# Maths question

Discussion in 'Community Discussion' started by 835153, Nov 27, 2013.

1. ### 835153 Guest

Joined:
Aug 5, 2013
#1
I've put this in community section as its more of a maths question than fully relating to graphic design.

Im terrible with numbers and I need to know how many layout permutations a 32 page magazine containing 74 editorial sections and 18 adverts are. I need to quantify my work.

i.e. if each one of the adverts was placed next to each one of the editorial sections on each one of the pages what would be the total sum of all the possible layouts.

Would the correct formula be 74x18x32?

I think I need a 'to the power of x' on the 32 though (i have no idea what 'to the power of' does or means though) I really should of listened in school.

2. ### robbieduncan Moderator emeritus

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Jul 24, 2002
Location:
London
#2
There are no constraints? You can have all the adverts in one block at the end?

3. Nov 27, 2013
Last edited: Nov 27, 2013

### Staff Member

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Aug 16, 2005
Location:
New England
#3
Sounds like an attempt at a constraint to me!

EDIT:

Let's restate the problem. You have 32 pages (I assume these are all the same size), 74 articles (editorial section) and 18 ads.

This means, assuming each ad is the same size you have just over one ad for every two pages. (16 ads would mean every other page has an ad.

You can't place one next to each article because you have roughly 4x as many of them.

This also means that, on average there are ~4.5 articles per 2 page spread.

EDIT 2: So let's simplify the problem to get some easier to deal with numbers.

32 pages, 16 ads, 80 articles all the same size. Assume a uniform distribution of 1 ad and 5 articles per 2 page spread.

On the first two page spread alone, you can choose the 5 articles from the 80 and one of the 16.

That means you have 80*79*78*77*76 choices of article and 16 choices from the ads.

So a total of 80*79*78*77*76*16=46,156,830,720 choices for the first 2 page spread alone.

The next 2 page spread would have 75*74*73*72*71*15=31,066,902,000 choices.

and so on.

If you don't have this uniformity constraint the number of choices only go up!

Big numbers!

B

4. Nov 27, 2013
Last edited: Nov 27, 2013

Joined:
Aug 5, 2013
#4
It doesn't have to be EXACT. Really just a rough idea of how many combinations there could be. So lets not get bogged down with, no adverts on the front cover etc. as the 18 advert count is a variable in itself. Just an example of a formula to give an estimate of the permutations without constraints.

I have people suggesting its very simple laying out a magazine when I have constraints of this advert needs to go next to such and such and this advert goes next to such and such but this advert can't go there and these two need to be kept apart. They don't see how complicated the thought process behind these layouts can get nor the problems that arise when there is a conflict. They seem to think everything just gets dropped in to the layout. :roll eyes:

EDIT: Just to add these adverts are all different sizes too but again lets not get bogged down with that.

While your laying these things out you kind of mentally have to go through all these possible layouts. Most of the time you hit upon a successful layout without having to exhaust all the options. Sometimes the constraints make it impossible and thats what they are not seeing.

Probably easier to work out how many options there are with just one advert thinking about it. So 1 advert can be placed next to 72 different bits of edit. Thats 72 possible layouts there.

5. Nov 27, 2013
Last edited: Nov 27, 2013

### Staff Member

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Aug 16, 2005
Location:
New England
#5
See my second edit. Tons.

EDIT: This is why it's more of an art than a science.

Take this example.

Take a standard 52 card deck. Shuffle it and reveal 5 cards at random. If the order of those cards matters (as it might in a magazine layout). You originally have 52 choices in the deck. Once you pick one you only have 51 choices, etc...

That's where the terms like 52*51*50*49*48=311,875,200 come from.

If you choose 5 cards at random from a 52 card deck and the order of the cards matters there are 311 million possible combinations.

B

6. ### 835153 thread starter Guest

Joined:
Aug 5, 2013
#6
Exactly my point haha. That will do. Cheers.

Obviously common sense and experience can rule a lot of those permutations out but when people start dictating the layout for you without knowledge of the rest of the layout its can only be bad news.

7. ### mobilehaathi macrumors G3

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Location:
The Anthropocene
#7
Yes, you should have, but you know there isn't anything stopping you from reviewing basic mathematics now, right?

