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Math Help - combination

  1. #1
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    combination

    why is "nCk + nCk-1" equal to " n+1Ck " ?

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  2. #2
    Senior Member jakncoke's Avatar
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    Re: combination

    Algebriacally, well Assume  k \geq 1
     n C k + n C (k-1) = \frac{n!}{(n-k)!k!} + \frac{n!}{(k-1)!(n-k+1)!}

    Observe that  (n-k+1)! = (n-k)!*(n-k+1) (1)

    so  \frac{n!}{(n-k)!k!} + \frac{k * n! }{k*(k-1)!(n-k+1)!} (i can do this since k/k = 1)

     \frac{n!}{(n-k)!k!} + \frac{k * n! }{k!(n-k)!(n-k+1)} (im using (1) )

    multiply the first equation by  \frac{(n-k+1)}{(n-k+1)} to get both denominaters equals

     \frac{(n-k+1)*n!}{(n-k+1)*k!(n-k)!} + \frac{k*n!}{k!(n-k)!*(n-k+1)} = \frac{(n-k+1)*n! + kn!}{k!(n-k)!(n-k+1)}

    which equals  \frac{n!( (n-k+1) + k )}{k!(n-k)!(n-k+1)} =\frac{n!*(n+1)}{k!(n-(k+1))!} = \frac{(n+1)!}{k!*(n+1-k)!}

    which is n+1 C k.
    Last edited by jakncoke; February 2nd 2013 at 01:06 AM.
    Thanks from miki
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