PMP plz answer...plz...?

This problem interested me a lot and I couldn't figure it out so I went to the Dr. Math website and asked them, here's the answer:

You consider the 24 symmetries of a cube and sum all those symmetries that keep colours fixed. The number of non-equivalent configurations is then the total sum divided by the order of the group (24 in this case).

We first find the cycle index of the group of FACE permutations induced by the rotational symmetries of the cube.

Looking down on the cube, label the top face 1, the bottom face 2 and the side faces 3, 4, 5, 6 (clockwise)

You should hold a cube and follow the way the cycle index is

calculated as described below. The notation (1)(23)(456) = (x1)(x2)(x3) means that we have a permutation of 3 disjoint cycles in which face 1 remains fixed, face 2 moves to face 3 and 3 moves to face 2, face 4 moves to 5, 5 moves to 6 and 6 moves to 4. (This is is not a possible permutation for our cube, it is just to illustrate the notation.) We now calculate the cycle index.

(1)e = (1)(2)(3)(4)(5)(6); index = (x1)^6

(2)3 permutations like (1)(2)(35)(46); index 3(x1)^2.(x2)^2

(3)3 permutations like (1)(2)(3456); index 3(x1)^2.(x4)

(4)3 further as above but counterclockwise; index 3(x1)^2.(x4)

(5)6 permutations like (15)(23)(46); index 6(x2)^3

(6)4 permutations like (154)(236); net index 4(x3)^2

(7)4 further as above but counterclockwise; net index 4(x3)^2

Then the cycle index is

P[x1,x2,.x6] =(1/24)[x1^6 + 3x1^2.x2^2 + 6x2^3 + 6x1^2.x4 + 8x3^2]

and the pattern inventory for these configurations is given by the above generating function:

With up to 6 colours available we replace x1, x2, ... etc with 6. (This means we could have only one colour or only 2 colours or 3 or 4 or 5 or 6. It DOES NOT mean that we must have 6 colours for each

configuration).

= (1/24)[6^6 + 3(6^2).(6^2) + 6(6^3) + 6(6^2).(6) + 8(6^2)]

= (1/24)[46656 + 3888 + 1296 + 1296 + 288]

= 53424/24

= 2226 <-----

So there are 2226 non-equivalent configurations using up to 6 colours.

If we wanted the number of non-equivalent configurations for EXACTLY 6 colours on each cube then we proceed as follows

= (1/24)[(a+b+c+d+e+f)^6 + 3(a+b+..+f)^2.(a^2+b^2+..+f^2)^2 + 6(a^2+b^2+..+f^2)^3 + 6(a+b+..+f)^2.(a^4+b^4+..+f^4) + 8(a^3+b^3+..+f^3)^2]

and we require the coefficient of abcdef in this expansion.

Using Maple we get the term (1/24)[720abcdef]

And so the number of non-equivalent configurations using exactly 6 colours is

(1/24) x 720 = 30

Since we only wanted power 1 for each of the 6 letters we could have got this result more quickly by simply expanding (a+b+c+d+e+f)^6.

- Doctor Anthony, The Math Forum

i dont understand your question. u have 6 colors or 3?

if its 3, and paint twice of each color, u have

1 way for the doubles are opposite each other

1 way for all 3 doubles touch each other

3 ways for 1 opposites and 2 touching doubles

that give u 5

if u have to use 1/2/3 colors

3 of 1 color, say it's (6 0 0) and (0 6 0) and (0 0 6)

(1 5 0)=> 3 ways, 3C2

(2 4 0)=> u can hv 1 doubles opp. or 1 touching doubles

=2*3C2=6

(3 3 0)=> all triples touch each other(vertexing) or 3 continuous triples

=2*3C2=6

(1 1 4)=> 2 singles touch(1)+opp.(1)

=2*3C1=6

(1 2 3)=> triples made vertex(1)+continuous/triples have a middle(2)=3*3!=18

(2 2 2)=> 5

3+3*6+18+5=44

36

The correct answer to your question is thirty-six.