There are I think nine possible permutations(P) of test tubes without the variable of the lighting (lit) represented in the following matrix. column a represents first organism (org) column b represents second org I split it up into three groups(G) representing the value of three possibilities(pos) of the first org there are then three ordered pairs in each G based on the three pos of the second org.
A,B
G 1
0,0
0,1
0,2
G 2
1,0
1,1
1,2
G 3
2,0
2,1
2,2
To add to this matrix I added the third variable of lit I added sets (S) in between the individual level and the G level (Oh and if someone could tell me what the technical term for the individual level is I would really appreciate it)
A,B,C
G 1
S 1
0,0,0
0,1,0
0,2,0
S 2
0,0,1
0,1,1
0,2,1
G 2
S 1
1,0,0
1,1,0
1,2,0
S 2
1,0,1
1,1,1
1,2,1
G 3
S 1
2,0,0
2,1,0
2,2,0
S 2
2,0,1
2,1,1
2,2,1
I think I got it right but the basic math behind what I did (I think) is org 1 pos * org 2 pos * lit or in simple terms G 1 * G 2 * G 3= total # of P