i=1}^{m}(\cdot)p_{i}^{\omega_{i}}$
where $(\cdot)$ is meant to represent a particular combination. I would have thought that the binomial coefficient would have been a good idea, but I have been told it is rather the multinomial coefficient, but I do not understand why.
A binomial distribution arises from the sum of independent and identically distributed Bernoulli trials. The outcome of a Bernoulli trial is either $0$ or $1$; or more generally, a binary set of outcomes.
A multinomial distribution arises from the sum of independent and identically distributed categorical trials. The outcome of a categorical trial is some number in $\{0, 1, 2, \ldots, m-1\}$, where $m \ge 2$; or more generally, some discrete and finite set of outcomes.
I chose $m$ in such a way that it matches the $m$ in your problem. If you have only $2$ boxes, then the resulting distribution of balls is binomial, because for any given ball, there are only two choices for which box to place it. If you have more than $2$ boxes, then it is not possible to characterize the distribution of balls using only a single binomial variable. you'd need to describe whether a given ball is put into the first, second, or third box, for example.