How many different ways can you colour each side of a Tetrahedron?

Question

Does any one know how many different ways you can colour each side of a Tetrahedron, using the same 4 colours each time.

This is quite hard to explain, and if anyone else can put it in a simpler sentence, please do! Also, any websites on this subject would be helpful.

Thanks.

Can you please explain the process as well please?

Also, the same colour cannot be repeated in the same Tetrahedron (eg. you cannot have 2 red sides, 1 yellow side and 1 green side, they must all be different)

Answer 1

If you are compelled to use four colours, then there are 4! (4x3x2x1) = 24 different combinations. If you are allowed to use the same colour on more than one side then you can have 4^4 = 256 combinations.

You have four colours. The first colour can be applied to any of the four sides. Then go on to the next colour - which can be applied to any of the three remaining sides. That's 4 x 3 = 12 combinations of sides to be coloured from just the first two colours. Then you use the third colour and apply it to one of the two remaining sides. You multiply the 12 you already had by 2, giving 24. When you get to the fourth colour, there is only one side left, so that doesn't increase the number of possibilities. So the total combinations is 4x3x2 = 24. This is known as the factorial of 4, or 4!.

The same methodology applies to any other sequencing question or system of mutually exclusive permutations, such as how many different possible finishing orders there are for 6 horses in a race (6! = 720), or the order bingo balls can be selected from a bag.

Answer 2

If you start with a "regular" tetrhedron with 4 identical faces and you apply a different color to each face, how many different "ways" could you paint it?

only 2:

all 4-color tetrahedrons fall into exactly 2 groups of indistinguishable tetrahedrons

see below:



if you can "rotate" the tetrahedron to evaluate whether it is "different" from another color-scheme, there are exactly 2 ways to color a tetrahedron (assuming each side is a different color)


consider: Red, green, blue, yellow

looking dowm, place the "RED" side on bottom [facing where you can't see it that is]

leaves G, B, Y that you can see.

place the Greeen side at "base" where you see an equilateral triangle and green on the lower portion of it

there's only two ways the Blue and Yellow can be arrangedv ...

so there are exactly TWO color schemes

{ you could call them "Levo" and "dextro" is you wanted to I supose ..

3-D space is like that
:)

Answer 3

There are 24 ways to paint a T with three unique colors. Four possibilities for the two that are the same, three for the first unique, and two for the second unique = 4*3*2 or 24. It is not possible to create duplicates, since duplicates are taken care of by successively reducing the number of possible colors for the next consecutive side.

Answer 4

Since you have exactly 4 colors and 4 sides, the question becomes how many ways you can order the 4 colors. In other words, how many permutations are there on 4 colors? There are always n! permutations on n elements, so there are 4! = 24 permutations on 4 colors.

Answer 5

4! = 24

Answer 6

Call the colours a,b,c d
then colouring one particular side 'a' then you can have the following perms.
a,b,c,d a,b,d,c a,c,b,d a,c,d,b a,d,b,c a,d,c,b
by placing 'a' on each of the other three faces you can repeat these six combinations four times in all.
So the answer is 24 times.

Answer 7

These kinds of questions are called "permutations".