+118687 That seems a reasonable starting point dragon. 
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.
+379 Hi Melody!
Thank you for the guidance. Could you explain how you got to 2! * 3! * 3! = 24 ways?
Thanks again for the help!
+118687 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
+118687 Over 23 housr later and no response from SinclairDragon.428
Why does this not surprise me.
+379 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.
+118687 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.
