# Permutation Calculation - help needed!

Rhymester
Science 22 Oct '12 08:16
1. Rhymester
and RedHotTed
22 Oct '12 08:16
Hi, I am selling a toy car system called Battle Deck on eBay.

Each car has three separate parts: A chassis, an engine and a bumper.

I want to include eight cars in the bundle and want to tell people how many possible different combinations are possible.

However, it isn't quite that simple... Each battle involves two cars so how many different one-on-one confrontations are possible and what's the formula?

Cheers

Andrew
2. 22 Oct '12 09:2712 edits
Originally posted by Rhymester
Hi, I am selling a toy car system called Battle Deck on eBay.

Each car has three separate parts: A chassis, an engine and a bumper.

I want to include eight cars in the bundle and want to tell people how many possible different combinations are possible.

However, it isn't quite that simple... Each battle involves two cars so how many different one-on-one confrontations are possible and what's the formula?

Cheers

Andrew
how many possible different combinations are possible

I find your post unclear and confusing; “combinations” of what exactly?
Exactly what category of attributes or things do you want permutations/combinations actually of?
Permutations/combinations of particular cars X Y or Z or what exactly? Give a couple of examples.
I will gladly be able to give you the answers for I enjoy puzzles like this and I am extremely good at solving this sort of problem! but first you must clearly define the complete problem!

Also, is this a 'permutation problem' or a 'combination problem'? -because it is important not to confuse the two!
See http://www.mathsisfun.com/combinatorics/combinations-permutations.html
In other words, does the order ( if you can have such a meaningful 'order' ) of the elements within each combination matter ( and 'matter' because you also want to count the number of possible orders within each combination and not just the number of combinations themselves ) or is any such order, even if you can have such a meaningful 'order', irrelevant?
Because if the order matters then it is a permutation problem else combination problem and the maths is different in each.
3. Rhymester
and RedHotTed
22 Oct '12 10:29
OK I'll try and explain...

A car is made up of a bumper, a chassis and an engine. The three pieces are stacked to create a car.

There are eight cars so 8 x bumpers, 8 x chassis, 8 x engines.
All are different.

A battle is between two cars.

So, if the cars were made up in their original form am I right in thinking there would be 64 different pairings?

However, each car can take one or more parts from the others so there are 24 car parts available.

Now how many pairings are possible?
4. sonhouse
Fast and Curious
22 Oct '12 11:032 edits
wouldn't it be 8 factorial, or 40,000 and change?
Maybe not, that would factor in 3 bumpers as one car and so forth.
5. Rhymester
and RedHotTed
22 Oct '12 11:15
Originally posted by sonhouse
wouldn't it be 8 factorial, or 40,000 and change?
Maybe not, that would factor in 3 bumpers as one car and so forth.
Yes each car can only have one of each part... Three engines would be good but not possible LOL
6. 22 Oct '12 13:4210 edits
Originally posted by Rhymester
OK I'll try and explain...

A car is made up of a bumper, a chassis and an engine. The three pieces are stacked to create a car.

There are eight cars so 8 x bumpers, 8 x chassis, 8 x engines.
All are different.

A battle is between two cars.

So, if the cars were made up in their original form am I right in thinking there would be 64 different pa ...[text shortened]... rts from the others so there are 24 car parts available.

Now how many pairings are possible?
I am afraid I find your post a bit unclear again but I think I can at least now guess what you might mean:
So, if the cars were made up in their original form am I right in thinking there would be 64 different pairings?

“pairings” between what exactly? are you referring to the number of possible pair options you can make out of pairing two of the eight cars i.e. how many different possible selections of two cars you can make out of eight cars?
If so, put simply and without getting any more deep into maths than we have to, if we label the eight cars: A, B, C, D, E, F, G, H,
then the list of all possible pairs you can select from those eight are:
AB AC AD … AH ( 7 pairs )
BC BD BE … BH ( 6 pairs )
CD CE CF … CH ( 5 pairs )

GH ( 1 pair )

so that is 7+6+5+4+3+2+1 possible pairs which is 28 pairs in total and not 64.

However, each car can take one or more parts from the others so there are 24 car parts available.

yes: if each car can take one or more parts from the others then, since there are 8 cars each of which have 3 parts, there must be 8*3=24 parts available of making each car.

Now how many pairings are possible?

are you now asking how many ways you can select 6 parts from those 24 parts to make two cars i.e. how many ways to make one pair of cars?
Working on the assumption that this is what you meant and I apologize in advance if this was not what you meant: I can see why you are stuck with the maths because that would require a pretty complicated formula! That's only in part because it is complicated by the fact that once you have made one car with 3 parts, you cannot use those same 3 parts to make another car to pair up with that first car so you have not 24 car parts available to make that second car but 24-3=21 parts available to make that second car. Also complicating this is the fact that there are three types of parts and each car must have and only have one of each type of part.
Fortunately I have found a way of calculating this without getting bogged down with the maths by avoiding proper formulas and by keeping it more conceptional than mathematical:

First I work out how many ways to make just one car from selecting 3 parts from the 24 car parts available and taking into account that there are 3 types of parts and 8 of each:

number of ways to make just one car = 8*8*8 i.e. 8^3 = 512.

now, if we have made just one car, we work out how many ways of making a second car to pair with that first car and take into account that we have now one less part available in each of the three categories of parts to make that second car i.e. we now have just 7 of each type of part available to make the second car because 3 parts have been used up making the first car:

number of ways to make a second car after making just one first car = 7*7*7 i.e. 7^3 = 343.

now, that means that for each of the 512 different ways we can make a first car, there are 343 ways of making the second car which give 512*343 = 175,616 ways of making a pair of cars where which car has which of the 3 parts ( out of the 6 parts used to make that pair of cars ) matters. BUT that is not quite what we want because it does NOT matter which of the 3 parts out of the 6 parts used to make that pair of cars makes the 'first' car because we are unconcerned the totally arbitrary distinction we made here between the 'first' car and the 'second' car ( unless you ARE concerned with which car is which within each pair of cars? are you? ) .
In other words, if the 'first' car has parts a, b and c and the 'second' car has parts d, e and f then that car pair is equivalent to the combination where the 'first' car has parts d, e and f and the 'second' car has parts a, b and c.
This means the actual answer is going to be less than 175,616 combinations.
Now to work out exactly how much less: this is simply a matter of logic! For every pair where there is a car with 'first' car having parts a, b and c and the 'second' car having parts d, e and f, there must be one and only one equivalent pair where the 'first' car has parts d, e and f and the 'second' car has parts a, b and c.
Therefore, half of those 175,616 combinations are going to be equivalent to the other half and thus the actual answer we want must be half of 175,616
i.e. 175,616/2 = 87,808 ( anyone: have I got that reasoning exactly right? )

so 87,808 is the answer ( I think ) and comes from: 8*8*8*7*7*7/2 which equals 87,808

Hope I got that right and also hope this is what you wanted!
7. Rhymester
and RedHotTed
22 Oct '12 14:49
Great stuff!

I think I will just say, "Over 80,000 different one-on-one battles are possible with this set!".

Thanks a lot!
8. Kewpie
since 1-Feb-07
23 Oct '12 02:271 edit
If you're challenged, you can just copy'n'paste that beautiful lucid explanation to justify your 80,000. Our friend humy should get a royalty for that, though. ðŸ™‚
9. Rhymester
and RedHotTed
23 Oct '12 13:31
Originally posted by Kewpie
If you're challenged, you can just copy'n'paste that beautiful lucid explanation to justify your 80,000. Our friend humy should get a royalty for that, though. ðŸ™‚
I don't think eBay listings can be that long LOL