In how many ways can a necklace be made using $6$ identical red beads and $2$ identical blue beads?

Question

I got $(8-1)! /2 \cdot 2! 6!$ But it's a decimal number…

Also need to find the general formula that counts the total number of distinct necklaces made using $n$ identical red beads and $2$ identical blue beads.

Answer 1

HINT: The blue beads can be adjacent, separated by $1$ and $5$ red beads, separated by $2$ and $4$ red beads, or diagonally opposite each other with $3$ red beads between them in both directions. That’s just $4$ distinct necklaces. Can you generalize that idea to $n$ red beads?

Answer 2

There are $\binom{8}{2}$ ways to arrange the 6 red beads and 2 blue beads along a circle. The dihedral group $D_{16}$ of order 16 acts on the set of arrangements. Let $\omega$ be a rotation about the center by angle $\frac{2\pi}{8}$, $\sigma$, reflection about the line joining the mid points of a pair of opposite sides (there are 4 such reflections), $\tau$ a reflection about the line joining a pair of opposite vertices (there are four such reflections).

For $\omega, \omega^2, \omega^3, \omega^5, \omega^6, \omega^7$ there are no fixed elements. For $\omega^4$ four elements are fixed. For any $\sigma$, 4 elements are fixed and for any $\tau$, 4 elements are fixed. Thus by Burnside Theorem, the number of distinct necklaces is $$\frac{1}{16}(28 + 4 + 16 + 16) = 4$$

Answer 3

The number of possible necklace using $6$ identical red beads and $2$ identical blue beads are same as the number of coloring $D_8$, a dihedral group, with $6$ reds and $2$ blues. Find the cycle index. There is $8$ rotations(including identity: $360^\circ$ rotation), 4 reflections across vertexes, and 4 reflections across lines. Thus $$ P_G(z_1,z_2,\cdots,z_8)=\frac{1}{16} (z_1^8 + 5z_2^4 + 4z_1^2 z_2^3 + 2z_4^2 + 4z_8) $$ But it is restricted to use $6$ red and $2$ blue. Substitute $z_n=r^n+b^n$ and find the coefficient of $r^6 b^2$. By Polya enumeration theorem, the answer is $$ \frac{1}{16}\left(\binom{8}{6}+5\binom{4}{3}+4\left(\binom{2}{0}\binom{3}{3}+\binom{2}{2}\binom{3}{2}\right)\right)=4. $$

Answer 4

There are 4 possible ways

1. Two B TOGETHER that's they are separated by 0 red. RRRRRR BB

2. Separated by 1 and 5 red

RRR RR B R B

Or B RRRRR B R

3

Seperated by 2 and 4

RRRR B RR B

Or B RRRR B RR

4 Last Seperated by 3 and 3

RRR B RRR B