How does Pascal triangle apply to (a+b+c)^n and (a+b+c+...+d)^n?

[MathematicalPhysicist]

does pascal triangle use in this equation (a+b+c)^n i know it is used in (a+b)^n?

and how could you solve for m number of numbers to the power n?
(a+b+c+...+d)^n
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m numbers.

 

[HallsofIvy]

No, Pascal's triangle give binomial coefficients.

What you need are "multinomial" coefficients.

The binomial coefficients are given by _nC_m= n!/(m!(n-m)!) because there are that many ways of arranging m x's and n-m y's to give the product x^my^n-m.

The "trinomial" coefficient for xⁱy^jz^k would be (i+j+k)!/(i! j! k!)

If you have "m" numbers to the "n" power: (x₁+ x₂+...+x_m)ⁿ then the "multi-nomial" coefficient for x₁ⁱx_j^j...x_m^k would be

(i+ j+ ...+ k)!/(i! j! ... k!).

 

[MathematicalPhysicist]

thanks