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Math Help - Need help crafting a Combination formula...

  1. #1
    Oct 2007

    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?
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  2. #2
    Eater of Worlds
    galactus's Avatar
    Jul 2006
    Chaneysville, PA
    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:


    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?.
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  3. #3
    Oct 2007
    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...
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