We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

+0

# How many ways are there to arrange beads of distinct colors in a grid if reflections and rotations are considered the same? (In other

+1
778
2

How many ways are there to arrange 6 beads of distinct colors in a 2x3 grid if reflections and rotations are considered the same? (In other words, two arrangements are considered the same if I can rotate and/or reflect one arrangements to get the other.)

Mar 20, 2018
edited by Guest  Mar 20, 2018

### 2+0 Answers

#1
+638
0

Here is my solutions, not sure if it is correct.

we could put A anywhere we want on the top row, 3!

and any where on the bottom row, 3!

3! * 3! = 36

We then have to divide by two because of repeats.

36/2 = 18

18 is my final answer.

I am not sure this is correct, but I hope this helped.

Supermanaccz

Mar 20, 2018
#2
0

I'm going to represent the beads using the numbers 1-6 (bead number 1, beads number 2....)

Without considering rotations and reflections, we can arrange the beads in 6*5*4*3*2*1=6! (6 factorial) ways to arrange the beads (because we have 6 exactly 6 ways to pick the location of the first bead, then 5 (because we already placed one bead) ways to pick the location of the second bead...)

Now we have to find the number of arrangements we can get from a combination of the 6 beads by reflecting and/or rotating the combination-

Let's find all the operations we can perform on a conbination of the beads-

1. We can rotate the combination

(Like this: $$\begin{bmatrix} 1 && 2 && 3\\ 4 && 5 && 6 \end{bmatrix}\rightarrow \begin{bmatrix} 6 && 5 && 4\\ 3 && 2 && 1 \end{bmatrix}$$

2. We can reflect the combination horizontally:

$$\begin{bmatrix} 1 && 2 && 3\\ 4 && 5 && 6 \end{bmatrix}\rightarrow \begin{bmatrix} 3 && 2 &&1\\ 6 && 5 && 4 \end{bmatrix}$$

3. We can reflect the combination vertically:

$$\begin{bmatrix} 1 && 2 && 3\\ 4 && 5 && 6 \end{bmatrix}\rightarrow \begin{bmatrix} 4 && 5 && 6\\ 1 && 2 && 3 \end{bmatrix}$$

Performing any of those operations twice will give us the same combination we started with, meaning we can perform any of the operations once to get a different conbination, or not perform that operation at all. So can get 2*2*2=8 different areangements by applying any of the operations to a combination (2 different options for each operation: performing it or not performing it.

Let the set of arrangements when reflections and rotations are considered the same be X. For each member y of X, we can define a set Ay that consists of 8 arrangements (the 8 arrangements we can get from y). This means that for every 2 different members y and z, Ay and Az are disjoint sets (no artangement from Ay appears in Az). Therefore, the number of arrangements we can get from X by applying any of the operations to an arrangement from X is 8*(number of arrangements from X)

The set that contains those arrangements (B) must contain every single arrangement from the beginning (every arrangement we can get by arranging the different beads in the grid). Why? Because if there's an arrangement we CAN'T get by taking an arrangement from X and applying any of the operations to it, then that arrangement must be in X! (Prove it!)

And finally, we get that (number of arrangements in X)*8=total number of artangements=6*5*4*3*2*1=720

Dividing by 8 will give us the answer-

(Number of arrangements in X)=720/8=90

And that is the answer (90)

Guest Mar 25, 2018