Combinations vs. permutations: knowing when and how to account for order Part 2 In this question, I tried to get at some underlying confusion I'm experiencing when it comes to the multiplication principle and order: right now it seems to me that sometimes multiplying expressions account for order, and sometimes they do not.
The original problem essentially had 6 distinct choices from group $A$ and 5 distinct choices from group $A'$, and asked for the number of distinct groups of 4 objects that had exactly 2 members from $A$. It was established that the correct reasoning was:
$$
\Big(
\underbrace{{~ \color{blue}{6} ~}}_{A_1} \cdot 
%
\underbrace{{~\color{blue}{5} ~}}_{A_2} 
\Big)
%
 \cdot 
%
\tfrac{1}{2!}
%
%
  \; \cdot \;
%
%
\Big(
\underbrace{{~ \color{green}{5} ~}}_{A_1'} \cdot 
%
\underbrace{{~ \color{green}{4} ~}}_{A_2'} 
\Big)
%
 \cdot 
%
\tfrac{1}{2!}
$$
which corresponds to the expression in terms of combinations: 
$$
_6C_2 \; \cdot \; _5C_2
$$
What is bothering me is that, in the first case, the multiplications $\color{blue}{6 \cdot 5}$ and $\color{green}{5 \cdot 4}$ apparently account for order (hence division by $2!$, for each), whereas in the second case the multiplication of $_6C_2$ and $_5C_2$ does not account for the number of ways 2 objects are ordered. How come? Is it simply because we have already "taken the ordering" out of combinations and so multiplying them doesn't account for order? 
To further clarify, what if I want to make a group from three sets, where $A$ has 8 elements, $\color{orange}{B}$ $\color{orange}{\text{has 6 elements}}$ and $\color{purple}{C}$ $\color{purple}{\text{has $12$ elements}}$, and I want to have 3 elements from $A$, $\color{orange}{\text{4 elements from $B$}}$ and $\color{purple}{\text{5 elements from $C$}}$. Following the above, it seems the answer would just be: 
$$
\boldsymbol{_8C_3} \; \cdot \; \color{orange}{_6C_4} \; \cdot \; \color{purple}{_{12}C_5}
$$
But when I look at that, what I see is a sequence of decisions, where we're multiplying the number of possibilities at each step, which leads me to think that it is a permutation, so I really want to "divide out the number of ways of permuting three objects," i.e. I'm tempted to write:
$$
\big(
\boldsymbol{_8C_3} \; \cdot \; \color{orange}{_6C_4} \; \cdot \; \color{purple}{_{12}C_5}
\big) \cdot \tfrac{1}{3!},
$$
which I believe is actually wrong. Can anyone provide a systematic way of knowing $\color{red}{\text{when multiplying options accounts for ordering, and when it does not?}}$ 
 A: Think of the multiplication principle as counting the number of ways there are to make a particular sequence $C_1, C_2, C_3,\dots$ of choices for an $s$-stage decision process so that $C_0,C_1,\cdots C_s$ produces one of the outcomes you want to count. If there are $n_i$ ways in which you can make the specific choice $C_i$, where $n_i$ is independent of the particular choices made before the $i$-th stage, the number of ways to make the entire sequence of choices will be $\prod n_i$. If your multi-stage process generates each thing you want to count exactly once, the number of things you wanted to count is $\prod n_i$. However, sometimes, it’s easiest to come up with a multi-stage process (perhaps depending on order) that generates each thing you want to count multiple times, in which case you can (if the multiple is a constant) correct your answer by division.
In your example of choosing from sets $A$, $B$, and $C$, you set up your stages so that you always chose from set $A$ first, then $B$, then $C$. So you did not generate results from each ordering of $A$, $B$, and $C$, which is what you are wondering whether you should divide by.
In fact, it would have been hard to use the multiplication principle if your stages were 


*

*Choose one of the three sets to choose elements from.

*Choose the required number of elements from that set.

*Choose another of the three sets.

*Choose the required number of elements from that set.

*Choose the required number of elements from the remaining set.


because, while you would produce every outcome six times, the number of choices at some stages would depend on the particular choices made at other stages.
