Bracelets (1 Viewer)

[glittergal96]

How many distinct bracelets can you make out of a string and beads of colour k for k=1,2,...,m?

 

[FrankXie]

are beads of same colour identical ?

 

[InteGrand]

are beads of same colour identical ?

I think m different colours: colour 1, colour 2, colour 3, ..., colour m.

And the number of beads of colour k is .

And every bead is identical in shape/size etc.

I think this problem was solved by Pólya?

 

[GoldyOrNugget]

https://en.wikipedia.org/wiki/Pólya_enumeration_theorem

 

[glittergal96]

I think m different colours: colour 1, colour 2, colour 3, ..., colour m.

And the number of beads of colour k is .

And every bead is identical in shape/size etc.

I think this problem was solved by Pólya?

I am not sure if this particular question was solved by Pólya, but he did a lot of stuff in this area. I just made this one up, but it has almost certainly been asked and answered before.

https://en.wikipedia.org/wiki/Pólya_enumeration_theorem

This isn't quite the same question, the theorem you link provides a formula for the number of n-bead bracelets from m colours, assuming we have plenty of each.

My question has a finite number of beads of each colour, and asks you to count all bracelets you can make using all of these beads. This is less than the total number of m coloured bracelets with n beads.

That's the right way to think about the problem though (counting orbits of group actions), I used Burnside's lemma which is just a stones throw away from the unweighted version of your linked theorem.

By the way, the resulting answer will not be in closed form, there will be summations involved.

 

[RealiseNothing]

The coincidence that I saw something very similar to this (Pólya stuff) in a combinatorics book yesterday at Kinokuniya.