1. ## Alternating Necklaces

Hi, bit of help needed:

A necklace has 10 beads around it. If half the beads are pink and the other half are grey, in how many ways can they be arranged so that the colours are alternate? Assume that the beads are distinguishable.

I'm stuck deciding between:

4!/2 * 4!/2 = 144

or

(4!*5!)/2 = 1440

Help?

2. ## Re: Alternating Necklaces

EDIT: I got the question wrong.

3. ## Re: Alternating Necklaces

Originally Posted by eskimogenius
Hi, bit of help needed:

A necklace has 10 beads around it. If half the beads are pink and the other half are grey, in how many ways can they be arranged so that the colours are alternate? Assume that the beads are distinguishable.

I'm stuck deciding between:

4!/2 * 4!/2 = 144

or

(4!*5!)/2 = 1440

Help?
Let's fix the location of Pink #1 to account for the fact that the necklace can be rotated. Then proceeding counterclockwise, we have 5 ways to pick the next (white) bead, then 4 ways to pick the next (pink) bead, and so on until we have 1 way to pick each of the last two beads. So the total number of ways is 5 * 4 * 4 * 3 * 3 * 2 * 2 * 1 * 1.

But necklaces can not only be rotated, they can be flipped, so we have counted each necklace twice. Divide the previous answer by 2 to compensate.

4. ## Re: Alternating Necklaces

Hello, eskimogenius!

I agree with awkward . . .

A necklace has 10 beads around it: 5 pink and 5 gray.
In how many ways can they be arranged so that the colours are alternate?
Assume that the beads are distinguishable.

We have 10 distinct beads: .$\displaystyle P_1, P_2, P_3, P_4, P_5, G_1, G_2, G_3, G_4, G_5$

The beads will be placed like this:

. . $\displaystyle \begin{array}{cccccccccc} &&& P \\ & G &&&& G \\ P &&&&&& P \\ \\ G &&&&&& G \\ & P &&&& P \\ &&& G \end{array}$

$\displaystyle P_1$ can be placed anywhere.

Then the other 4 P's can be placed in $\displaystyle 4!$ ways.

Then the 5 G's can be placed in $\displaystyle 5!$ ways.

. . Hence, there are:$\displaystyle (4!)(5!)$ ways.

Since a necklace can be flipped, mirror-images must be eliminated.

Therefore, there are: .$\displaystyle \dfrac{(4!)(5!)}{2} \:=\:1,\!440$ necklaces.