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