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Calcmathlete

  • 3 years ago

Very confused on these particular problems...no clue what it's talking about...

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  1. Calcmathlete
    • 3 years ago
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  2. Calcmathlete
    • 3 years ago
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    @ganeshie8 I figured out C(x), but I don't get P(x)

  3. ganeshie8
    • 3 years ago
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    For 16th question, i think P(X) = 4C2

  4. hartnn
    • 3 years ago
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    the two element group...say there are 4 elements,ABCD,there will be 16 2 element grp,AA,AB,AC,....DD....but if repetition like AA are not allowed,then there will be 12...and if AB and BA are considered same ,then there will be 6....

  5. Calcmathlete
    • 3 years ago
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    I am getting B for the first one.

  6. Calcmathlete
    • 3 years ago
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    Because I did xP2 = P(x) and (n - 1)! = C(x)

  7. Calcmathlete
    • 3 years ago
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    Or would it be xC2 = P(x)?

  8. ganeshie8
    • 3 years ago
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    oh i see... but i guess forming 2 element groups involves just choosing elements, order doesnt matter...

  9. Calcmathlete
    • 3 years ago
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    Oh ok...then I get D for 16 and B for 17. I ddin't realize that, I was still thinking about the other part of the problem.

  10. hartnn
    • 3 years ago
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    if order doe not matter,then AB and BA are same,then use xC2

  11. Calcmathlete
    • 3 years ago
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    Ok, then it would be: 16. D 17. B ?

  12. ganeshie8
    • 3 years ago
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    yeah i guess that it.... .

  13. Calcmathlete
    • 3 years ago
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    Alright. Thanks :)

  14. ganeshie8
    • 3 years ago
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    np... :)

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