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Coefficients of trinomial theorem

  1. Nov 18, 2013 #1
    There is a trinomial theorem relationed with the Pascal's tetrahedron, so that...

    [tex](x+y+z)^5=[/tex]
    [tex]
    \\ +01x^5y^0z^0+05x^4y^1z^0+10x^3y^2z^0+10x^2y^3z^0+05x^1y^4z^0+01x^0y^5z^0
    \\ +05x^4y^0z^1+20x^3y^1z^1+30x^2y^2z^1+20x^1y^3z^1+05x^0y^4z^1
    \\ +10x^3y^0z^2+30x^2y^1z^2+30x^1y^2z^2+10x^0y^3z^2
    \\ +10x^2y^0z^3+20x^1y^1z^3+10x^0y^2z^3
    \\ +05x^1y^0z^4+05x^0y^1z^4
    \\ +01x^0y^0z^5
    [/tex]

    Well, when it comes the binomial coefficients, they are easily determined so:

    [tex]\binom{5}{0}...\binom{5}{1}...\binom{5}{2}...\binom{5}{3}...\binom{5}{4}...\binom{5}{5}[/tex]
    But this sequence is for a linear development (binomial theorem). And when it comes to a bilinear development (trinomial theorem), how will be the coefficients? Certainly, there will a product between binomial numbers, but how will be such scheme?
     
  2. jcsd
  3. Nov 18, 2013 #2

    tiny-tim

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    Hi Jhenrique! :wink:

    The coefficient of xaybzc wil be the number of ways of choosing a of this b of that and c of the other out of a+b+c,

    which is … ? :smile:
     
  4. Nov 18, 2013 #3
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