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Misunderstanding part of pascals identity algebraic proof

  1. Aug 12, 2013 #1
    I understand the combinatorial proof and the common sense behind why it works but lately I am trying to play around with proofs since I am still new to them. So I understand part of this:


    Where I am getting confused is the step where we combine terms. The denominator of the first term is (n-k+1)k!(n-k)! , and the denominator of the second term is k(k-1)!(n-k+1)!

    Then on the next line the denominator is k!(n-k+1)!.

    How is that a common denominator for those two terms? I see there is a (n-k+1)! in both terms.....but in the second term there is a k(k-1)! and in the first term there is a k!(n-k)!....how do those two terms somehow both reduce to k!?
  2. jcsd
  3. Aug 12, 2013 #2


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    It works because of the way the factorial is defined:
    $$k! = k \cdot \underbrace{(k - 1) \cdot (k - 2) \cdots 2 \cdot 1}_{{} = (k - 1)!}$$
    Similarly, you can show that (n - k + 1) * (n - k)! = (n - k + 1)!.

    This is actually the reason for taking the first step: multiplying denominator and numerator by the same number so that the denominators become equal.
  4. Aug 14, 2013 #3
    Ok, that makes sense. But how does [(n-k+1)n!kn!]/[k!(n-k+1)!] simpylify to [(n+1)n!]/[k!((n+1)-k)!]

    Totally lost as to how that step happened
  5. Aug 14, 2013 #4
    For the numerator,
    [tex](n-k+1)n!kn![/tex]can you see that this is:
    [tex][(n-k+1)+k]n![/tex] ?

    ...which equals:
    For the denominator, surely you can see that [itex]n-k+1=(n+1)-k[/itex]
  6. Aug 14, 2013 #5
    How? (n−k+1)n!kn! = [(n-k+1)n!]*kn!, whereas [(n−k+1)+k]n! = [(n-k+1)n!]+kn!

    Yeah, that part is obvious
  7. Aug 14, 2013 #6
    Sorry, I made the mistake of quoting your misquote. :redface:

    The numerator is actually [itex](n-k+1)n!+kn![/itex] which is [itex][(n-k+1)+k]n![/itex] which is [itex](n+1)n![/itex]
  8. Aug 14, 2013 #7
    Ah, I understand now, thank you so much! That was bugging me for hours
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