#### Parkie

##### New member
I have 4 marbles coloured Red, Green, Blue, Yellow.

I have a board of 16 cells comprising 4 subcells of 2 by 2 in which my marbles R,G,B,Y must (all) be placed.

The added constraint is that each marble must occupy the same relative position in all four subcells of the table so that

R=TopLeft, G =TopRight

Y=BottomLeft, Blue=BottomRight

This means that

RG xx
YB xx
xx xx
xx xx

and

Rx xG
Yx xB
xx xx
xx xx

and

xx xx
xB Yx
xG Rx
xx xx

Are legal positions. But

GY xx
xx xx
BR xx
xx xx

Or

xx xx
xR Yx
xG Bx
xx xx

are not legal positions.

So the question is - how many legally filled tables do I have here? And how do I extend this to bigger tables, and larger colour sets?

P

#### Parkie

##### New member
Did you read this site's rules?

I suggest you change your letters to numbers:
12 xx
34 xx
xx xx
xx xx

Using letters gets confusing.

Thanks Dennis, you are right! Here we are

Legal positions

12 xx
43 xx
xx xx
xx xx

and

1x x2
4x x3
xx xx
xx xx

and

xx xx
x3 4x
x2 1x
xx xx

Illegal positions

24 xx
xx xx
31 xx
xx xx

and

xx xx
x1 4x
x2 3x
xx xx

As for how far I got, I figured out that most legal subcells (but not all) must have 4 legal 90-degree rotations just by drawing them... which is not very far unfortunately... and doesn't help me calculate bigger sets. I'm not very good at this.

#### pka

##### Elite Member

Legal positions

xx xx
x3 4x
x2 1x
xx xx

Illegal positions
xx xx
x1 4x
x2 3x
xx xx
This is one of the most confusing posts I can remember.
Why is one of those legal & one illegal? They look the same to me? Second & third row then the second & third column.

What is this about subcells? Are they all the same? Where does a rotation come into it?

#### Ishuda

##### Elite Member
Just thinking out loud (to make sure I know what the problem is), you 4 cells arranged as
C11 C12
C21 C22
where each cell has 4 positions arranges as a 2X2 matrix, i.e
C:$$\displaystyle \begin{bmatrix} x_{11}& x_{12}\\ x_{21}& x_{22} \end{bmatrix}$$
Choosing cell C11 (without loss of generality) you now number each of those positions, i.e.
C11 = $$\displaystyle \begin{bmatrix} 1& 2\\ 3& 4 \end{bmatrix}$$
Note that there are 24 (=4!) ways of doing this.

The question then is in how many ways can these numbers be redistributed through the cells with the restriction that the number keep its position, i.e. in the example the number 1 must occupy the x11 spot of some cell, the number 2 must occupy the x12 spot, the number 3 the x21 spot and the number 4 the x22 spot.

Each number has four places it can go independent of the other numbers [choose a cell for the number and there is only one place it can go in that cell]. There are four numbers.