understanding when "order" is implied by the multiplication principle I feel really stupid posting this, but I keep turning in circles, so I'm really hoping someone can help elucidate the correct reasoning for the following problem:
Statement of the Question:

An agricultural researcher has produced multiple varieties of different strains of a particular plant: $\color{blue}{\text{six varieties of strain}}$ ${\color{blue} A}$, $\color{green}{\text{three varieties of strain}}$ $\color{green}{B}$ and $\color{orange}{\text{two varieties of strain}}$ $\color{orange}{C}$. For a particular experiment she has to select four different varieties, (exactly) two of which must be varieties of strain $A$, and no variety is repeated. How many possible groups are there to choose from? 


My Attempt & the Source of My Uncertainty:
What I can't settle for myself is whether (or how) order is accounted for. From my understanding of the multiplication principle, we effectively have 4 choices to make: 
$$
\underset{\small{ A_1}}{\underline{\hspace{1cm}}} \quad
%
\underset{\small{ A_2}}{\underline{\hspace{1cm}}} \quad
%
\underset{\small{ A_1'}}{\underline{\hspace{1cm}}} \quad
%
\underset{\small{ A_2'}}{\underline{\hspace{1cm}}} \quad
$$
where there are 6 choices for the first variety of strain $A$; 5 choices for the second variety of strain $A$; then 5 choices for the third variety (as it can be either strain $B$ or strain $C$) and 4 choices for the fourth variety, i.e.
$$
\underset{\small{ A_1}}{\underline{~6~}} \; \cdot \;
%
\underset{\small{ A_2}}{\underline{~5~}} \; \cdot \;
%
\underset{\small{ A_1'}}{\underline{~5~}} \; \cdot \;
%
\underset{\small{ A_2'}}{\underline{~4~}} \; \;
$$
This is where I get confused: does the above give me the number of permutations or the number of combinations?

My Attempt to Resolve My Uncertainty:
Combinations $_nC_r$ can be thought of as an adjustment to the value of a permutation $_nP_r = \frac{n!}{(n-r)!}$, by dividing out the number of ways of $r$ distinct objects can be organized ( viz. $r!$): 
$$\; _nC_r = \frac{_nP_r}{\color{red}r!} = \frac{n!}{(n-r)!\color{red}{r!}},$$
so I reasoned that if the above expression did account for order, then I ought to divide it by $4!$, as there are 4 distinct varieties, so I'd end up with: 
$$ (6 \cdot 5 \cdot 5 \cdot 4 ) \cdot \tfrac{1}{4!}; $$
however, when I try reasoning through the problem using combinations, instead, I don't get the same result. In the case of considering combinations, there seem to be only two choices: 


*

*choose 2 types of strain $A$: $\; _6C_2 = \frac{6!}{4!2!} = \frac{6\cdot 5}{2!}$;

*choose 2 types that are not of strain $A$: $\; _5C_2 = \frac{5!}{3!2!} = \frac{5 \cdot 4}{2!}$;
the total number of possible outcomes is then their product (again by the multiplication principle): 
$$ _6C_2 \cdot _5C_2 = \frac{6 \cdot 5}{2!} \cdot \frac{5 \cdot 4}{2!}  = (6\cdot 5 \cdot 5 \cdot 4) \cdot \frac{1}{2!2!},$$ 
which is clearly not equal to the expression found, above. 
What is really bothering me here is not only that I'm not certain about the final expression, but that I don't understand why I'm getting completely different values. 
I can think of ways to resolve the differences (e.g. the permutations should each be divided by $2!$ because they are indistinguishable as $A$'s or $A'$'s, respectively), but doctoring the expression and reasoning backwards doesn't feel like I'm clarifying the actual source of my confusion. 
I don't care about the final answer -- I just really want to clarify the reasoning here, as clearly there is something I'm not appreciating in the way these expressions work. Any help would be greatly appreciated

$\color{red}{\text{EDIT}}:$
I started to edit the post to clarify further, but think that the clarification constitutes a separate question, so will post it separately & link. Thank you kindly to all those who have responded so far.

The solution you posed using combinations is correct.
As for your question, as you stated, you first selected a variety of strain $A$, then selected another variety of strain $A$, then selected a variety selected from strains $B$ or $C$, and finally selected another variety selected from strains $B$ or $C$.  That is, you selected the strains in that particular order.  Therefore, it does not make sense to divide by the $4!$ orders in which you could have picked the same strains had you picked two varieties of strain $A$ and two varieties from strains $B$ and $C$ in any order.  
What we do care about is which strains we pick.  Picking varieties $A_1$, $A_2$, $B_1$, and $C_1$ in that order results in the same set of varieties as selecting varieties $A_2$, $A_1$, $C_1$, and $B_1$ in that order.  Thus, selecting two varieties of strain $A$ and then selecting two varieties from strains $B$ or $C$ in that order counts each possible selection four times since there are $2!$ orders in which the same varieties of strain $A$ could be selected and $2!$ orders in which the same varieties of strains $B$ or $C$ could be selected.  Notice that 
$$\frac{6 \cdot 5}{2!} \cdot \frac{5 \cdot 4}{2!} = \frac{6!}{2!4!} \cdot \frac{5!}{2!3!} = \binom{6}{2}\binom{5}{2}$$
