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In how many ways can seven beads of distinct colors be put on the hexagonal grid shown, if reflections and rotations of an arrangement are considered equivalent?

Mathgenius Dec 14, 2018

#1

#8**-1 **

The first answer is correct! You came up with the same solution, yet you say it is incorrect...

PartialMathematician
Dec 16, 2018

#2**0 **

There are 7 ways to choose the dot in the middle. You can place a random dot down (that way this takes care of the reflection).

Then, there are 5! ways to order the remaining dots.

5!=840

You then need to take into consideration reflections, you can divide by two.

840/2=420

There are 420 ways to arrange the seven beads.

guoiris458 Dec 15, 2018

#4**0 **

The goal of counting arrangements: Count everything ONCE and ONLY ONCE.

Understand that once we choose the outer six, the center dot is automatic.

There are 6! ways to put the colors in any arrangement without considering overcounting. 6! = 720.

We must consider overcounting. Since there are 6 axes of symmetry, we divide 720 by 6 to get 120. The rotations of the arrangements also matter (Eg: [1,2,3,4,5,6, 7 in the center] is the same as [4,5,6,1,2,3, 7 in the center]), so we divide 120 by 2 to get 60. There is your answer, \(\boxed{60}\).

Hope this helps, (do not get confused by the answers above)

- PM

PartialMathematician Dec 15, 2018

#5**+1 **

There are** seven **colours MP

If you are not going to consider **any **over counting then there are 7! ways to put thise in place.

Where do you get 6! from?

If you were going to put 7 coloured beeds in a circle (rotations classed as the same) then the answer would be 6!

But this is not a circle **AND** you said that you were counting each rotation as different for this part of the question.

Melody
Dec 15, 2018

#6**+1 **

I've changed my mind a bit.

I think the very first answer might be correct.

There are 7 ways to choose the colour in the middle

Then you can put one colour any other place and there are 5! ways to place the rest.

This allows for rotation but it does not allow for reflectionl

There are 6 axes of symmetry for divide by 6 and you might have athe correct answer.

\((7*5!)/6\)

\(\frac{7*5!}{6} = \boxed{140}\)

Thanks for reminding me of that \boxed command PM

Melody
Dec 15, 2018

#7**+1 **

I'm a little confused, by PM's answer because would you not have 7 different options for the dot in the middle? I'm sorry, I'm new to this subject...

guoiris458
Dec 15, 2018

#9**-1 **

Sorry, I forgot about the dot in the middle. At first, I had your solution, but since Melody said it was incorrect, I tried something else, but I happened to forget about the middle dot.

PartialMathematician
Dec 16, 2018