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Binomial Proof

  1. Jul 17, 2008 #1
    1. The problem statement, all variables and given/known data
    Let l, m, and n be positive integers with l [tex]\leq[/tex] m and l [tex]\leq[/tex] n. Prove the identity.
    ([tex]\stackrel{m + n}{l}[/tex]) = ([tex]\stackrel{m}{0}[/tex])([tex]\stackrel{n}{l}[/tex]) + ([tex]\stackrel{m}{1}[/tex])([tex]\stackrel{n}{l-1}[/tex])+...+([tex]\stackrel{m}{l}[/tex])([tex]\stackrel{n}{0}[/tex])

    2. The attempt at a solution
    I have no clue, I see proof and my brain goes dead. I thought I could just start writing the definition of the parts out on both sides and maybe something would make sense but it got messy quick and I didn't see any light at the end of that tunnel.
  2. jcsd
  3. Jul 18, 2008 #2


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    Homework Helper

    How about a proof by induction? The integers l, m, and n have to be positive, so we'd start from l = 1:

    (\stackrel{m+n}{1}) = (\stackrel{m}{0})(\stackrel{n}{1}) + (\stackrel{m}{1})(\stackrel{n}{0}) = m + n

    which plainly works. (It doesn't work for l = 0, but it doesn't have to, under the specified conditions.)

    Now assume the proposition

    ([tex]\stackrel{m + n}{l}[/tex]) = ([tex]\stackrel{m}{0}[/tex])([tex]\stackrel{n}{l}[/tex]) + ([tex]\stackrel{m}{1}[/tex])([tex]\stackrel{n}{l-1}[/tex])+...+([tex]\stackrel{m}{l}[/tex])([tex]\stackrel{n}{0}[/tex]) .

    What happens when we advance to the case for ([tex]\stackrel{m + n}{l+1}[/tex]) ?
    Last edited: Jul 18, 2008
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