|

Algebra

Cool goIITian

 Joined: 9 Oct 2011 Post: 92
17 Jan 2012 19:08:36 IST
5 People liked this
25
1569
how many ways can we colour a cube with six different colours
Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Algebra

how many ways can we colour a cube with six different colours

• 1
• 2
• GO
• Go to Page...

Blazing goIITian

Joined: 25 Jul 2011
Posts: 322
17 Jan 2012 19:12:29 IST
0 people liked this

36...Right?

Forum Expert
Joined: 1 Jun 2009
Posts: 1330
18 Jan 2012 16:11:08 IST
2 people liked this

let the colors be A,B,C,D,E and F

now I did fix A opp to F

I am left with 4 colors and 4 places (on a round table as seen from top A)

now to sit 4 people on a round table  we have 3! ways if the clockwise and anticlockwise is diff. but in our case they are same so 3! / 2 ways.............

and now this can be done 5 times with B, C, D, E, and F as opp. to A..............

all the other opp are taken care in round table!!!!!!

so ans is 5* 3! / 2 = 15!!!( if clockwise and anticlockwise is taken same) and it'll be 30 if taken diff.

New kid on the Block

Joined: 18 Jan 2012
Posts: 1
18 Jan 2012 16:24:50 IST
0 people liked this

i think it is 720

Hot goIITian

Joined: 18 Jan 2012
Posts: 147
18 Jan 2012 16:42:06 IST
3 people liked this

Hey Abhishek, if you know the answer by any chance; is it 30?

If that's right, i can tell you the logic...

First fix 2 colours opposite to each other say C1 and C2.

Therefore the remaining 4 colours are arranged as if around a round table in (4-1)! ways=3! ways=6 ways

6 ways for C1,C2

Similarly 6 for C1-C3,C1-C4,C1-C5,C1-C6.

Total ways=6*5=30.

Blazing goIITian

Joined: 25 Jul 2011
Posts: 322
18 Jan 2012 18:44:46 IST
0 people liked this

Sorry the ans is 720!!!!!!6Ci x 5Ci x4Ci x 3Ci x 2Ci(i=1)

Blazing goIITian

Joined: 27 Dec 2010
Posts: 1508
18 Jan 2012 19:20:04 IST
0 people liked this

well i agree with Arjun Virmani..

and @sathyaram : it is not 720!!!!

Cool goIITian

Joined: 7 Jul 2011
Posts: 59
19 Jan 2012 04:27:15 IST
0 people liked this

i think  your quest. is incomplete you should give some condition like we have to colour each face with only one colour or something because in your quest there is no restriction in colouring hence there are infinite ways

if your quest. is to colour each  face with only  one colour and we can select each colour many times then there is total

6^6 ways

if your quest is to colour each face with only one colour but each colour can be selected once then there is total

6! ways ,gautham is correct

if faces are identical(in same case above)

only 1 way

Blazing goIITian

Joined: 27 Dec 2010
Posts: 1508
19 Jan 2012 09:14:26 IST
0 people liked this

well i would like to make it clear.

even if the colours are repeated, we don't get 6^6 as the solution.

visit this....

http://ken.duisenberg.com/potw/archive/arch99/991203sol.html

New kid on the Block

Joined: 20 Jan 2012
Posts: 6
20 Jan 2012 06:33:01 IST
0 people liked this

it's very simple. there are 6 faces so those 6 can be coloured in 6! means 6 factorial waya i.e 720 ways

New kid on the Block

Joined: 31 Dec 2011
Posts: 8
20 Jan 2012 18:36:33 IST
0 people liked this

there are 6 different faces and if each face is to be coloured with different colour i.e. the colours are not to be repeated, then it will be 6!=6*5*4*3*2*1

Hot goIITian

Joined: 18 Jan 2012
Posts: 147
20 Jan 2012 19:34:21 IST
0 people liked this

It cannot be 6 factorial.

Because rotating a cube does NOT give a different arrangement.

Also if we change the face we are viewing we do NOT get a different arrangement.

So 6 factorial fives us some cases repeated.

Hot goIITian

Joined: 18 Jan 2012
Posts: 147
20 Jan 2012 19:36:32 IST
0 people liked this

The second last line-

6 factorial gives us some cases repeated...

Cool goIITian

Joined: 7 Jul 2011
Posts: 59
20 Jan 2012 22:50:27 IST
0 people liked this

HEMANG    I THINK YOU DONT HAVE HEARD ABOUT COIN PROBLEMS

ONE COIN HAS 2 OUTCOME

WHEN THERE ARE N COINS THEN NO OF OUTCOMES IS 2^N

DICE PROBLEM, ONE DICE HAVE 6 OUTCOME

WHEN THERE  ARE N DICE

THEN TOTAL NO OF OUTCOMES  6^N

SIMILARLY  1 FACE CAN BE COLOURED IN 6 WAYS

SO  SIX FACE CAN BE COLOURED IN  6^6 WAYS

AND IF YOU KNOW THESE  THINGS ,SO I THINK YOU ARE CONFUSED BECAUSE

I HAD CONSIDERED THE CASE WHEN ALL FACES ARE NOT  IDENTICAL.

IF I AM RIGHT  THEN OTHERS PLEASE COMMENT ALSO ON THIS.

Blazing goIITian

Joined: 27 Dec 2010
Posts: 1508
23 Jan 2012 10:03:09 IST
0 people liked this

:)

i am not confused. you can read my earlier post again.

i have provided a link. that guy had runned a program. that program showed that outcome..

and you certainly got it wrong..

i am giving two solutions to the problem now..

