# Difference between revisions of "Multinomial Theorem"

(→Proof by Induction) |
Twod horse (talk | contribs) m |
||

(2 intermediate revisions by 2 users not shown) | |||

Line 18: | Line 18: | ||

<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> | ||

Line 29: | Line 29: | ||

=== Combinatorial proof === | === Combinatorial proof === | ||

− | |||

{{stub}} | {{stub}} | ||

Line 45: | Line 44: | ||

===Olympiad=== | ===Olympiad=== | ||

− | |||

− | |||

− | |||

[[Category:Theorems]] | [[Category:Theorems]] | ||

[[Category:Combinatorics]] | [[Category:Combinatorics]] | ||

[[Category:Multinomial Theorem]] | [[Category:Multinomial Theorem]] |

## Latest revision as of 02:59, 12 February 2021

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

## 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

*This article is a stub. Help us out by expanding it.*

## 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)