Melody

No he is not kidding you he is the same m***n that we have both been dealing with.

Thanks EP

I answered this question before.  I even displayed the pic for him/her.

I finally get it.

Thanks Alan.   

Where is your justification?

You found 0,0,0,2,1?  There are only 4 numbers and I already included 0,0,2,1

But it is nice that you have thought about the question.    

ok, do you know what the answer is?

I started with Alan's technique.  

Thanks Alan I would not have thought of it myself.

I am thinking of the numbers as four distinct zeros, 4 distinct ones and 4 distinct twos.

I am going to say it is 3 columns and 4 rows with each row adding to a multiple of 3 because that will be much easier for me to work with.

Here is one possibility

0,0,0,0     4! = 24

1,1,2,2     4*3*4*3  * 4!/(2!2!) = 144*6 = 864

1,1,2,2     4!                     =24

So far I get 24*864*24 = 497664 ways

But the rows can be in any order. I think that is 3 ways =3 ways

So that is  497664*3 = 1 492 992 ways.

Another possibility is

2,2,2,0    4*3*2*4*4 = 384 ways

1,2,0,0    4*1*4*3*(4!/2!) = 48*12 = 576 ways

1,1,1,0    3*2*4 = 24 ways

384*576*24 = 5308416

Times by 6 = 31 850 496 ways

Another possibility is 

1,2,0,0    4*4*4*3*(4!/2!) =   193*12 =  2304

1,2,0,0    3*3*2*1*12      =    18*   12=  216

1,2,1,2     2*1*2*1*(4!/4) =     4*    6 =     24

2304*216*24= 11 943 936

Times by 3 = 35 831808

1 492 992+31 850 496+35 831808 = 69 175 296 ways

This does seem like a lot and maybe I have double counted

but if there were no restrictions at all then there would be 12! ways = 479 001 600.

69175296/497001600 = 0.1391852581561106

So my answer is only 14% of all possible answers with no conditions at all.

So it may be correct. It passes my reasonableness test.

Thanks Alan,

I am still not understanding this statement

"The non-remainder part vanishes when x=0, 1 or 2, "

Could you expand upon that please?