How many ways can you arrange $2$ blue beads, $1$ green bead, $1$ red bead, and $1$ yellow bead in a circle?

Question

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?? $$3!=6\;\text{or}\;\dfrac{4!}{2!}=12$$

Answer 1

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 $3+3=6$.

Answer 2

The answer is $\frac {4!}{2!}$. The reason is this:

Suppose your 2 blue beads are distinct. Then how many ways are there to arrange these 5 beads in a circle? Well, by the formula it's $\frac {5!}{5}=4!$. But because earlier on we assumed our blue beads are distinct, we permuted them in $2!$ ways. Hence, we need to reverse this order and therefore, the answer is $\frac {4!}{2!}$. Hope it helps!