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

• It's only six beads...why not write out all the cases by hand? – lulu Mar 13 '17 at 0:19
• Because I want to know why the other "solution" is incorrect. Also, it's only 5 beads – suomynonA Mar 13 '17 at 0:20
• At a guess, the question is whether reflection symmetry counts or not. For a necklace, which can be flipped over, I'd say that two arrangements that differ by reflection coincided. – lulu Mar 13 '17 at 0:21
• To be clearer: does $YBBGR$ coincide with $YRGBB$ or not? At a dinner table, I'd say no. It matters who is on my left and who is on my right. For a necklace (which I assume is the case, given the beads) I would say yes. – lulu Mar 13 '17 at 0:26
• For counting the number of circles (not necklaces), put the yellow bead down at some arbitrary spot in the circle. Use this as a point of reference. Pick which two spaces out of the remaining $4$ are used by the blue beads. Then pick whether the red occurs before or after the green bead clockwise from the yellow bead. There are $6\cdot 2=12$ ways to do this. – JMoravitz Mar 13 '17 at 0:40

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 $3+3=6$.

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!