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

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SUMMARY

The discussion clarifies that Pascal's Triangle is applicable for binomial coefficients in the equation (a+b)^n but not for multinomial expansions like (a+b+c)^n or (a+b+c+...+d)^n. Instead, multinomial coefficients are required for these cases, defined as (i+j+k)!/(i! j! k!) for three variables and generalized for m variables. The conversation emphasizes the distinction between binomial and multinomial coefficients, providing the formula for calculating multinomial coefficients when expanding expressions with multiple terms raised to a power.

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  • Understanding of binomial coefficients and their calculation using Pascal's Triangle.
  • Familiarity with factorial notation and operations.
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  • Study the derivation and applications of multinomial coefficients in combinatorial problems.
  • Explore the concept of generating functions and their relation to polynomial expansions.
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  • Investigate advanced topics in combinatorics, such as the multinomial theorem and its proofs.
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Students and professionals in mathematics, particularly those focusing on combinatorics, algebra, and polynomial expansions, will benefit from this discussion.

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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
||
\/
m numbers.
 
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No, Pascal's triangle give binomial coefficients.

What you need are "multinomial" coefficients.

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

The "trinomial" coefficient for xiyjzk would be (i+j+k)!/(i! j! k!)

If you have "m" numbers to the "n" power: (x1+ x2+...+xm)n then the "multi-nomial" coefficient for x1ixjj...xmk would be

(i+ j+ ...+ k)!/(i! j! ... k!).
 
thanks :smile:
 

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