How to determine the number of combinations in a conditional group? I have a probability/combinatorics question that is giving me trouble specifically I don't understand why there is a need for the denominator $6$ "the different orderings of $3$ people" was the explanation but I'm not sure why that is necessary. I've put the question below and my thought process as well.
Question:

If a committee of $3$ people is to be selected from among $5$ married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

Thought Process:


*

*First person of the $3$ selected can be any one among the $10$ folks $$\frac{10!}{(1!)(9!)}$$

*Second Person of the $3$ selected must not be married to the first person so lowers availability to $8$ folks to choose from $$\frac{8!}{((1!)(7!)}$$

*Third Person of the $3$ selected must not be married to either the first or second person - lowers availability to $6$ folks to choose from $$\frac{6!}{(1!)(5!)}$$

*Number of possible ways a committee can be formed: $10 \times 8 \times 6 = 480$
***So I'm able to get to the $480$ but the answer explanation says that I need to divide this $480$ by $6$ and I don't know why because it sounds illogical to do so. Any help would be amazing.
 A: Suppose the three people who are selected are Anne, Barbara, and Charles.
There are six orders in which those same three people could be selected:
Anne, Barbara, Charles
Anne, Charles, Barbara
Barbara, Anne, Charles
Barbara, Charles, Anne
Charles, Anne, Barbara
Charles, Barbara, Anne
However, all six choices constitute the same committee.  Therefore, you need to divide your answer by the $3! = 6$ orders in which you could obtain the same three people, which yields the answer
$$\frac{10 \cdot 8 \cdot 6}{3!} = \frac{10 \cdot 8 \cdot 6}{6} = 10 \cdot 8 = 80$$
Alternate Approach: Choose three of the five couples from which to choose the committee members in $\binom{5}{3}$ ways, then choose one of the two members from each selected couple, which can be done in $2^3$ ways.  Thus, the number of possible selections is 
$$\binom{5}{3} \cdot 2^3 = 10 \cdot 8 = 80$$
A: you can do this way too
trio committee with no couples togeather= (total no. of committees - number of committees with couples togeather)
$$=(^{10} C_{3}-^5 C_{1}.8)=80  ways$$
foot note: 
$^5 C_{1}:-$ for choosing $1$ pair or couple(i.e, $2$ peoples) and after we've chosen $2$ peoples we are left with $8$ choices for $3rd$ person to form committee that is why we multiply them to get          
$^5 C_{1}.8$
