Question

U have a colorless cube and 6 distinct paints. How many different ways can it be painted?

Answer 1

theres 6 faces . so 6 times 6 is 36 ?

Answer 2

Nope..Actually I've got the ans and it's matching but I'm not sure if I have used the right method..

Answer 3

Might be easier to think of each distinct paint as a colour from 0=>5. Now we have to find every single combination of those from
[0 0 0 0 0 0] to [5 5 5 5 5 5] with all the ones in between.

Answer 4

That can be done using basic permutations...but the problem is, there is a high level of symmetry for a cube so many of the counted combinations will actually produce an identical cube...I had to use a bit of a rigourous method...want to know if there is any easier way out..

Answer 5

@hea

Answer 6

Yeah that is a very good point, give me a minute

Answer 7

Sure!

Answer 8

I think this has something to do with Burnside's lemma?

Answer 9

I guess it can approached by that but I'm not in a course that has taught Burnside's lemma...Don't have much knowledge about it..

Answer 10

I wouldn't worry too much about that, that's what lemma's are for. What course is it in?

\[ \frac{1}{24}(n^6 + 3n^4 + 12n^3 + 8n^2) \]in our case n = 6 therefore we have 2226 istinct solutions

Answer 11

There might be a solution slightly more course specific of course

Answer 12

I'm preparing for college entrances for undergrad programme...
But the answer given is 30 :O Are u sure the lemma is applicable here? No deviations?

Answer 13

Thanks anyways btw! :)

Answer 14

haha, no probs, sorry I couldn't help more

Answer 15

Birkhoff might know

Answer 16

I just read an article for class that had to do with a determinant formula for counting the number of ways of coloring a map with a given set of colors

Answer 17

it goes a little something like this:

if given a set of colors \(\lambda\), (\(\lambda=1,2,3,4,...\)), and a map of \(n\) regions; let \(m_i\), (i=1,2,3,...,n), be the number of ways of coloring the map with exactly \(i\) colors disregarding permutations. Then \(m_i\cdot\lambda\cdot\lambda(\lambda-1)\cdot\lambda(\lambda-1)(\lambda-2)\cdot\cdot\cdot\lambda(\lambda-1)...(\lambda-i+1)\) is the number of ways to color the map in \(i\) colors allowing for permutations

Answer 18

this then creates a polynomial:\(P(\lambda)=m_1\lambda+m_2\lambda(\lambda+1)+\cdot\cdot\cdot+m_n\lambda(\lambda-1)...(\lambda-n+1)\) for the number of ways of coloring the map

Answer 19

spose to be a l-1 in that second term, not a l+1 :)

Answer 20

Hmmm...Ok let's see if i can somehow try and use that for the cube..

Answer 21

http://www.jstor.org/stable/1967597

im sure the original article does a better job at explaining it than I can :)

Answer 22

Haha :) U have any links?

Answer 23

Oh sorry just saw the link!

Answer 24

have fun ;)