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.
 A: 


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 to Resolve My Uncertainty:
Combinations $_nC_r$ can be thought of as an adjustment to the value of a permutation $_nP_r = \tfrac{n!}{(n-r)!}$, by dividing out the number of ways of $r$ distinct objects can be organized ( viz. $r!$):

I find it more intuitive to think of it the other way around.  Counting permutations involves counting ways to select the items, then to arrange them. Thus we have ${^n\mathrm P_r}={^n\mathrm C_r}\cdot{r!}$
All we need do here is select two from six varieties of A, and two from the five other varieties. We have no need to arrange the selections. The answer is thus simply ${^6\mathrm C_2}{^5\mathrm C_2}$.
A: $6.5.5.4$ gives you the number of permutations of four varieties satisfying the given condition in the question, where first two varieties are of strain A and rest two are of strain B.
Note: $(A_{1}, A_{2}, B_{1}, C_{1)}$ is a permutation considered in '6.5.5.4' and so is $(A_{2}, A_{1}, B_{1}, C_{1)}$ but $(A_{2}, B_{1}, A_{1}, C_{1)}$ isn't. Clearly, as the four chosen varieties cannot permute amongst themselves in 'any' order, dividing it by $4!$ will be incorrect. The method you used in your last approach is alright.
A: 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 
$$\frac{6 \cdot 5}{2!} \cdot \frac{5 \cdot 4}{2!} = \frac{6!}{2!4!} \cdot \frac{5!}{2!3!} = \binom{6}{2}\binom{5}{2}$$
