I have to find ALL possible combinations of the following.

I have 8 questions and all 8 can have an answer of "A" (Any) or "V" (Value) only.

So

1. (A or V)
2. (A or V)
3. (A or V)
4. (A or V)
5. (A or V)
6. (A or V)
7. (A or V)
8. (A or V)

I need all possible combinations that could be chosen using the 8 questions.
So all 8 answers could be A or all could be V and need every other combination possible in between i.e 4. = V and the rest are A OR 7,8 = A and rest are V etc. There are obviously quite a few :)

1. Probably be good to know how many possible combinations there are
2. What all the combinations are

How many possible combinations are there for one question? What are they?
How many possible combinations are there for two questions?
See any pattern yet?

So where you have "V" it will actually have a real value of 0-9?
Then each of your 8 questions will have a possible 11 answers
If you have more values for "V" then you'll still have that number plus 1 answers for each.
But if the value of "V" is unlimited then your possible outcomes will be unlimited also.
How does that fit your problem?

0

ALTO999Author Commented:

Please read the question.

There are only 8 questions and only have 2 possible answers.

Its ALL possible combinations possible as per question explains.

so some examples of the MANY.

1.A
2.A
3.A
4.A
5.A
6.A
7.A
8.V

OR

1.A
2.A
3.V
4.A
5.A
6.A
7.A
8.V

There are lots of the above its knowing ALL of them and how to work out ALL of them.

0

GDPR? That's a regulation for the European Union. But, if you collect data from customers or employees within the EU, then you need to know about GDPR and make sure your organization is compliant by May 2018. Check out our preparation checklist to make sure you're on track today!

calculating is the easy part, drawing all of them out is the pain.

you have the important data, knowing that each of the eight questions have two possible answers. so now, account for that... remember that how you answer one question has no bearing on any of the others.

so in this case, you figure out the number of possible combinations for each question, then multiply all of those numbers. ( combinations for #1 x combinations for #2 x ... )

if you are familiar with decision trees, that would make a good way for you to draw out the combinations and also make it easy to count. here is a wikipedia link, just let me know if you feel that it does not clarify it enough: http://en.wikipedia.org/wiki/Decision_tree

If you had one question with 2 possible answers you have 1 to the second = 2 or 1 squared
If you had two questions with 2 possible answers you have 2 to the second = 4 or 2 squared.

Since you have 8 questions with 2 possible answers you have 8 to the second = 64 or 8 squared.

if you had three possible answers it would be 8 to the 3rd, or 8 x 8 x 8 = much bigger number.

Ha! This is so simple... you have 256 possibilities there (2 to the power of 8).
All possible combinations are drawn out by counting up from 0 to 255 in binary, like this:

00000000
00000001
00000010
00000011
...

Just replace (0,1) with whatever your two possibilities are.
You can do that in just about any programming or scripting language.

0

ALTO999Author Commented:

Ha! This is so simple... you have 256 possibilities there (2 to the power of 8).
All possible combinations are drawn out by counting up from 0 to 255 in binary, like this:

00000000
00000001
00000010
00000011
...

Just replace (0,1) with whatever your two possibilities are.
You can do that in just about any programming or scripting language.

anyone know how to do this in excel??

0

ALTO999Author Commented:

LESS COMBINATIONS

Eight bits have 256 combinations, but if you know that eight out of eight bits are zero, than there's just one combination possible: 00000000. Thus knowing how many bits are 0 and 1, can reduce the amount of possible combinations extremely. The more information there is about a data block, the less combinations are possible. Not just knowing how many zeros and ones there are but knowing more; how many groups of each: 00, 01, 10, and 11 there are (or larger groups), reduces the amount of possible combinations even more.

Next are all 256 combinations of 8 bits, in order of the ratio 1/0

So... what you are arriving at is that what you needed was not all possible combinations (that was answered in my 1st reply), but the number of possible ocurrences of each element (1/0, A/V, or whatever) in a group of 8, regardless of order.

0

Featured Post

Your question, your audience. Choose who sees your identity—and your question—with question security.

How many possible combinations are there for two questions?

See any pattern yet?