How to interpret this combination formula

Question

If you have $m$ and $n$ number of distinct items, and you get to choose $p$ and $q$ things respectively out of them, then the number of ways you can permute $p+q$ items is as follows:

$m$ items can be chosen for $p$ places like $^mC_p$

$n$ items can be chosen for $q$ places like $^nC_q$

and $p+q\:$ items can be permuted among themselves as $(p+q)!$

so the permutation will be $^mC_p\times{}^nC_q\times(p+q)!$

Now, I more or less have the intuition behind this formulation. But a problem arose when I tried to describe it to my twelfth grader student. He understood the reasoning behind that. But he insisted me on writing this formulation in terms of $P$. I had no clue at first. After some trial and error with some data, I found the following term which is numerically same as the previous one.

$^mP_p\times{}^nP_q\times{}^{p+q}C_{p\text{ or }q}={}^mC_p\times{}^nC_q\times(p+q)!$

Though it spits out the same numbers, I'd like to grasp some insight into it. How should this term be interpreted $^mP_p\times{}^nP_q\times{}^{p+q}C_{p\text{ or } q}$?

Answer 1

$$ \eqalign{ & \left( \matrix{ m \cr p \cr} \right) = {{m^{\,\underline {\,p\,} } } \over {p!}}\quad \Rightarrow \cr & \Rightarrow \quad \left( \matrix{ m \cr p \cr} \right)\left( \matrix{ n \cr q \cr} \right)\left( {p + q} \right)! = {{m^{\,\underline {\,p\,} } } \over {p!}}{{n^{\,\underline {\,q\,} } } \over {q!}}\left( {p + q} \right)! = \cr & = m^{\,\underline {\,p\,} } n^{\,\underline {\,q\,} } {{\left( {p + q} \right)!} \over {p!q!}} = m^{\,\underline {\,p\,} } n^{\,\underline {\,q\,} } \left( \matrix{ p + q \cr q \cr} \right) \cr} $$

That means that the alternative scheme is:
You take a first group of $p$ items from $m$, forming all possible $p$-tuples (order considered).
You do the same with the items from $n$.
For each combination of a $p$-tuple and a $q$-tuple, you permute the items.

Answer 2

Here's a way to think of things. In the ${}^mC_p{}^nC_q(p+q)!$ way of counting things, you are picking $p$ out of the $m$ objects from the first set, then picking $q$ out of the $n$ objects from the second set, then taking these $p+q$ objects and lining them up in a row, while in the second way of counting things, you start with an empty row with $p+q$ slots, first decide which $p$ slots will be filled with objects from the first set, then fill these $p$ slots, and finally fill the remaining $q$ slots with objects from the second set. The initial decision has ${}^{p+q}C_{p\text{ or }q}$ possibilities, followed by ${}^mP_{p}$ possibilities for filling the chosen $p$ slots, followed by ${}^nP_{q}$ possibilities for filling the remaining $q$ slots.