Any painting of the cube can be oriented so that a particular color, say red, is on the bottom, so we can assume that we start with a cube that has five blank faces with the bottom painted red. Pick one of the 5 colors remaining to paint the top (5 options), and arrange the 4 remaining colors in a circle and paint the sides accordingly – there are 3! ways to do this. Thus there are a total of 5*3! = 30 ways to paint the cube.

If we fix the orientation of the cube and number its faces, we will get a permutation of the six colors for any particular painting. There are 6! permutations of six colors. We want to regard as equivalent permutations which correspond to different orientations of the same cube. A cube can be uniquely oriented by picking one of the 6 faces to be on the bottom, and then rotating a particular one of the 4 side faces to be in front, thus there are 24 distinct orientations and our equivalence classes all have 24 elements.

there are 6! / 24 = 5*3! = 30 distinct ways to paint the cube

Blazing goIITian

Joined: 27 Dec 2010
Posts: 1508
23 Jan 2012 10:11:31 IST
1 people liked this

and yeah abhishek, i don't really like your tone man..

try to be calm when you write your posts

. you can read my earlier posts on this website. i am not a loser. and if you think that i have not heard of coin problems then it is really amusing..

In general, the number of rotationally distinct colorings of the faces of a cube in n colors is given by

$\frac{1}{24}\left(n^6+3n^4 + 12n^3 + 8n^2\right).$

this formula is derived from burnside's lemma.

Blazing goIITian

Joined: 27 Dec 2010
Posts: 1508
23 Jan 2012 10:22:52 IST
0 people liked this

Let us colour a cube, Red, Orange, Yellow, Green, Cyan and Blue..
We can describe the cube by listing first the base colour, then the top, followed by the colour facing us, the left face, the back face and finally the right.

So a cube with Red on the bottom Orange on top and Yellow, Green, Cyan and blue clockwise round the sides would be written: ROYGCB

The possible colour patterns are

ROYGCB, ROYCGB, ROYCBG, ROYBCG, ROYBGC, ROYGBC
RYOGCB, RYOCGB, RYOCBG, RYOBCG, RYOBGC, RYOGBC
RGOYCB, RGOCYB, RGOCBY, RGOBCY, RGOBYC, RGOYBC
RCOGYB, RCOYGB, RCOYBG, RCOBYG, RCOBGY, RCOGBY
RBOGCY, RBOCGY, RBOCYG, RBOYCG, RBOYGC, RBOGYC

Blazing goIITian

Joined: 27 Dec 2010
Posts: 1508
23 Jan 2012 10:25:14 IST
0 people liked this

http://en.wikipedia.org/wiki/Polya_enumeration_theorem

this is best known to solve such problems...

Blazing goIITian

Joined: 27 Dec 2010
Posts: 1508
23 Jan 2012 11:07:30 IST
0 people liked this

8 colourings with 2 colors
30 colourings with 3 colors
56 colourings with 4 colors
45 colourings with 5 colors
30 colourings with 6 colors

this above thing is when we have no repition of colours..

moreover i checked with the fomula i had written above,...

and guess what the answer is not 6^6!!!!!!!!!!!!!!!!!!

Blazing goIITian

Joined: 27 Dec 2010
Posts: 1508
23 Jan 2012 13:30:43 IST
0 people liked this

.

this solution is provided by a student on another forum..

am assuming the colorings must be rotationally distinct. (If not, it's $6!$ and $6^6$.)

1. There are $6!$ ways to paint the sides, but we divide by $24$ (the number of rotational symmetries) to get $30$.

2. We apply Burnside's Lemma. Let me generalize to $n$ colors. There are $n^6$ colorings of the cube, ignoring rotational distinctness. Of the rotations of the cube:

There is $1$ identity, which leaves $n^6$ rotationally distinct colorings.
There are $6$ rotations (face rotations) by $\pm90^\circ$ about the axes connecting midpoints of opposite faces ($1$ per each of the $2$ directions of rotation per each of the $3$ axes), each of which leaves $n^3$ rotationally distinct colorings.
There are $3$ face rotations by $180^\circ$ ($1$ per axis), each of which leaves $n^4$ rotationally distinct colorings.
There are $8$ rotations (vertex rotations) by $\pm120^\circ$ about the axes connecting opposite vertices ($1$ per each of the $2$ directions of rotation per each of the $4$ axes), each of which leaves $n^2$ rotationally distinct colorings.
There are $6$ rotations (edge rotations) by $180^\circ$ about the axes connecting midpoints of opposite edges ($1$ per each of the $6$ axes), each of which leaves $n^3$ rotationally distinct colorings.

So the polynomial yielding the number of colorings is $\frac{1}{24}(n^6+6n^3+3n^4+8n^2+6n^3)=\frac{1}{24}(n^6+3n^4+12n^3+8n^2)$.

Substituting $n=6$ yields $2226$.

 Some HTML allowed. Keep your comments above the belt or risk having them deleted. Signup for a avatar to have your pictures show up by your comment If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team

## For Quick Info

Name

Mobile

E-mail

City

Class

Vertical Limit

Top Contributors
All Time This Month Last Week
1. Bipin Dubey
 Altitude - 16545 m Post - 7958
2. Himanshu
 Altitude - 10925 m Post - 3836
3. Hari Shankar
 Altitude - 9960 m Post - 2185
4. edison
 Altitude - 10815 m Post - 7797
5. Sagar Saxena
 Altitude - 8625 m Post - 8064
 Altitude - 6330 m Post - 1979

Physics

Topics

Mathematics

Chemistry

Biology

Institutes

Parents Corner

Board

Fun Zone