Number Of Possibilities?

[davidandkaze]

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.

[Don Blazys]

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.

 

You will need a total of 8 test tubes.

 

2 for testing what happens when no organism is placed in dark and light areas,

 

2 for testing what happens when organism A is placed in dark and light areas,

 

2 for testing what happens wnen organism B is placed in dark and light areas, and

 

2 for testing what happens when a mixture of organisms A and B is placed in dark and light areas.

 

Don.

[Qfwfq]

No, if he wants all combinations it is 14.

 

Clearly light vs. dark doubles them. The number of species is irrelevant for a count of zero ([imath]2^0=1[/imath]), for a count of 1 it adds [imath]2=2^1[/imath] while the count of 2 adds [imath]4=2^2[/imath] more, totalling seven which then doubles into 14 QED. Seven of course is three ones in binary.

[JMJones0424]

Qfwfq- I don't understand how you came to an answer of 14. What are the seven combinations that are doubled?

I count (A or NOT A) x (B or NOT B)x (LIGHT or DARK) = 8

[Qfwfq]

Actually my only flaw, which hit me after posting, was in distinguishing the case of A + B from B + A, which means I should have said twice six is twelve, but I must have thought that when the count is two one must be male and the other female, so they could produce either a mule or a hinney. :doh: :lol:

 

More seriously, perhaps I'm the only one to have thought this guy means 0, 1 or 2 individuals (not quite too microscopic) but of course maybe you two guessed it better and, if he meant microorgansims, there's no point counting the case of two acquatic bugs of the same species. :shrug:

[davidandkaze]

There are I think nine possible permutations(P) (I learned a new word when googling how to do this for the millionth time) 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

Though based on what you guys are saying I may be way off. I can't seem to find where or how I got those extra P, I don't see them anywhere but maybe I'm just going blind. Let me know what I did wrong.

[davidandkaze]

By the way the what type of org these things are seems to be somewhat of a big deal so, the first one I have to add is some fresh water snails(sn)(I think, not quite sure what type of sn they are, all I know is they're small and they do well in fresh water) and the second org is elodea(e). The lab is supposed to demonstrate the cycle of O2 and CO2, thus the lit factor, for photosynthesis. I'm supposed to put this BTB(possibly TBT but I think it's BTB it changes colors when it comes in contact with CO2)solution to test tubes and put a varying amount of sn and e in a number of test tubes and see how the amount of sn and e affects CO2 levels.

[JMJones0424]

Your matrices would be correct if order mattered, and if you were to include two quantities of the same organism as a permutation. So, re-writing your G matrices, lets say 0= no organism, 1 = snail, and 2 = elodea.

 

G1

0,0

0,1

0,2

 

G2

1,0 same as 0,1 in G1

1,1 * Is this a valid test? Need clarification from your teacher if you are intended to test two quantities of snails. If not, this is same as 0,1 in G1

1,2

 

G3

2,0 same as 0,2 in G1

2,1 same as 1,2 in G2

2,2 * Like 1,1 are two quantities of elodea an intended test?

 

Because order does not matter, you are left with either four test tubes in each of light and dark, or six (if the double quantities are intended). When doubled for light/dark, this gives eight or twelve tests.

 

I suspect you will be using BTB, and I hope you will also be using distilled water. Planted aquarium enthusiasts are familiar with the intertwined measurements of dissolved CO₂, pH, CO₃^2-, and HCO₃^-. It may be too advanced for your situation, but here's a link that goes over the equilibrium reactions: http://www.chem.usu.edu/~sbialkow/Classes/3600/Overheads/Carbonate/CO2.html . The presence of carbonates in the water will skew your pH measurements, but using distilled water will prevent this.

[CraigD]

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.

I’m thinking you’re right, David.

 

Using () to show a test tube, A and B to show individual organisms, and * to indicate light, you’d have 18 test tubes as follow:

() (A) (AA) (B) (AB) (AAB) (BB) (ABB) (AABB) *() *(A) *(AA) *(B) *(AB) *(AAB) *(BB) *(ABB) *(AABB)

 

You already know how to simply count the number of test tubes needed – multiply the number of allowed numbers of organism A (3) by that of B (3) by the number of allowed lights on/off conditions (2), to get 3x3x2=18. All you need to do to produce a list like this is something like this little MUMPS program,

for L="","*" for B="","B","BB" for A="","A","AA" write L,"(",A,B,") "

which actually produced the list above.

 

You can produce lists using this approach for practically and kind of rules, even complicated one like “you can have 0 to 3 of 3 organisms, but no organism can outnumber the sum of the counts of others unless the lights are off”. :)

 

Previous responses seem to me to be overcomplicating this, because due to the special way the question is stated, this isn’t a permutation or combination problem at all. Because its stated as counts of kinds of items, it’s a simple counting problem.

[JMJones0424]

I'm glad Craig's reading comprehension was sufficient to notice what the rest of us missed. I simply misread the original post and I hope the confusion I helped to cause wasn't too great.

