I got the the part where they divided $(m+n+p)$ things in groups containing $m$ and $(n+p)$ things respectively and then they proceeded to divide $(n+p)$ things in groups containing n and p things respectively. But, why did they multiply the two ways?
Any sort of help or assistance is much appreciated. Thanks in advance!
First you divide the $m+n+p$ things in groups containing $m$ and $n+p$ things respectively in one way. Then you divide the $n+p$ things in the second group into new groups containing $n$ and $p$ things respectively, which you do in all possible ways. At this point you have found $\frac{(n+p)!}{n!p!}$ ways to subdivide the original $m+n+p$ things.
Next you divide the $m+n+p$ things in groups containing $m$ and $n+p$ things respectively in a different way. Then you divide the $n+p$ things in the second group into new groups containing $n$ and $p$ things respectively, which you do in all possible ways. At this point you have found another $\frac{(n+p)!}{n!p!}$ ways to subdivide the original $m+n+p$ things. None of these ways is the same as any of the first $\frac{(n+p)!}{n!p!}$ ways, because the set of $m$ things in the first group is different. So you have a total of $2\frac{(n+p)!}{n!p!}$ ways to subdivide the original $m+n+p$ things so far.
You continue to repeat the first step, each time using a new way to divide the $m+n+p$ things in groups containing $m$ and $n+p$ things respectively. You continue this until you run out of ways to divide the $m+n+p$ things in groups.
At the point when you run out of ways to divide the $m+n+p$ things in groups, you have done the first step $\frac{(m+n+p)!}{m!(n+p)!}$ times, since that is the number of ways to divide the $m+n+p$ things in groups of $m$ and $n+p$ respectively. Each time you did this, you added $\frac{(n+p)!}{n!p!}$ ways to subdivide the original $m+n+p$ things in groups of $m,$ $n,$ and $p$ things respectively. A sum with $\frac{(m+n+p)!}{m!(n+p)!}$ terms each equal to $\frac{(n+p)!}{n!p!}$ is equal to $$ \frac{(m+n+p)!}{m!(n+p)!} \times \frac{(n+p)!}{n!p!}. $$
This is just an example of a more general rule: if you can do one task $Q$ different ways, and for each way you did the first task you can do a second task $R$ different ways, then the total number of different ways you can do the two tasks together is $Q \times R.$
Short answer: the multiplications rule of probability.
Not quite so short answer: The first thing to do is find out how many ways exist for the initial split into two groups. Second, they find how to split the second group. To get the total number of ways you multiply the two. Consider a tree diagram for the scenario where you choose one of two colours red or green, and once you have your desired colour, you choose one of three shapes, circle, square, triangle. the possibilities are
There are six ways to choose here. This was also available to us by multiplying the choices of colour by the choices of shape, 2 x 3.