All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 FIGURE it out ! (Posted on 2004-09-10)
1. With an unlimited supply of toothpicks of n different colors, how many different triangles can be formed on a flat surface, using three toothpicks for the sides of each triangle?
(Reflections are considered different, but rotations are not.)

2. How many different squares?

 No Solution Yet Submitted by SilverKnight Rating: 3.3333 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: My take | Comment 9 of 12 |
(In reply to My take by Tristan)

Nice approach and nice writeup.  However, something isn't right in your counts for the 4-bead necklaces since (n^4-n-n*(n-1)/2)/4 = (16-2-2/2)/4=13/4 when n=2.

From http://mathworld.wolfram.com/Necklace.html comes the formula (my notation, however)

N(b,n)=(1/b)(p(b1)n^(b/b1)+...+p(bD)n^(b/bD)

for the number N(b,n) of b-bead necklaces with beads of n possible colors, where two necklaces are "the same" only if one can be rotated to coincide with the other (no flipping), and where

D=number of divisors (including 1 and b) of b,

b1,...,bD=the divisors of b,

p(k)=number of whole numbers j such that 1 <= j <= k and gcd(j,k)=1 (Euler's totient function).

For 3 beads, n colors this gives

N(3,n)=(1/3)(n^3+2n)

and for 4 beads, n colors

N(4,n)=(1/4)(n^4+n^2+2n).

This N(4,n) agrees with Charlie's numbers.

Edited on September 11, 2004, 1:16 pm
 Posted by Richard on 2004-09-10 23:11:18

 Search: Search body:
Forums (0)