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Combination problem

  1. Dec 28, 2003 #1
    Can anyone explain to me why k!/(k!*(k-k)!)+(k+1)!/(k!*(k+1-k)!)+(k+2)!/(k!*(k+2-k)!)+...+(n-1)!/(k!*(n-1-k)!)=n!/((k+1)!*(n-k-1)!) please. Thanks a lot!
     
  2. jcsd
  3. Dec 29, 2003 #2

    Hurkyl

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    Look at the case where n = k+1... then the case where n = k+2...
     
  4. Dec 29, 2003 #3
    logarithm problem help

    can you help me with the four following problems by showing me the right procedures of doing it even though it's so troublesome? thanks alot and i would happily accept any recommended good sites from you guys for this topic.

    1)4(2^2x)=8(2^x)-4
    2)8(2^2x)-10(2^2x)+2
    3)3*2^2x-18(2^x)+24=0
    4)9^x-4(3^x)+3=0
     
  5. Dec 29, 2003 #4
    1)4(2^2x)=8(2^x)-4
    2)8(2^2x)-10(2^2x)+2
    3)3*2^2x-18(2^x)+24=0
    4)9^x-4(3^x)+3=0


    1) Substitute 2^x with t and the solve the quadratic equation

    2) Substitute 2^x with t and then solve the quadratic equation

    3) Substitute 2^x with t and the solve the quadratic equation

    4) Substitute 3^x with t and the solve the quadratic equation
     
  6. Dec 29, 2003 #5
    Hurkyl,

    Sorry, but I didn't quite get where you are going with n=k+1, etc. Could you please explain in more detail?
     
  7. Dec 29, 2003 #6

    NateTG

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    How about this:

    Assume that [tex]n-k > 1[/tex] and simplify:
    [tex]\frac{n!}{(k+1)!(n-k-1)!}-\frac{(n-1)!}{(k+1)!((n-1)-k-1)!}[/tex]

    Then compare it to the terms in your series.
     
  8. Dec 29, 2003 #7
    Now I see, thanks.
     
  9. Dec 29, 2003 #8
    I have just one more question:

    Why does n!/(0!*(n-0)!)+n!/(1!*(n-1)!)+...+n!/(n!*(n-n)!)=2^n ?
     
  10. Dec 30, 2003 #9

    Hurkyl

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    Use the binomial theorem on [itex](1+1)^n[/itex].
     
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