# Thread: Need help crafting a Combination formula...

1. ## Need help crafting a Combination formula...

This actually isn't for an assignment or anything. I'm just using my free time to try to come up with a formula for something. Here's the lowdown.

After about an hour of trying different things, I was able to create a formula that can be used to determine the # of possible combinations for a set of 2 things such that: One may choose the same thing twice, and the formula accepts things such as A,B and B,A and ONE combination.

Here's what I got: (((X^2)+X)/2)

Now, I'm onto something much more challenging.

I've been trying to create a formula with the same conditions as the first, but with one difference: A set of 3 things.

After a long time of trying various things, I am very close to finding the formula but I just can't seem to get it. I do have some useful data though. Apparently a "long" way of finding the possible combinations is to take your number of types of things "n" and find the "n combinations in a 2-set" and add it to the "(n-1) combinations in a 2-set", and so on until your final term's (n-x) = 1. It's sort of hard to explain, lemme give an example.

Possible combinations in a 2-set with 2 objects, using my previously stated conditions: 3 (AA, AB, BB)

Possible combinations in a 2-set with 1 object: 1 (AA)

Now, I'll list the combinations of a 3-set with 2 objects:
(AAA,AAB,ABB,BBB) Now as you can see there are 4 combinations. Which just so conveniently is the sum of 3(Poss. Combos with 2-set 2-obj.) and 1(Poss. Combos with 2-set 1-obj.)

Now say you were trying to find 3-set 3-obj. It'd just be 2S3O + 2S2O + 2S1O. You just keep adding until your final term is the Poss. Combos of one object.

SO ANYWAY. Does anyone know, or can anyone help me create a formula that can do this? Anyone?

2. If I am understanding you correctly, the number of ways to arrange 2 objects 3 at a time is $2^{3}=8$, not 4.

You are using A and B:

AAA, BBB, BBA, BAB, ABB, AAB, ABA, BAA

If you want to see how many are possible using 3 different items 3 at a time is: $3^{3}=27$

And so on.

How many ways can you arrange 2 objects 10 at a time: $2^{10}=1024$

Is that what you were getting at?.

3. Okay, I can understand that.

But that's not what I'm getting at.

2^3 = 8, yes, there are 8 combinations.

But my formula counts things such as ABA and AAB and BAA as the same combinations. I'm trying to make a formula for a set of 3 that tell every Completely Different combination.

If that helps... yea, it's sort of confusing...