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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

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It would be silly to ask "How many possible orders" and at the same time insist that order does not matter, so you may safely assume that order does matter here. If really you wanted to count the number of ways to visit those cities if order does not matter, then there is exactly one solution; indeed there is one combination of $7$ out of $7$ candidates that can be chosen. –  Marc van Leeuwen Mar 22 at 22:56
    
She has to begin her trip in a "specified city", meaning the initial value is the same. (The first city will not be rearranged into another.) –  Don Larynx 6 hours ago

3 Answers 3

up vote 3 down vote accepted

Note that the question says any order she wishes, not that the order does not matter. Hence, we are looking at a permutation (the specific ordering of the cities make up her route).

And note that she starts in a specified city, i.e. there is no decision here. After that there are $8-1=7$ cities to visit, which can be visited in a certain order in $7!$ ways.

I find it difficult to give a more general answer about when to use permutations vs combinations other than what you already seem to know, that it depends on whether order matters or not. However, if you have specific questions, I will be glad to help.

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Ok thanks! So if a question says order doesn't matter.. its 100% a combination equation? I posted a question a few days ago and someone kind of clarified the confusion, but I'd love to hear your feedback if you don't mind: math.stackexchange.com/questions/718899/… –  Cozen Mar 22 at 22:48
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@Cozen Yes, that is true. And to comment on your link, when selecting 6 animals, we only care about what the selection will be, not in the actual order. If someone asks you about the six animals you've selected, he/she won't care if the zebras or the elephants were selected first. –  naslundx Mar 22 at 22:52
    
Ok thanks! Could you possibly provide hints for those on my other link? That other user didn't completely explain to my understanding. –  Cozen Mar 22 at 23:00
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@Cozen Edited answer to explain why we use $7!$ instead of $8!$ here. –  naslundx Mar 22 at 23:00
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@Cozen I will try :) –  naslundx Mar 22 at 23:00

If there is any hint that the answer depends on the order in which the objects are counted, then you use permutations, else combinations are good. This can be remembered by defining as follows:

Definition: A Permutation is an ordered Combination.

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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.

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