here is what i think (i'm really bad at combinatorics, so this might not be right):

suppose we label the positions where we put our beads as 1-12. we select 5 beads from a box of beads of number 2 to put in the 12 possible positions. what we are essentially doing is choosing 5 of the 12 positions to be "color number 2 beads". thus there are "12 choose 5" ways to do this:

12!/[(5!)(7!)] = (8*9*10*11*12)/(5!) = (8*9*10*11*12)/(2*3*4*5) = 6*11*12 = 792 possible necklaces, not counting duplicates.

let's call this set of 726 necklaces X. we're going to act on this set with a cyclic group of order 12 (the rotation group of the dodecagon). thus the number of non-duplicate necklaces is the number of orbits of X. we will call an necklace fixed by g in our rotation group, if g maps a color one bead to a color one bead, and a color two bead to a color two bead.

to do this properly, we need to invoke Burnside's Lemma:

where X/G is the partition of X into its orbits, and X^{g}is the number of elements of X fixed by g.

let's call a rotation by 1/12 of a circle (shifting one bead over) r. so G = {e,r,r^{2},...,r^{11}}.

first, the easy one: how many of our 792 configurations are fixed by e? all of them.

how many are fixed by r? well, if they are fixed by r, we must have only beads of ONE color, but we have TWO, so: none of them. similar reasoning applies to r^{-1}= r^{11}.

how about r^{2}? if a coloring is fixed by r^{2}, it must have period 2 (it repeats every two beads), so we only need to consider the possible colorings of the first 2 beads. suppose we use B (for black) for color number 1, and W (for white) for color number 2. here are our possible colorings:

BB <---leads to 12 black beads

BW <---leads to 6 black, and 6 white beads

WB <---see above

WW <---leads to 12 white beads

so no coloring can be fixed by r^{2}, or similarly, by r^{10}.

ok, on to r^{3}(of order 4). possible colorings of the first 3 beads:

BBB <---no, all black

BBW <---no, 8 black 4 white

BWB <---no, 8 and 4 again

BWW <---no, 4 black and 8 white

(verify that none of the 4 remaining colorings of the first 3 beads will work). this means no coloring can be fixed by r^{3}or r^{9}.

how about r^{4}? again, we only need to consider the first four beads:

if we have 0 W beads in the first 4, and r^{4}fixes the coloring, all the beads are black.

if we have 1 W bead in the first 4, we wind up with 3 W and 9 B.

if we have 2 W beads in the first 4, we wind up with 6 W and 6 B.

if we have 3 W beads in the first 4, we wind up with 9 W and 3 B.

if we have 4 W beads in the first 4, all the beads are W.

so none of the colorings can be fixed by r^{4}, or r^{8}.

let's skip ahead and look at r^{6}next. since we have an odd number of each color, and since any pattern of period 6 will contain an even number of each color, no coloring is fixed by r^{6}.

that leaves just r^{5}, and r^{7}.

5 B beads is obviously not going to work (the first 10 beads would have to be B).

4 B beads is "too many" we'd wind up with at least 8 B beads total, and we only have 7.

5 W beads wouldn't work, either, it winds up giving us 10 W beads in the first 10, and we only have 5.

4 W beads is too many, we'd have at least 8 W beads, and we only have 5.

3 W beads is also too many, we get 6 W beads in the first 10.

so the only viable combination is 3 B beads, 2 W. suppose the first bead is W. then we have 4 W beads in the first 10, and the last W bead we can use in the 11th place. so if the first bead is W, the second bead CAN'T be.

this means we might have:

WBWBB

WBBBW

WBBWB

furthermore, for the pattern to "match up" after it wraps around, the 3rd bead must be W, and the 4th bead B. so of the 3 we've narrowed it down to, only WBWBB is still a possibility. let's see what happens:

WBWBB WBWBB WB rotate by 5:

WBWBB WBWBW BB <---not a match.

so the first bead cannot be W. so now assume that the first bead is B. if the 2nd bead is also B, we get B beads in place 11 & 12, leading to 8 B beads in all. so the second bead has to be W. so we have:

BWBBW

BWBWB

BWWBB. just as with having the W bead first, the 3rd bead has to be B. this gives us 2 to choose from:

BWBBW BWBBW BW rotate by 5:

BWBBW BWBWB BW <--not a match.

BWBWB BWBWB BW rotate by 5:

BWBWB BWBWB WB <--not a match.

i leave it to you to show that r^{7}also does not fix any colorings.

so by burnside's lemma, we have:

|X/G| = (1/12)(792) = 66 distinct necklaces.