Math Is Fun Forum

wintersolstice

Real Member

Registered: 2009-06-06

Posts: 128

combination misconception

A common misconception (I think) is that if you've got a set of permuatuations (which are open to variation that could mean that they're not all distinct) the simple way to find out how many there are (ignoring what they are) is to take the number of permuations P (where variation are classed as different) and divide by the number of variations V :these could be "rotations and reflections" (for shapes) cyclic combinations of colours (for couloured patturns etc)

but this doesn't always work Proof:

Lets say that there are P permuations and V variations then if you take "P/V" distinct possibilities and then write them down in a list

1a,1b,1c ... 1V
2a,2b,2c ... 2V

            .
            .
            .

Ra,Rb,Rc, ... RV (where R=P/V)

then if all these possibilities are different! then the number of possibilties is R X V (=P)

but if some have variations which are the same (such as them having isometry themselves which matches the isometry of the shape for e.g.) then some permuations will be mapped on to themselves (instead of another permuation) this means that this list (containing P possibilties in total) has duplicates, once filtered there are less than P (and that must mean there's more distinct possibilties to find)

Conclusion:

The number of dinstinct possibilities can end up being more than "P/V"

this can apply to a lot of things:

Polyforms
Coloured grids
and much more:D

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Why did the chicken cross the Mobius Band?
To get to the other ...um...!!!

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: combination misconception

Maybe Burnside's Lemma will help for some of these examples.
He and some others did this in the 1800's.

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