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Hi could someone help me? I don't know how to solve this.

 

 

Find the number of ways of arranging the numbers 1-12 in a  array (with three rows and four columns), so that the sum of the numbers in each column is divisible by 3. Each number must appear exactly once in the array.

 

Thank you so much!

 Mar 12, 2020
 #1
avatar+23566 
+1

here is one

 

1  2  3

5  4  6

7  8  9

11 10 12       there are many more....see Alan's answer below:

 Mar 12, 2020
edited by ElectricPavlov  Mar 12, 2020
 #2
avatar+30038 
+3

Think of each number in terms of their remainder when divided by 3.

1 has rem 1

2 has rem 2

3 has rem 0

4 has rem 1

5 has rem 2

6 has rem 0

7 has rem 1

8 has rem 2

9 has rem 0

10 has rem 1

11 has rem 2

12 has rem 0

 

In order that each column sums to a number divisible by three we must have combinations such that each number with remainder 0 is combined with one of the numbers with remainder 1 and one with remainder 2.

 

There are 4*4*4 = 64 ways of doing this. This is the result if the columns are distinguishable. If they aren't distinguishable this number should be divided by 4 (I think!).

 Mar 12, 2020
 #4
avatar+109509 
+1

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.

 Mar 13, 2020
edited by Melody  Mar 13, 2020
edited by Melody  Mar 13, 2020
edited by Melody  Mar 13, 2020
 #5
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0

Sorry melody, thats incorrect

Guest Mar 13, 2020
 #6
avatar+109509 
0

ok, do you know what the answer is?

Melody  Mar 13, 2020
 #7
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0

477,757,400

Guest Mar 13, 2020
 #8
avatar
0

lol i was wrong too, the result is 477,757,400/24=19,906,560

Guest Mar 13, 2020
 #11
avatar+109509 
0

Where is your justification?

Melody  Mar 14, 2020
 #9
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0

I found that a few other combinations of remainders are three 0 rems and two 2 rems and one 1 rem. I hope this will help...

 Mar 13, 2020
 #10
avatar+109509 
0

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.    laugh

Melody  Mar 13, 2020
edited by Guest  Mar 14, 2020

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