How many ways to colour 20 triangular faces with 5 colours, each used 4 times?

  • Thread starter Thread starter BrownianMan
  • Start date Start date
  • Tags Tags
    Combinatorics
BrownianMan
Messages
133
Reaction score
0
So I found a formula for the number of ways of coloring a shape with 20 triangular faces, 30 edges, and 12 vertices: (1/60)*(k^20+15*k^10+20*k^8+24*k^4).

Now I need to find the # of ways of coloring the faces with exactly 5 colors each with each color used exactly 4 times. I know how to find the # of ways of coloring the faces with exactly 5 colors (just plug k=5 in the formula) But the part about "each color used exactly 4 times" is throwing me off. How do I do this?
 
Physics news on Phys.org
Anyone?
 
How did you arrive at that formula? Maybe it can be adjusted to your new colouring problem.

Meanwhile, if you see all of the 20 faces as different (you count rotations of your object as separate colourings), you have
$$
\left( \begin{array}c 20 \\ 4 ~~ 4 ~~ 4 ~~ 4 ~~ 4 \end{array}\right) = \frac{20!}{4!\cdot4!\cdot4!\cdot4!\cdot4!}
$$
possibilities. This is similar to the number of ways you can choose ##k## objects from a set of ##n## objects, i.e. ##\binom nk = \frac{n!}{k!\cdot(n-k)!}##.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top