+0  
 
0
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avatar+379 

deleted.

 Nov 16, 2019
edited by sinclairdragon428  Nov 16, 2019
edited by sinclairdragon428  Nov 19, 2019
 #1
avatar+118667 
+1

That seems a reasonable starting point dragon.    laugh

I do not know if you did it yourself but it is a better start than i had when I tried to do it with no hints.

I had no idea where to start.

 

However:

 

I am not convinced about your continued logic.

 

Just for one of these configurations I think you have to start by placing one number. I chose to place one of the A which is a multiple of 3

 

So I think it would be  2! * 3! * 3!   = 24 ways

But you have

 

C B A C B A C B A          YES THAT WILL WORK   24 ways

B C A B C A B C A          THAT WILL WORK TOO    24 ways

 

I can't see any other way to do it.  But maybe you can.  

 

So I only get 48 ways but I could easily be wrong.

 Nov 16, 2019
 #6
avatar+379 
-1

Hi Melody!

Thank you for the guidance. Could you explain how you got to 2! * 3! * 3!   = 24 ways?

 

Thanks again for the help!

sinclairdragon428  Nov 18, 2019
 #7
avatar+118667 
+1

I am not sure that it is right but this is my logic.

 

It is a circle and rotations are considered the same.

In order to deal with this I have fixed on number in place. 

I do not think it makes any difference which initial number you use so the chance of getting this first place right is 1.

Say that first place was filled with an A number. Then the position of the other two As is already fixed. there are 2!=2 ways the other A's can be placed.

 

There are three fixed places for the Bs  so that is 3! ways and the same for the C's.

so that is where I got  2!3!3! from

Melody  Nov 18, 2019
 #8
avatar+118667 
0

Over 23 housr later and no response from SinclairDragon.428

Why does this not surprise me.

Melody  Nov 18, 2019
 #9
avatar+379 
0

thank you for explaining. could you show me where my reasoning went wrong? i would like to know how i could fix it next time.

sinclairdragon428  Nov 19, 2019
 #10
avatar+118667 
+1

This line makes no sense at all

 

\(3^3 \cdot 2^3 \cdot 1^3 = 27 \cdot 8 \cdot 1 = 216.\)

 

There are 3 choices for the first A then 2 choices left for the second A then only on spot left fot the third A  that is 3! places not 3^3 places.

 

Normally powers are only appropriate for replacement questions.

Melody  Nov 19, 2019
 #11
avatar
0

Do you do that on purpose? I wont tell anyone, I promise, I just want to know.

Guest Nov 20, 2019

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