Given 2 red beads, 2 blue beads, 1 yellow bead, and 1 green bead, how many different necklaces can be made? I need to write a Matlab code to determine the answer (which was given as 16) and I need to utilize loops to remove flips and circular shifts


We use the Polya Enumeration Theorem (PET).
The cycle index $Z(D_6)$ of the dihedral group $D_6$ is given by
$$Z(D_6) = \frac{1}{12}
\left(\sum_{d|6} \varphi(d) a_d^{6/d}
+ 3 a_2^3 + 3 a_1^2 a_2^2\right).$$
This is
$$\frac{1}{12}
\left(a_1^6 + a_2^3 + 2 a_3^2 + 2 a_6
+ 3 a_2^3 + 3 a_1^2 a_2^2\right)
\\ = \frac{1}{12}
\left(a_1^6  + 2 a_3^2 + 2 a_6
+ 4 a_2^3 + 3 a_1^2 a_2^2\right).$$
We seek
$$[R^2 B^2 Y G] Z(D_6; R+B+Y+G).$$
The only contribution comes from
$$\frac{1}{12}
\left(a_1^6  + 3 a_1^2 a_2^2\right).$$
This is because with the Polya substitution the terms $2 a_3^2 + 2 a_6
+ 4 a_2^3$ produce powers of three, six, and two, exclusively.
Continuing, we get
$$\frac{1}{12}
[R^2 B^2 Y G] (R+B+Y+G)^6
\\ + \frac{1}{4} [R^2 B^2 Y G]
(R+B+Y+G)^2 (R^2+B^2+Y^2+G^2)^2.$$
This is
$$\frac{1}{12} {6\choose 2,2,1,1}
+ \frac{1}{4} \times 2 \times 2 = 16.$$
BTW  the convention  at the  OEIS is  to use  the term  necklace for
cyclic symmetries and bracelet for dihedral ones.
