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Binomial theorem

  1. Jun 26, 2012 #1
    1. The problem statement, all variables and given/known data

    Screenshot2012-06-26at12743AM.png


    What am I supposed to do with the 3 over 2 in the parentheses? It can be divide and it can be take the factorial. So what do I do with it?
     
  2. jcsd
  3. Jun 26, 2012 #2
    Hi g.lemaitre :smile:

    It is the number of ways of choosing 2 items out of 3 different items. In other words, combinations.

    3C2.
     
  4. Jun 26, 2012 #3
    Hi Infinitum! Call me Georges.

    Does that mean you take 3!/2!? That work for the 3rd and 4th term but not for the second term and for the first term I think it's undefined.
     
  5. Jun 26, 2012 #4
    Okay, Georges then.

    The binomial coefficient is given as,

    [tex]\binom{n}{r} = \frac{n!}{r!(n-r)!}[/tex]

    Where, [itex]0 \leq r \leq n[/itex]

    Why do you think this isn't defined for the first term??
     
  6. Jun 26, 2012 #5
    Man, infinitum, you're such a big number it takes me like forever just to count you.

    I understand the binomial coefficient and can get the right answer for terms 2 3 and 4 but I'm still having trouble with the first term.

    if
    [tex]\binom{n}{r} = \frac{n!}{r!(n-r)!}[/tex]

    then

    [tex]\binom{3}{0} = \frac{3!}{0!(3-0)!} = \frac{6}{0}[/tex]
     
  7. Jun 26, 2012 #6

    SammyS

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    The binary coefficient, [itex]\displaystyle \binom nk[/itex] is defined as follows.

    [itex]\displaystyle \binom nk = \frac{n!}{k!\,(n-k)!}\ , \quad \mbox{for }\ 0\leq k\leq n[/itex]
     
  8. Jun 26, 2012 #7
    0! (zero factorial) is not equal to 0....

    See the summary of this article : https://www.physicsforums.com/showthread.php?t=530207 [Broken]
     
    Last edited by a moderator: May 6, 2017
  9. Jun 26, 2012 #8
    thanks, i got it now.
     
  10. Jun 27, 2012 #9
    zero factorial equals one.
     
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