Are you good at math (or maybe not math..)? I need some help!

Discussion in 'General Chat' started by NwS, Feb 12, 2009.

  1. #1
    Heya,

    So i am trying to figure out how many combinations i can have from 33 objects if i separate them into 10 size groups. The order doesn't matter.

    So for example i could have:
    1st option - obj1, obj2, obj3, obj4, obj5, obj6, obj7, obj8, obj9, obj10
    2nd option - obj1, obj2, obj3, obj4, obj5, obj6, obj7, obj8, obj9, obj11
    etc etc

    wrong option would be - obj10, obj2, obj3, obj4, obj5, obj6, obj7, obj8, obj9, obj1 (since it is the same like 1st option but in a different order)

    Any help would be really appreciated! :)

    Thanks!
     
    NwS, Feb 12, 2009 IP
  2. krolbiz

    krolbiz Peon

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    #2
    I think it's not impossible for someone to succeed in internet marketing although you're not expert in math...
     
    krolbiz, Feb 12, 2009 IP
  3. gr8liverpoolfan

    gr8liverpoolfan Notable Member

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    #3
    How many objects are there in each size group?

    Correct me if I'm wrong, but you don't have any restriction on the way the groups are arranged right?

    The object thingy didn't make much sense to me- you could arrange it as Obj 1, 2, 3, 4 ..9, 11 but can't arrange as Obj 10, 1, 2, 3, 4?
     
    gr8liverpoolfan, Feb 12, 2009 IP
  4. NwS

    NwS Well-Known Member

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    #4
    Thank you for your time :p
    Ok so we got a total of 33 objects. No restriction on the way the groups are arranged you are correct.

    You can arrange it as obj1, 2, 3, 4, 5 .., 10 you can also arrange it as obj10, 1, 2, 3, 4.., 9 (you have exactly the same objects and that's why i said this won't count as a different combination, each combination must be unique thats the key here).

    I hope you can follow me here :p almost 5am so yeah :/


    PS: The first reply was spam?.. :/
     
    NwS, Feb 12, 2009 IP
  5. ActiveFrost

    ActiveFrost Notable Member

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    #5
    How many objects allowed ? I mean - I can add different count of objects in each variation ( like 4 in the first, 6 in the second, etc. ), or for example, 1 option contains 10 objects ?
     
    ActiveFrost, Feb 12, 2009 IP
  6. NwS

    NwS Well-Known Member

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    #6
    Each combination must contain 10 objects in it. :)
     
    NwS, Feb 12, 2009 IP
  7. gr8liverpoolfan

    gr8liverpoolfan Notable Member

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    #7
    If each combination was unique ( meaning not more than one object in one group), then this could be your solution.

    There are 33 objects.

    To make a group of 10, you have 33 choices for your first object ( anything between 1 and 33)

    For your second object in your group of 10 ( you have a choice of 32 objects- because the first object cannot be repeated again)

    So your answer should be 33*32*31*30*29*28*27*26*25*24 ( Going so on and so forth till 10 objects)?
     
    gr8liverpoolfan, Feb 12, 2009 IP
  8. NwS

    NwS Well-Known Member

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    #8
    Each combination is unique (no less or more than 10 objects per group though).

    If each object was a group i think it would be 33*33 hmm :/

    Example:
    123456789 10
    123456789 11
    123456789 12
    etc etc
    123456789 33
    13456789 10 11
    23456789 10 11

    10 123456789 = wrong, this is not a unique combination (same combination as the first one in a different order)


    PS: Sorry for being a pain..

    PS2: 33*33*10 ? :/
     
    NwS, Feb 12, 2009 IP
  9. NwS

    NwS Well-Known Member

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    #9
    Is it a simple 33*10 or is it just my idea now?
     
    NwS, Feb 12, 2009 IP
  10. NwS

    NwS Well-Known Member

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    #10
    Dang you are correct so i got 92,561,040 results lol!! :/
    Oh well.. Ty for the help! Rep added! :)
     
    NwS, Feb 13, 2009 IP