make 11 distinct necklaces from 3 different types of beads - without polya this is a home work question.
how many distinct necklaces can be made with 11 beads, each can be either circle, square or triangle?
necklaces are the same under rotation but not the same under mirror (turnover).
the thing is - i need to solve this without Pólya enumeration theorem or Burnside's lemma.
thanks in advance,
Yaron.
 A: Let there be $N=3^{11}$ ways of arranging 11 beads of 3 different type in a string (not formed a necklace yet).
The question is, when we connect them up to form a necklace, how many of these necklaces are repeats of each other?
Given an arbitrary necklace, we could rotate it $k$ spaces, and get another necklace. It likely is a different necklace, but could be the same.
Lemma: If a necklace rotated by $1\leq k \leq 10$ places yields the same necklace, then all beads are of the same type. The converse is trivially true.
Hint: Let's consider what it means, if the rotated necklace is the same. This means that the bead in the 1st place, is the same as the bead in the $1+k$ place, is the same as the bead in the $1+2k$ place, etc. 
As such, this tells us, given any string, it will (very likely) form the same necklace as 10 other strings (for a total of 11).
The only exception are the 3 necklaces, which are beads all of the same type. Hence, for each of these $N-3$ strings correspond to a set of 11 strings which yield the same necklace, thus there are $\frac{N-3} {11} $ distinct necklaces. Adding on the 3 strings whose beads are of the same type (which clearly give us distinct necklaces), we get $\frac{N-3}{11} + 3$ distinct necklaces.
Note: Burnside is the best way to solve this; you can see how the ideas above are expressed in Burnside.
