How many different seven-bead necklaces are possible, assuming each bead is one of four different colors and each necklace contains exactly one bead of one color and exactly two beads of the three remaining colors?
I know that the necklaces are actually $D_7$ in disguise and that I have to use Polya's Enumeration Theorem, but I am having trouble
So for example, how many ways can we make unique necklaces with 1 red, 2 blue, 2 green, 2 yellow and so on and so on
Since the OP mentions $D_7$, I am going to assume that we are allowed to flip the necklace over as well as rotate it. In that case, the group of symmetries is $D_7$. We give a solution via the Polya Enumeration Theorem.
The cycle index of $D_7$ is $$Z = \frac{1}{14} (z_1^7 + 6 z_7 + 7 z_1 z_2^3)$$ For the four colors red, blue, green, and yellow, we take $r+b+g+y$ as the figure inventory. "Substituting" the figure inventory into the cycle index, we have $$Z = \frac{1}{14} \bigl( (r+b+g+y)^7 + 6(r^7+b^7+g^7+y^7) + 7 (r+b+g+y)(r^2+b^2+g^2+y^2)^3 \bigr)$$ The number of distinct colorings with one red, two blue, two green, and two yellow beads is the coefficient of $r b^2 g^2 y^2$ when $Z$ is expanded. Using $[\cdot]$ as the "coefficient of" operator, $$\begin{align} [r b^2 g^2 y^2]Z &= \frac{1}{14} \bigl([r b^2 g^2 y^2](r+b+g+y)^7 +7 [b^2 g^2 y^2](r^2+b^2+g^2+y^2)^3 \bigr) \\ &= \frac{1}{14} \left( \binom{7}{1 \quad 2 \quad 2 \quad 2} + 7 \binom{3}{1 \quad 1 \quad 1} \right) \qquad \text{(multinomial coefficients)}\\ &= \frac{1}{14} (630 + 7 \cdot 6) \\ &= \boxed{48} \end{align}$$