I'm doing a biology lab and was wondering, a basic version is this I can either put 0 1 or 2 of two types of organisms in a test tube and the test tube can then either be put into the dark or the light. I was wondering how many test tubes I will need to conduct this experiment. I'm thinking that you would multiply 3 by 3 by 2 because there are 3 possible number of each organism and their is 2 places they could be put. If someone could explain how to do this type of problem I would be very grateful. I don't actually need to fill all possibilities, just enough to answer some lab questions, which I have answered before I did anything. I was just wondering what the formula would be to find out how many possibilities there are as I like going above and beyond the requirements, and it seems like a fun lab and want to exploit it to the fullest.

Your statement is unclear. Are you mixing organism in test tubes or do have 3 test tubes for each organism separately? Also test tube with 0 is a separate case, since it doesn't matter which organism it is.

Your statement is unclear. Are you mixing organism in test tubes or do have 3 test tubes for each organism separately? Also test tube with 0 is a separate case, since it doesn't matter which organism it is.

 

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

You are correct in that there are 3x3x2=18 possibilities. Your immediate thought is correct. You seem to be looking for something more complicated, but there isn't any.