[Math] What does “order matters” regarding permutations refer to

Question

I psychoanalyze EVERYTHING and permutations/combinations are frustrating me. Sorry for posting so many questions lately but I really appreciate all of the help!

Ok so I know the permutation formula: $\frac{n!}{(n-r)!}$ and combination formula: $\frac{n!}{(n-r)!r!}$

I don't understand how to be certain if a question has a permutation or combination answer. I seriously can convince myself that both make sense.. I look at book examples and I don't understand!

I know that in general.. permutations are larger than combinations.. order matters with permutations but NOT combinations. I try using this knowledge after reading a question but never know for certain. Again, I look at book examples and this permutation example has me confused:

Suppose that a saleswoman has to visit eight different cities. She
must begin her trip in a specified city, but she can visit the other
seven cities in any order she wishes. How many possible orders can the
saleswoman use when visiting these cities?

I get how the solution says 7! because there are 8 cities, the first city is where she starts.. so 8-1=7 obviously. But if the order of those other 7 cities don't matter.. wouldn't those 7 be a combination?

Also.. Idk how the formula would apply. I thought n would be 8 since there are 8 cities.. and r would be 7 since there are 7 more cities to travel to. But clearly that isn't correct.

Ugh can someone please help me again? 🙁

Thanks

Answer 1

My advice is that you should focus on the nature of the outcomes you are enumerating, rather than specific keywords in the language of the problem.

In the saleswoman question, ask yourself if it matters if the cities she visits are $A, B, C, D, E, F, G, H$ versus, say, $A, H, G, F, B, D, E, C$. If you are to interpret these as different outcomes, then you are looking at a permutation.

In the following example:

There is a bag of 9 marbles. 2 are red, 3 are blue, and 4 are green. How many ways are there to select a subset of 6 marbles from the bag such that there is at least one each of the three colors?

Does it matter if the marbles chosen are, say, $(r, r, g, g, b, b)$ versus $(r, g, b, g, r, b)$? Then a permutation on the individual outcomes is not applicable. But note that this question is a bit more complicated than a simple binomial coefficient computation, too. I mention it because counting methods ultimately rely not just on the question of "is it a permutation or combination" but other considerations as well.