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Math Help - n choose k proof

  1. #1
    Newbie
    Joined
    Sep 2010
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    1

    n choose k proof

    Hi, I have to hand in my homework in two days and I have no idea how to prove the following (freely translated, English is not my native language so hope it makes sense):

    "Show that for all positive whole numbers n and k, when n is bigger than or equal to k, the following applies:

    (n
    choose (k-1))+(n choose k)=((n+1) choose k)

    My textbook uses the word choose, not sure if that's universal so "n choose k"=n!/k!(n-k)! so you know what I'm talking about.

    Any help greatly appreciated!
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  2. #2
    Senior Member
    Joined
    Oct 2009
    Posts
    340
    Just calculate the sum directly. Get a common denominator using the recursion identity j! = j (j - 1)!. Everything falls out in two lines at the most.
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