8. ### 835153 thread starter Guest

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Aug 5, 2013
#8
Yeh well Im busy working out billions of combinations laying out magazines for a living.

9. ### mobilehaathi macrumors G3

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Aug 19, 2008
Location:
The Anthropocene
#9
You'll never finish even your first project at just about any humanly possible rate.

Seriously, why not take a little time to bone up?

### Staff Member

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Aug 16, 2005
Location:
New England
#10
That's a bit of a different problem, but let's take it as another one you can work out quantitatively.

Assume this is a linear webpage like MR.

You have 72 posts and thus 71 interfaces between posts. From the 71 interfaces, you need to pick 18 of them to place ads.

So once again I get BIG numbers like 71*70*...*(70-18), but since it's having an ad at interfaces 2 and 5 is equilvalent to having an article between 5 and 2, I remove the choices of order.

So in this case, just selecting spots where to distribute 18 ads between 72 articles whose linear order has already been selected you have > 3,200,000,000,000,000 choices.

And that does not include anything about which ad goes where. There are 18 factorial choices there, and that's equivalent to making order important again. This takes the number of choices to 6x10^28.

The constraints are the only thing that keep this type of problem manageable.

EDIT; I ended up using this online tool for some of these later numbers. http://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html

B

11. Nov 27, 2013
Last edited: Nov 27, 2013

Joined:
Aug 5, 2013
#11
What?! I've produced over 700 of these publications in the last 5 years thanks. I don't need to know the maths to do my job.

Agreed obviously. My bone of contention is people think its simple to my job but when the constraints are physically impossible i.e. 2 adverts in the same spot then you end up rejigging the whole thing only to find another problem crops up and so on. Funeral parlours next to old people homes for example is not good hehe.

12. Nov 27, 2013
Last edited: Nov 27, 2013

### Staff Member

Joined:
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Location:
New England
#12
Of course not, but it helps you put your skill/art in context for those who don't have the same skill/experience and why you don't just leave it up to a roll of the die.

B

13. ### mobilehaathi macrumors G3

Joined:
Aug 19, 2008
Location:
The Anthropocene
#13
You said you were 'busy working out billions of combinations.' Hence my statement that you'd never finish your first project.

I suspect math would help you in general. Oh well. Have fun!

14. ### 835153 thread starter Guest

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Aug 5, 2013
#14

Thanks for the info guys, interesting reading - I was thinking it might be a few hundred thousand haha.

Er... that doesn't even make sense? I've already said I've completed 700 projects but I'm still figuring out 'billions of combinations' for more projects so don't have time to learn the math and also I don't need to.

15. ### mobilehaathi macrumors G3

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Aug 19, 2008
Location:
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#15
Ok buddy. Have fun and good luck.

### Staff Member

Joined:
Aug 16, 2005
Location:
New England
#16
mobilehaathi's point is that if you were doing it blind, and at random you would not have been able to finish your first one, much less 700.

I posted this as an edit to my last one, but move it here since the thread has moved on.

Here's a seemingly manageable example you can work out/demonstrate with a standard deck of cards.

Select out the face cards K-A of Hearts, then the numbered cards (2-10) of Clubs.

There are 3,628,800 possible orders for the Clubs alone, but let's ignore that.

You can spread the Hearts among the Clubs in 210 different ways (order being unimportant for both suits), If the order of the Hearts matters, that makes for 5040 different ways.

So if both the order of the Clubs and Hearts matters you have ~ 18,000,000,000 ways they could be ordered without constraints.

Thus, even for a reduced scale problem, the numbers are big.

B

17. ### 835153 thread starter Guest

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Aug 5, 2013
#17
Fair point to both of you but obviously I wouldn't be doing it blind and at random. The 'no constraints' was to make things easier to work out.

18. ### iMacFarlane macrumors 65816

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Apr 5, 2012
Location:
Adrift in a sea of possibilities
#18
This whole discussion seems a little off-kilter to me. What is the drive to figure out this rough astronomical number of possible permutations/combinations? To satisfy your curiosity? Your boss's curiousity? At the end of the day, knowing that there are quintillions of combinations doesn't really help you any more than guessing there were only trillions of combinations. Right?