1. There could be some additional learning in the thread below which I thought will be a relevance and value
2. Permutations is described as nPr or n!/(n-r)! and combinations is expressed as nCr or n!/r!(n-r)!. Note in both cases r<=n.
3. The way to read the problem under discussion is that => from 8 it gets to 6 and from 6 it gets to 3. We will take the same flow for easy logic, hence it is 6 out of 8 (which is 8C6) and 3 out of 6 (which is 6C3) => totalling to 8C6*6C3 ways.
You would notice that complementary logic is best used in those situations when that is more convenient without loosing the crux of the logic.
In the couples proble, when 5 people are to selected and 2 of them are married. We dont need to really know which are these two. Why? because we can read the question simply
as 2 out of 11 are couples and 9 others are not couples. If we need to choose 4 of these 11 people, I now know that I can select any 3 out of 9 and the couples OR any 5 out of the 9 and not the couples. Here the couples is the restraint added to the question and we take that into careful consideration. However, again in what order we select the 5 people also does not matter.