Puzzles Riddles Galore!
Here are the rules:
1. You CANNOT put down a puzzle or riddle of your own unless you are the one to correctly answer the latest puzzle or riddle first.
2. Follow rule #1.
I'll start. This is a good one for all you math sharpies out there. (Extra points if you can figure out a formula that will solve this for you without having exhausted all possibilities.)
How many different tetrahedrons can be produced by coloring each face a solid color and using n different colors? (Two tetrahedrons are the same if they can be turned and placed side by side so that corresponding sides match in color.)
How many cubes with n colors?
1. You CANNOT put down a puzzle or riddle of your own unless you are the one to correctly answer the latest puzzle or riddle first.
2. Follow rule #1.
I'll start. This is a good one for all you math sharpies out there. (Extra points if you can figure out a formula that will solve this for you without having exhausted all possibilities.)
How many different tetrahedrons can be produced by coloring each face a solid color and using n different colors? (Two tetrahedrons are the same if they can be turned and placed side by side so that corresponding sides match in color.)
How many cubes with n colors?
Comments
How many different tetrahedrons can be produced by coloring each face a solid color and using n different colors? (Two tetrahedrons are the same if they can be turned and placed side by side so that corresponding sides match in color.)
How many cubes with n colors? <!--QuoteEnd--></td></tr></table><div class='postcolor'><!--QuoteEEnd-->
12 possibilities (the question is a bit vague so it may or may not be right depending on interpretation)
edit: I took a really weird interpretation on this one so answers may vary.
Riddle:
Top to bottom. What is the next row?
1
11
21
1211
111221
312211
13112221
Answer:
it is in my profile listed as my website.
My mistake if the question is a little vague. Supposing you have n colors, how many different variations of tetrahedrons can you have such that you cannot take another tetrahedron and rotate it to look like it. Assuming repeating colors is possible. In others painting one tetrahedron half yellow and half blue will yield 1 unique tetrahedron. Paint another one half blue and then half blue rotated around will be the same one, thus it is not unique and isn't counted. Make sense?
*EDIT * I hope I explained this better */EDIT*
Case 1: All colours are different
You can choose C(n,4) colours. (which is n!/((n-4)!(4)!)).
Case 2: 3 different colours are chosen (1 repeating colour).
You can choose C(n,3) colours. The last colour can be any one of the 3, so you have a total of 3C(n,3)
Case 2: 2 different colours are chosen: 2 sub possibilities:
Subcase i: 1 colour is used for the other faces
In this case, you have 2C(n,2), since there are different colours you can use.
Subcase ii: 2 colours are used for the other faces
In this case, there is only C(n,2) possibilities, since you have no choice as to which colours you use.
The total for this case is then 3C(n,2).
Case 3: Only 1 colour is picked
In this case, you have C(n,1) or just n possibilities.
So the sum is C(n,4) + 3C(n,3) + 3C(n,2) + C(n,1).
I'm guessing, though that this can be written a little differently, since it looks like a binomial expansion.
I'm guessing it will look a little like:
s
Sigma [C(s-1,j-1)][C(n,j)]
j=1
where s is the number of sides.
This should give the right sum for 4 sides. For more than that, it probably won't work unless you have every side toughing each other. So you'll need to have an s-1 dimensional solid.
I'm not sure about the cube yet, but I think I have a rough idea of how it works. If it comes to me, I'll be to post it.
EDIT: fixed some spelling.
2nd EDIT: Since it seems that you can have mirror images of the tetrahedron (which will yield identically coloured tetrahedrons but will not match in orientation), it appears my answer is wrong.