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Trinomial and Multinomial theorem

  1. Oct 13, 2011 #1
    I can't understand the sum notation shown in Wikipedia or in this article:

    http://mathforum.org/library/drmath/view/53159.html

    I want to find the sum notation for (a+b+c)^n
    however I can't understand the sum notation:

    a3b5ea89c1417ff4ef0c05ea0b7a43a6.png

    I don't understand the use of brackets or what they mean here and in the binomial theorem.

    I think they are what Wikipedia calls the binomial coefficient:

    http://en.wikipedia.org/wiki/Binomial_theorem

    I also don't understand, is that they are taking the some from i,k,j to what? Shouldn't their be a number on top of the sigma?

    I would appreciate the help!
     
  2. jcsd
  3. Oct 13, 2011 #2
    The notation here is a so-called multinomial coefficient, which is a generalization of a binomail coefficient. The definition is

    [tex]\binom{n}{a,b,c}=\frac{n!}{a!b!c!}[/tex]

    A binomial coefficient

    [tex]\binom{n}{k}[/tex]

    then equal the multinomial coefficient

    [tex]\binom{n}{k,n-k}[/tex]
     
  4. Oct 13, 2011 #3
    Adding to Micromass's remarks, the summation is meant to be over all triples i, j, k where [itex]i \ge 0, j \ge 0, k \ge 0[/itex] and [itex]i+j+k=n[/itex]
     
  5. Oct 16, 2011 #4
    Okay, thanks but I still don't understand the multi variable summation notation-

    What does the i,j,k under the sigma represent and why is there nothing on "top" of the sigma?

    Does this notation also assume that:

    "the summation is meant to be over all triples i, j, k where i≥0,j≥0,k≥0 and
    i+j+k=n"

    are must this be stated separately from the Summation?
     
  6. Oct 18, 2011 #5
    There should be something in the text saying that the summation is over i,j,k where i+j+k=n. Sometimes you will see i+j+k=n written under the summation symbol instead.
     
  7. Dec 3, 2011 #6
    It does'nt matter for i+j+k=n is over or under the summation symbol. It means how many ways to make i+j+k=n, where i,j,k≥ 0.
    For example: (a+b+c)3
    i+j+k=3
    How many ways to make i+j+k=3?
    3+0+0=3 => i=3,j=0,k=0
    0+3+0=3 => i=0,j=3,k=0
    0+0+3=3 => i=0,j=0,k=3
    1+1+1=3 => i=1,j=1,k=1
    2+1+0=3 => i=2,j=1,k=0
    2+0+1=3 => i=2,j=0,k=1
    1+2+0=3 => i=1,j=2,k=0
    1+0+2=3 => i=1,j=0,k=2
    0+2+1=3 => i=0,j=2,k=1
    0+1+2=3 => i=0,j=1,k=2
     
  8. Dec 4, 2011 #7
    Another way to expand (a+b+c)n is using Pascal's Triangle.
     
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