so the dice has the numbers 1,2,3,4,5 and 6
now we have to colour each number in a different colour (from 6 available colours)
so you can paint the first number in 6 ways
you can paint the second numbber in 5 ways
can choose 4 colours to paint the 3rd number
from this the total number of possibilities is 6x5x4x3x2x1 = 6! =120
another explanation is that since this is an ordered selection : you are arranging 6 different numbers to 6 different colours the total number of arrangements possible is 6P6 = 6! = 120
Its asking for unique cubes, so lets say we have our cube where red is top and white is bottom.
Another arrangement (according to your solution) is white on top red on bottom, but they are the same cube.
First lets arrange the top and bottom faces of the cube.
We pick 2 colours from the 6, so that is
ways to do so.
we still need to colour in the sides, there are 4 colours remaining. Now for these 4 colours, a, b, c, d.
It is similar to arranging around a round table, the total arrangements is (4-1)! = 3! = 6
For example, looking at the cube from the top (the net)
The four sides could be:
...a
b.....c
...d
...a
c.....b
...d
Can you see how these 2 arrangements are the same if we put it in a cube?
I have no elegant way of doing this, but first draw out the 6 cases that we can have for the sides. Then eliminate all that is not unique.
You will have 2 remaining.
From the beginning, 6C2 * 2 = 30
