If I have a box that has 8 sides. Each of the sides has a different picture. How many configurations can there be if there are 8 different pictures, no repeats? (Not counting the different orientations the pictures may have).

Can someone help me figure this out please?

Thank you

Most boxes have six sides, if the top and bottom are counted as sides. If they are not, is the box octagonal? If they are is the box a hexagon? Where are the pictures? On the outside, inside, top, bottom ? You need to provide mor information for me to hazard a guess of any sort.

I'm just as confused as frankvan about the given problem. If this is about the possible shapes of a convex (or concave?) polyhedron with 8 faces, then more info is needed.

On the other hand, if this is about a box with pictures on 8 sides, ignoring the top and bottom, then it is equivalent to asking how many distinct ways 8 guests may be seated at a round dinner table, not counting rotations of the table. The answer is 7! (7-factorial) = 7*6*...*2*1 = 5,040.

umn, technically it is n factorial
i.e. 8!
= 8x7x6x5x4x3x2x1

but do explain
how did the two extra sides enter into a cuboid or cube>?

or is it that one of the lengths is divided into two halves?
so eight separate slabs result for the pictures to be pasted on??

Are the sides otherwise distinguishable? I would guess they are not, in which case the answer would be less than 8-factorial (how much less, I'm not sure).

Tried to edit, but I apparently timed out. A more complete reply with a partial thinking through:


Are the sides otherwise distinguishable?

If they are, Pin~Jinx is correct.

I would guess, however, they they are not. In that case, you would need a correction factor to account for arrangements that are only rotations of other arrangements. For a round table with 8 seats, the correction would be to divide by the number of seats. That is 8!/8 = 7!, which is the answer Professor gave. However, I'm not sure that 8 is the appropriate factor for an 8-sided object. For the 8-sided object, you can rotate to 4 positions horizontally and to 4 positions vertically. I think those need to be multiplied rather than added (that is, I think there are 4 vertical rotations for each of the 4 horizontal rotations). That would give 8!/(4*4) = 2520.

I may be completely off about that, and I don't think I'm being particularly clear, but I thought I should put in my two cents, and I don't have time for three at the moment. Maybe Prof. and Pin~jinx can sort out if what I'm saying is correct and can better explain it.

methos, are you picturing this box as an octahedron? My image is of an octagonal prism, the end faces (top and bottom) of which are ignored. Another possibility is a hexagonal prism with the end faces counting, in which case there are (8*7/2!)*(5!)*2 = 6,720 permutations.

The fact is, we need the original questioner to explain what was meant by "a box that has 8 sides."

Yes, I was picturing an octahedron (I missed your comment about ignoring the top and bottom)... something like:


You're right that we need more info. Even if it's an octagonal prism, now that I think about it, it matters whether or not the top and bottom are interchangeable. If not, I'd agree with 8!. If so, I think it would be 8!/2.