A lettered cube has 6 sides with a different letter on each side. Each letter can be shown in 4 different positions. How many different ways?

answers:
atsuo say: First , assume that each letter must be shown on the center of its side ,
and the direction of a letter is disregarded .
Let the 6 letter be A,B,C,D,E and F .
Let type1 cube be the cube such that A and B are shown on opposite sides ,
and put type1 cube so that it becomes top side is A and bottom side is B .
4 letters C,D,E,F can be shown in other 4 sides with any combination ,
but we can rotate the cube around the vertical axis so 4 cubes become the same cube .
Therefore 4! / 4 = 6 different type1 cubes exist .
Let type2 cube be the cube such that A and B are shown on adjacent sides ,
and put type2 cube so that it becomes top side is A and front side is B .
4 letters C,D,E,F can be shown in other 4 sides with any combination ,
and we can not rotate the cube because it must be that top side is A and front side is B .
Therefore 4! = 24 different type2 cubes exist .
So 6(type1) + 24(type2) = 30 different cubes exist .
Next , think about "Each letter can be shown in 4 different positions" .
We can move each letter from the center to 1 of 4 different positions ,
so 1 cube can be modified into 4^6 = 4096 cubes .
Therefore total of 30 * 4096 = 122,880 cubes (ways) exist .
If "4 different positions" means "4 different directions" then the result is unchanged
because we do not move each letter but we can rotate each letter .
(I think the direction of a letter must be parallel to an edge . If the direction of a letter
may not be parallel to an edge then the problem becomes complicated .)

Nick say: Assuming you have 6 different letters each with no rotational symmetry e.g. A,B,C,D,E,F.
If this is a question about a *fixed cube* then each face is *distinct* to begin with and there are 6! ways to place letters on faces and 4^6 ways for them to be oriented: 6!*4^6 ways.
or
If this is a question about a *free cube* (i.e. equivalent under the symmetry group of rotations) then place A on a face to define it as the bottom face, let the orientation of this letter fix front, back, left and right faces, this then means that we may place the remaining 5 letters on 5 remaining *fixed* faces in 5! ways and in 4^5 orientations: 5!*4^5 ways.
Notice the second count is 1/24 of the first. This is because the assumed equivalence under rotation reduces the possible configurations.

Φ² = Φ+1 say: Can you please confirm:
1) How many letters there are from which the six are chosen?
2) Can you please complete the question "How many different ways?"? I suspect the two good answers you already have have been "thumbed down" because they (and I) are having to make assumptions which may be incorrect.

Puzzling say: This is known as the fundamental (or basic) counting principle:
"When there are m ways to do one thing, and n ways to do another, then there are m × n ways of doing both."
Read more at the link below.
There are 6 choices for which letter is facing up.
There are 4 orientations for that letter.
So, what's your conclusion?

rotchm say: For the first letter, there are 4 ways to display it.
For the second letter, there are 4 ways to display it.
Etc...
For the sixth letter there are 4 ways to display it.
Making a grand total of [left for you].
Done!
