Suppose you take a cube and color all the sides. Each side is colored either red or blue. If two different methods of coloring the sides can be rotated to match, they are considered the same.
In how many different ways can you color the cube?
Be systematic, and don't forget that all-red and all-blue are both coloring options.
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And there are plenty of folk out there who can juggle numbers, angles and graphs from here to kingdom come, but get all flustered when presented with a visual problem (I should introduce you to my son) - oh yes - as Jason has said.
Given only 53% of people got it correct as of this writing (in the early phases, where the more dedicated solvers tend to come out) I think a high cactus rating is appropriate (possibly 4, but it's impossible to know for certain until we try a problem - in some cases, we've ported problems from other parts of the site that had much different success rate in the other location).
A lot of people miscount this kind of puzzle - I think visualizing that you're doing is non-trivial for a fair chunk of our solvers.
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My initial thoughts on the problem were a little...harsh. I was just excited to see the new problem so I stayed up until midnight to see it posted. I saw 5/5 cacti and got excited for a challenge....then solved the problem in a minute. Additionally, I also enjoy looking at other solutions and seeing who came up with a creative/unique way of solving it. I gain a lot of insight from others' solutions. And this problem doesn't lend itself to that, everyone (so far) has just listed the 10 possibilities. That's why I mentioned Burnside's Counting Lemma, so that interested readers may be pointed in the direction of more general results. Still 59% success rate...I anticipated much higher for this problem.
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Down to 51%. Feel free to edit to expand on Burnside, if you want.
the cacti also denotes tediousness lol so i guess for a problem that requires listing all possibilities it's slightly tedious? definitely not 5 cacti tho
And then there are idiots like me - who did get it right, but only by looking at the options for answers, which pushed me into realising there could be all-red or all-blue
I can only praise people who understood correctly what is given in this problem. Sorry, but I did not, until I look at discussion of the solution.
@John Allums totally agree. I counted 1,2,3,4,5,6 red faces before realising there were different arrangements, but after that it was easy to find the answer :) I actually didn't want to click it because I thought it wasn't that easy and 5/5 cacti... I guess it was. :P
It should be immediately clear that you really only need to solve for four cases, since the solution for 0 red, 1 red and 2 red is the same as the solution for 0 blue, 1 blue and 2 blue.
So the solutions are as follows:
0 red (6 blue): 1 solution
1 red (5 blue): 1 solution
2 red (4 blue): 2 solutions (adjacent & non-adjacent)
3 red (3 blue): 2 solutions (touching-a-common-corner & in-a-row)
4 red (2 blue): 2 solutions (same as 2 red, except now blue are adjacent & non-adjacent)
5 red (1 blue): 1 solution (same as 1 red)
6 red (0 blue): 1 solution (same as 0 red)
I think these images cover the 10 general cases that I (and everyone else here) found. Please let me know if I made any errors!
The number of rotationally distinct colorings of the faces of a cube in n colors is given by
(n^6 + 3*n^4 + 12n^3 + 8n^2)/24
Substituting n = 2 yields: (64 + 3 16 + 12 8 + 8*4)/24 = (64 + 48 + 96 + 32)/24 = 240/24 = 10
The total is 10 ways.
Take a red colour as 1 and blue colour as zero
So to colour six phases, let the pattern be something like - RRRBBB
R- RED B-BLUE
So seven combinations to colour six sides of cube are-
111111-in this only one way to colour
111110-in this only one way to colour
111100-in this two ways to colour--opposite sides blue coloured and adjacent sides blue coloured
111000-in this two ways to colour--3 adjacent sides in plane and 2 in plane and one top or bottom
110000-in this two ways to colour--opposite sides red coloured and adjacent sides red coloured
100000-in this only one way to colour
000000-in this only one way to colour
It can be thought of as 6C2, analogous to how many ways are there to picking a group of 2 people from 6. Chris
Can you show how they are analogous?
i think the correct answer wd be 9. ... if each side must be coloured (which is clearly mentioned in the qs).
I only counted the scenarios from red's perspective and just inverted it where appropriate.
This is the way I thought about it:
All Red inv. All Blue
1 Red inv. 1 Blue
2 adjacent Red inv. 2 adjacent Blue
2 opposite Red inv. 2 opposite Blue
3 Red sharing vertex ( No Inversion)
3 Red in a line ( No Inversion)
Hence, if you count it all up, then there are 10 unique ways.
I have mentally counted just nine colorings.
The smallest answer is ten, greater than nine.
So I have picked ten as the closest.
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This should not be 5/5 cacti. Way too easy. Just literally count all the possibilities.
0 blue faces
1 blue face
2 adjacent blue faces
2 opposite blue faces
3 blue faces around a vertex
3 blue faces in a line
4 blue faces with 2 adjacent red faces
4 blue, 2 opposite red faces
5 blue faces
6 blue faces
The more formal solution would use Burnside's Counting Lemma , but this was so easy, just literally counting them the fastest thing to do.