Art of Problem Solving

Multinomial Theorem: Difference between revisions

1=2 (talk | contribs)
wrong category...
new notation, cleanup, better category
Line 1: Line 1:
The multinomial theorem states that
The '''Multinomial Theorem''' states that


<math>(a_1+a_2+\cdots+a_x)^n=\sum_{k_1,k_2,\cdots,k_x}\binom{n}{k_1,k_2,\cdots,k_x}a_1^{k_1}a_2^{k_2}\cdots a_x^{k_x}</math>
<cmath>(a_1+a_2+\cdots+a_k)^n=\sum_{\substack{j_1,j_2,\ldots,j_k \\ 0 \leq j_i \leq n \textrm{ for each } i \\
\textrm{and } j_1 + \ldots + j_k = n}}\binom{n}{j_1; j_2; \ldots ; j_k}a_1^{j_1}a_2^{j_2}\cdots a_k^{j_k}</cmath>


where
where <math>\binom{n}{j_1; j_2; \ldots ; j_k}</math> is the [[multinomial coefficient]] <math>\binom{n}{j_1; j_2; \ldots ; j_k}=\dfrac{n!}{j_1!\cdot j_2!\cdots j_k!}</math>.


<math>\binom{n}{k_1,k_2,\cdots,k_x}=\dfrac{n!}{k_1!k_2!\cdots k_x!}</math>
Note that this is a direct generalization of the [[Binomial Theorem]]: when <math>k = 2</math> it simplifies to
<cmath>(a_1 + a_2)^n = \sum_{\substack{0\leq j_1, j_2 \leq n \\ j_1 + j_2 = n}} \binom{n}{j_1; j_2} a_1^{j_1}a_2^{j_2} = \sum_{j = 0}^n \binom{n}{j} a_1^j a_2^{n - j}.</cmath>


==Problems==
==Problems==
===Introductory===
===Introductory===
===Intermediate===
===Intermediate===
*The expression  
*The [[expression]]


<math>(x+y+z)^{2006}+(x-y-z)^{2006}</math>
<math>(x+y+z)^{2006}+(x-y-z)^{2006}</math>
Line 25: Line 27:
{{stub}}
{{stub}}
[[Category:Theorems]]
[[Category:Theorems]]
[[Category:Number Theory]]
[[Category:Combinatorics]]

Revision as of 18:02, 29 April 2008

The Multinomial Theorem states that

\[(a_1+a_2+\cdots+a_k)^n=\sum_{\substack{j_1,j_2,\ldots,j_k \\ 0 \leq j_i \leq n \textrm{ for each } i \\ \textrm{and } j_1 + \ldots + j_k = n}}\binom{n}{j_1; j_2; \ldots ; j_k}a_1^{j_1}a_2^{j_2}\cdots a_k^{j_k}\]

where $\binom{n}{j_1; j_2; \ldots ; j_k}$ is the multinomial coefficient $\binom{n}{j_1; j_2; \ldots ; j_k}=\dfrac{n!}{j_1!\cdot j_2!\cdots j_k!}$.

Note that this is a direct generalization of the Binomial Theorem: when $k = 2$ it simplifies to \[(a_1 + a_2)^n = \sum_{\substack{0\leq j_1, j_2 \leq n \\ j_1 + j_2 = n}} \binom{n}{j_1; j_2} a_1^{j_1}a_2^{j_2} = \sum_{j = 0}^n \binom{n}{j} a_1^j a_2^{n - j}.\]

Problems

Introductory

Intermediate

$(x+y+z)^{2006}+(x-y-z)^{2006}$

is simplified by expanding it and combining like terms. How many terms are in the simplified expression?

$\mathrm{(A) \ } 6018\qquad \mathrm{(B) \ } 671,676\qquad \mathrm{(C) \ } 1,007,514\qquad \mathrm{(D) \ } 1,008,016\qquad\mathrm{(E) \ }  2,015,028$

-(2006 AMC 12A Problem 24)

Olympiad

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