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deleted.

Nov 16, 2019
edited by sinclairdragon428  Nov 16, 2019
edited by sinclairdragon428  Nov 19, 2019

#1
+108650
+1

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:

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
+397
-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
+108650
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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
+108650
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Over 23 housr later and no response from SinclairDragon.428

Why does this not surprise me.

Melody  Nov 18, 2019
#9
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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
+108650
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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
0

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

Guest Nov 20, 2019
#2
+117
0

Please do not discuss this problem!  This is an active homework problem.

To the original poster: I realize that homework may be challenging. If you wish to receive some help from the staff or other students, I encourage you to use the resources that the online classes provide, such as the Message Board.  Thanks.

Nov 16, 2019
#3
+108650
0

I understand the need to not answer but I am interested in this problem too.

I do not know the answer.

What is wrong with me discussing it here?

I know you are going to tell me it is against the rules.

Is it against the rules for students to discuss it with each other?

What is the difference, I, like the other students, do not know the answer.

Melody  Nov 16, 2019
#4
+117
-1

This problem is taken from an online class, and one of the rules is that the use of outside sources is strictly forbidden.

Students are actually quite free to discuss these problems, as long as they are limited to the class message board. This student has chosen to disregard this rule, which I think is unfortunate.  I appreciate your cooperation on this matter.

wonderman  Nov 17, 2019
#5
+108650
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You only have my partial co-operation.

I will not personally give a full answer to someone that I have good reason to think is cheating.

But, if I can learn myself and I believe they can learn with me then I will continue to discuss questions.

I have asked you why I should not and you have not given me a satisfactory answer.

You have just repeated. "It is against the rules"

It is against the rules for them to post.

It is against my moral code to knowingly provide full answers to cheats.

However it is not against my moral code to discuss questions and learn with students.

Melody  Nov 17, 2019