Permutations: Ball Bearings to be shared

Question

Nineteen identical ball bearings are to be randomly shared by two girls and three boys anyhow. What is the probability that the boys will receive one ball bearing each and the rest go to the girls?

Answer 1

By Stars and Bars, there are $\binom{19+5-1}{5-1}$ ways to distribute the ball bearings between the children. For we are counting the solutions of $x_1+\cdots+x_5=19$ in non-negative integers.

Let us assume that these $\binom{23}{4}$ possibilities are all equally likely. This is in my view an unreasonable model, but it is the model the problem-setter intends be used.

We now count the "favourables." Of these $\binom{23}{4}$ ways, there are $17$ where each boy has $1$, for we must distribute the remaining $16$ between the girls, and the older girl can have anywhere from $0$ to $16$.

Remark: If we are not given any information about the distribution process, I think it is more reasonable to assume that the bearings are given, one at a time, to a randomly chosen child. That produces a sample space of $5^{19}$ equally likely elements, and an answer different from the "official" $\frac{17}{\binom{23}{4}}$.