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Binomial coefficient

  1. Jun 1, 2008 #1
    1. The problem statement, all variables and given/known data

    How is possible this equality:

    [tex] {n \choose k} = \frac{n \cdot (n-1) \cdots (n-k+1)}{k(k-1)...1} = \frac{n!}{k!(n-k)!}[/tex]

    ? I mean where the second part [tex]\frac{n!}{k!(n-k)!}[/tex] comes from?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 1, 2008 #2
    [tex] n! = n \cdot (n-1) \cdot (n-2) \cdot \cdot \cdot (n-k-1) \cdot (n-k-2) \cdot \cdot \cdot 3 \cdot 2 \cdot 1 = n \cdot (n-1) \cdot (n-2) \cdot \cdot \cdot (n-(k-1)) \cdot (n-k)![/tex]

    [tex] k! = k \cdot (k-1) \cdot (k-2) \cdot \cdot \cdot 2 \cdot 1 [/tex]


    So we get:

    [tex] \frac{n!}{(n-k)!} = n \cdot (n-1) \cdot \cdot \cdot n-k+1 [/tex]
     
  4. Jun 1, 2008 #3
    multiply the first bit by (n-k)!/(n-k)! :smile:
     
    Last edited: Jun 1, 2008
  5. Jun 1, 2008 #4
    Ok, I understand about:

    [tex]
    \frac{n!}{(n-k)!} = n \cdot (n-1) \cdot \cdot \cdot n-k+1
    [/tex]
    But still I can't understand why n! is written like that..
     
  6. Jun 1, 2008 #5
    Note that:

    [tex] {n \choose k} \neq \frac{n!}{k!} [/tex]

    The definition is the one in your first post.
     
  7. Jun 1, 2008 #6

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Written like what?

    You started by saying that you understand that
    [tex] {n \choose k} = \frac{n \cdot (n-1) \cdots (n-k+1)}{k(k-1)...1}[/tex]
    but did not understand why
    [tex]\frac{n \cdot (n-1) \cdots (n-k+1)}{k(k-1)...1}= \frac{n!}{k!(n-k)!}[/tex]
    That is what was just explained. It is written "that way" in order to give a simple closed form expression to
    [tex]\frac{n \cdot (n-1) \cdots (n-k+1)}{k(k-1)...1}[/tex]
    "without the dots".
     
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