How many ways can you arrange $2$ blue beads, $1$ green bead, $1$ red bead, and $1$ yellow bead in a circle? > How many ways can you arrange $2$ blue beads, $1$ green bead, $1$ red bead, and $1$ yellow bead in a circle? Which of the following would the answer be and why is the other one wrong?? $$3!=6\;\text{or}\;\dfrac{4!}{2!}=12$$

I would compute the answer as

$$2(3+3)=12$$

Here's the idea: First place the Red, Green, and Yellow beads on the circle; there are $2$ ways to do this. Then place the two Blue beads either _together_ between two beads already on the circle, or _apart_ ; in either case there are $3$ ways to place the Blue beads.

If you consider mirror images as identical, then there is only $1$ way to place the Red, Green, and Yellow beads in the first step, which gives the answer $3+3=6$.