Question

Need help with a tricky question!

Answer 1

I think, it will depend on adjasent edges of each corner. So it will be a combination of the collors at each corner. Just have to think of the formula to calculate it, because to different corners share one color.

Answer 2

first, corners are 4. Choosing one collor that will be shared between first pair of corners, collors available for first corner 6-1. So first corner (6-1)C2. For the other corner (one that shares the color with 1º) (6-3)C2. And the rest is no choice.

Answer 3

So my guess would be: 5C2+3C2

Answer 4

I did not understand. Can you explain it again?

Answer 5

I am also not 100% sure. Just thinking, :). Here it goes again.
First notice that at each corner there are 3 adjasent colors. The order is not important, because by rotation you could get any.
Second notice that each 2 corners share one color.
1º Now lets start by choosing one color for one edge.
2º Calculate combinations of collors posible for one of the corners that have this adjasent edge. (6-1)C2.
3º Calculate combinations of collots left for the other corner that shares the same color edge (the one that we chose in step 1º). (6-3)C2
4º Notice that the other 2 corners left share one edge with the corners that we already colored. So that means that the only not collored adge left is the one that this 2 left edges share. And we have one color left for that. So this doesn't add any more combinations.
I think it is more or less logical, :)

Answer 6

I think I understand what you mean! :D
What are they asking here, 'Can you give a second argument?' ?

Answer 7

i think 6!
if i wanna deal with probability :)

Answer 8

wow that's quite the riddle, I don't know if i would be able to do this one off the top of my head...
hmm well it looks like we can make each length a single color and try a bunch of different combinations but if we are able to rotate it and match another color combination then that would mean those two combinations are really the same and its equilateral so there's going to be a bunch of repeated combinations when we rotate it so we need to find an equation that will give us the total possible combinations

Answer 9

Yes! But how? :D
I think what Myko said up there is logically correct. It makes sense, hopefully the answer would be right. :)

Answer 10

Yeah he is right for sure...

Answer 11

ok if we go with the edge approach then we have 3 total edges we'll call them e1 e2 and e3
each edge consists of 3 different colors so if we have n as the number of colors to chose from..