(In reply to
Solution for Squares by nikki)
First of all, after seeing Charlie’s solution, I realized I was a little too hasty when I "extrapolated" from my triangles solution. For the abcd group, I chose to mimic the "combinations times 2" thought process instead of the "permutations divided by 3 (or 4 for squares)." The two are equivalent for triangles, but not so for squares. I should have been more careful and chosen n!/(n4)!/4 when moving onto squares. So the answer should actually be:
S = (n^4 + n^2 + 2n)/4
Second, I realized that both of my solutions were a little incomplete. What I actually solved for was n>2 in the triangles, and n>3 in the squares. Here is the complete solution:
Triangles:
For n=1, T= n = 1
For n=2, T= n + n*(n1) = 4
For n>2, T= (n^3 + 2n)/3
Squares:
For n=1, S= n = 1
For n=2, S= n + 2[n^2  n] = 6
For n=3, S= n + 2[n^2  n] + [3/2*(n^3  3n^2 + 2n)] = 24
For n>3, S= (n^4 + n^2 + 2n)/4

Posted by nikki
on 20040910 16:45:14 