[Don Blazys]

The initial question was stated as follows:

 

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.

 

Since neither "types of organisms" nor "quantities" are specified,

we must (if we are to be logical) assume "generality" and

interpret the above question thusly:

 

We will be given a quantity of organism A and a quantity of organism B.

Then, either 0, 1 or 2 samples will be put in a test tube,

which will then be placed in either a dark or a light area.

How many test tubes will be needed to test what happens

to organism A, organism B and a mixture of both?

 

Now, 1 sample means 1 sample regardles of how large that sample is.

 

Thus, the only logical answer to that question is 8 test tubes.

 

Had the initial question been:

 

If I put either 0, 1 or 2 animals in a test tube, or 0, 1 or 2 plants in a test tube,

and place that test tube in either a dark or light location,

then how many test tubes will I need to cover all possibilities?

 

then the answer may be construed as 18,

but 10 of those possibilities will still be "redundant"

in the sense that whatever occurs to 1 animal or plant in

a dark and gloomy test tube will probably occur to

2 animals or plants in a dark and gloomy test tube.

 

If the experiment is meant to determine whether or not

1 animal or plant in a dark and gloomy test tube will become lonely and die

while 2 animals or plants in a dark and gloomy test tube might

give each other the moral support, strength and courage to carry on,

then such redundancy might still have a purpose.

 

Otherwise, it's just an unnecessary waste of animals and plants.

 

Don.

[Rade]

You will need 12 test tubes with no replicates to test all possibilities of placing two types for two organisms. Let T=empty test tube, water only, S=snail plus water, E=elodea plus water, L=light, D=dark :

 

1. T-L

2. T-D

3. S-L

4. S-D

5. E-L

6. E-D

7. S&E-L

8. S&E-D

9. S&S-L

10. S&S-D

11. E&E-L

12. E&E-D

 

But, of course, there must always be replicates in any scientific experiment to allow for statistical analysis of data.

[JMJones0424]

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.

 

 

Please reread the opening post. Davidandkaze had it correct from the beginning. There are three possible quantities of two different organisms in two lighting conditions.

 

Craig has listed correctly all of the permutations.

 

(0, 1, or 2 snails) x (0, 1, or 2 elodea) x (dark or light) = 18

[Rade]

Please reread the opening post. Davidandkaze had it correct from the beginning. There are three possible quantities of two different organisms in two lighting conditions.Craig has listed correctly all of the permutations.(0, 1, or 2 snails) x (0, 1, or 2 elodea) x (dark or light) = 18

The 0 snail and 0 elodea permutations do not require two different test tubes, they are two test tubes for the category "without any types of organisms", one placed in the light, one in the dark. Thus, there are not 18 test tubes needed, only 12. You do not put two test tubes with no organisms in the light (and dark) and say, one is for the permutation of 0 snail and the second is for the permutation 0 elodea. The correct formula is that there are 6 organism diversity possibilities x 2 light conditions, 6 x 2 = 12 test tubes needed (see my post above for the 6 organism diversity possibilities for a situation requiring 0,1,2 quantities of organisms).

[JMJones0424]

What? The duplicates you complain of do not exist. As written, the test requires 18 test tubes. We can certainly disagree with the necessity of some of the tests, or the need for repetitive tests, but that doesn't change the answer to the question asked.

 

If Craig's notation doesn't float your waterweed, I'll use yours.

 

1. T-L

2. T-D

3. S-L

4. S-D

5. E-L

6. E-D

7. S&E-L

8. S&E-D

9. S&S-L

10. S&S-D

11. E&E-L

12. E&E-D

---------

13. S&S&E-L

14. S&S&E-D

15. S&S&E&E-L

16. S&S&E&E-D

17. E&E&S-L

18. E&E&S-D

 

Total number of possible permutations with no duplicates remains (3 snail values) x (3 elodea values) x (2 lighting values) = 18

[Rade]

---------

13. S&S&E-L

14. S&S&E-D

15. S&S&E&E-L

16. S&S&E&E-D

17. E&E&S-L

18. E&E&S-D

OK, makes sense under your interpretation of "two types of organisms". My interpretation was that you could put up to 2 individuals of "two types of organisms", e.g., either a snail or elodea in any test tube. That is, never more than 2 organisms (one of each type) in any one tube. Under this interpretation only my first 12 tube categories needed.

[Don Blazys]

Here is the original question.

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.

 

Surely, that means:

 

Of two types of organisms, I can put either 0, 1 or 2 in a test tube

and the test tube can then be put either in the dark or the light.

I was wondering how many test tubes I will need to conduct this experiment.

 

In other words, given two types of organisms, he can put either 0, 1 or two types in a test tube.

 

Clearly, in a scientific experiment, the emphasis must be on the types of organisms

 

and not on the number of individual organisms!

 

Don't you see... any answer other than 8 test tubes implies that we are studying individual snails!

 

Next thing you know, we will be adopting them as pets and naming them!

 

Don.