Difference between revisions of "Multinomial Theorem"
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<cmath>(x_1 + x_2 + x_3 + ... + x_{k-1} + x_k)^n = \sum_{b_1 + b_2 + b_3 + ... + b_{k-1} + b_k= n}{\binom{n}{b_1, b_2, b_3, ..., b_{k-1}, b_k} \prod_{j=1}^{k}{x_j^{b_j}}}</cmath> | <cmath>(x_1 + x_2 + x_3 + ... + x_{k-1} + x_k)^n = \sum_{b_1 + b_2 + b_3 + ... + b_{k-1} + b_k= n}{\binom{n}{b_1, b_2, b_3, ..., b_{k-1}, b_k} \prod_{j=1}^{k}{x_j^{b_j}}}</cmath> | ||
− | + | <math>\bf Proof:</math> | |
When <math>k=1</math> the result is true, and when <math>k=2</math> the result is the binomial theorem. Assume that <math>k \ge 3</math> and that the result is true for <math>k=p</math> When <math>k=p+1</math> | When <math>k=1</math> the result is true, and when <math>k=2</math> the result is the binomial theorem. Assume that <math>k \ge 3</math> and that the result is true for <math>k=p</math> When <math>k=p+1</math> | ||
<cmath>(x_1 + x_2 + x_3 + ... + x_{p-1} + x_p)^n = (x_1 + x_2 + x_3 + ... + x_{p-1} + (x_p +x_{p+1})^n</cmath> | <cmath>(x_1 + x_2 + x_3 + ... + x_{p-1} + x_p)^n = (x_1 + x_2 + x_3 + ... + x_{p-1} + (x_p +x_{p+1})^n</cmath> |
Revision as of 23:53, 8 May 2019
The Multinomial Theorem states that where is the multinomial coefficient .
Note that this is a direct generalization of the Binomial Theorem: when it simplifies to
Contents
[hide]Proof
Proof by Induction
Proving the Multinomial Theorem by Induction
For a positive integer and a non-negative integer ,
When the result is true, and when the result is the binomial theorem. Assume that and that the result is true for When Treating as a single term and using the induction hypothesis: By the Binomial Theorem, this becomes: Since , this can be rewritten as:
Combinatorial proof
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Problems
Intermediate
- The expression
is simplified by expanding it and combining like terms. How many terms are in the simplified expression?
(Source: 2006 AMC 12A Problem 24)
Olympiad
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