solve this problem

[rajbahadur tomar]

In how many ways a cube whose all faces are identical can be painted using 6 different colors?
plz dont say 1,because in every book this is the given answer which is
not true.

[deepu]

Hi Raj,my answer is 90.
Reasoning - just consider the 4 sides of cube leaving upper and bottom sides.Now since the no of combinations of 4 colours out of 6 colours is 6c 4 i.e 15 and once u select 4 colours the no of ways they can be arranged ina circular order is (4-1)! i.e 6.therefore total no of ways of colouring 4 sides is 90.in each case 2 colours will be left out which can be coloured on upper and bottom sides of the dice and order doesnt matter.So total no of ways of colouring the dice in 6 diff colours is 90.
Plz correct me if i m wrong guyz.

[DINESH LALCHANDANI]

hello raj according to me the answer is
1 because M L KHANNA and RD SHARMA never be wrong

[keviv]

Originally Posted by deepu

Hi Raj,my answer is 90.
Reasoning - just consider the 4 sides of cube leaving upper and bottom sides.

The cube is identical. So no question of upper and bottom sides!

So I guess your answer is wrong, Deepu.

But the correct answer cannot not be 1. Because a colouring scheme in which colours 'a' and 'b' are on opposite faces is really different from the one in which they are on the adjacent faces.

My proposed answer: 5

Explanation:

I paint ay one face with one of the colours, say 'a'. On its opposite face I place another random colour, say 'b'. Remaining 4 colours can be arranged on rest of the faces in any way. This gives to one arrangement ( interchanging a and b or rearranging colours on the faces gives the same arrangement every time)

There can be 5 different colours opposite to 'a' and each of them will give only one unique arrangement.

So the answer might be 5.

Bye.

[deepu]

Yup i forgot the rule that all sides are same.
Seems Keviv's answer is right.His reasoning is quite correct.

[sachin_chatekar]

Originally Posted by deepu

Yup i forgot the rule that all sides are same.
Seems Keviv's answer is right.His reasoning is quite correct.

What about the following combinations?

A B C D E F

A X 1 2 3 4 5

B 2 X 6 7 8 9

C 3 6 X 10 11 12

D 4 7 10 X 13 14

E 5 8 11 13 X 15

F 6 9 12 14 15 X


X - impossible combination.
All similar sombinations are represented by same number.

So I think, in all 15 combinations are possible, if we think that way.

But ultimately the correct answer is 1.
simply because we have 6 different colours and 6 differnet facesw for them, so 6C6 gives us 1


Regards

[Mr.Incredible]

Lets consider the six colours as 1,2,3,4,5,6 painted on the respective faces of the cube.
As the cube is same changing adjacent colors would not make any difference but changing the color of opposite face certainly do.
Say the colors on opposite faces of the cube are :
1-3
2-4
5-6

Therefore the possible combinations on opposite faces taking each phase at a time are :
[P.S : (X) denotes a repeat combination and does not count.]

1- 2,4,5,6 = 4
2- 1(X),3(X),5,6 = 2
3- 2(X),4(X),5,6 = 2
4- 1(X),3(X),5,6 = 2
5- 1(X),2(X),3(X),4(X) = 0
6- 1(X),2(X),3(X),4(X) = 0

Total = 4+2+2+2+1(for initial combination)
= 11

Ans = 11 possible combinations

Guys plz let me know if the answer is correct ?????

[capricious]

But the correct answer cannot not be 1. Because a colouring scheme in which colours 'a' and 'b' are on opposite faces is really different from the one in which they are on the adjacent faces.

My proposed answer: 5

My age-long faith in ML Khanna is shred to pieces.
Keviv's answer is correct. There is no ambiguity left.

[keviv]

Originally Posted by keviv

I paint ay one face with one of the colours, say 'a'. On its opposite face I place another random colour, say 'b'. Remaining 4 colours can be arranged on rest of the faces in any way. This gives to one arrangement ( interchanging a and b or rearranging colours on the faces gives the same arrangement every time)

I was wrong.

I said that rearranging the colours on faces will not make any difference. But this is incorrect. here again the adjacent/ opposite thing will have to be considered.

So for a particular choice of colours 'a' and 'b' on opposite faces , the colours on remaining 4 faces can be arranged in 3 ways. Say the colurs are c, d, e, f. in one c,d will be opposite, in the second, c,e and in the third, c,f. This will also take care of the arrangement on the remaining pair of opposite faces.

So total no. of ways = 5 * 3 =15.

thus, i second Sachin's answer.

Correct me if I am still wrong.

Bye.

[deepu]

Originally Posted by sachin_chatekar

But ultimately the correct answer is 1.
simply because we have 6 different colours and 6 differnet facesw for them, so 6C6 gives us 1


Regards

sachin that 6c6 is true in normal circumstances but in the case of dice i dont think the formula works.not sure but correct me if i m wrong.

and yes keviv finally 15 seems more promising.