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.
i think it is 720
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.
well i agree with Arjun Virmani..
the answer is 30.
and @sathyaram : it is not 720!!!!
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
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
well i would like to make it clear.
even if the colours are repeated, we don't get 6^6 as the solution.
it's very simple. there are 6 faces so those 6 can be coloured in 6! means 6 factorial waya i.e 720 ways
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.
The answer is 30.
The second last line-
6 factorial gives us some cases repeated...
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.
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
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
this formula is derived from burnside's lemma.
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
this is best known to solve such problems...
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!!!!!!!!!!!!!!!!!!
this solution is provided by a student on another forum..
am assuming the colorings must be rotationally distinct. (If not, it's and .)
1. There are ways to paint the sides, but we divide by (the number of rotational symmetries) to get .
2. We apply Burnside's Lemma. Let me generalize to colors. There are colorings of the cube, ignoring rotational distinctness. Of the rotations of the cube:
There is identity, which leaves rotationally distinct colorings.
There are rotations (face rotations) by about the axes connecting midpoints of opposite faces ( per each of the directions of rotation per each of the axes), each of which leaves rotationally distinct colorings.
There are face rotations by ( per axis), each of which leaves rotationally distinct colorings.
There are rotations (vertex rotations) by about the axes connecting opposite vertices ( per each of the directions of rotation per each of the axes), each of which leaves rotationally distinct colorings.
There are rotations (edge rotations) by about the axes connecting midpoints of opposite edges ( per each of the axes), each of which leaves rotationally distinct colorings.
So the polynomial yielding the number of colorings is .
Substituting yields .
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