Difference between revisions of "Multinomial Theorem"

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(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}
 
(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}
 
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== Proof ==
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=== Using [[induction]] and the Binomial Theorem ===
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{{incomplete | section}}
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=== Combinatorial proof ===
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{{incomplete | section}}
  
 
==Problems==
 
==Problems==
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[[Category:Theorems]]
 
[[Category:Theorems]]
 
[[Category:Combinatorics]]
 
[[Category:Combinatorics]]
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[[Category:Algebra]]

Revision as of 10:50, 30 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 \[(a_1 + a_2)^n = \sum_{\substack{0\leq j_1, j_2 \leq n \\ j_1 + j_2 = n}} \binom{n}{j_1; j_2} a_1^{j_1}a_2^{j_2} = \sum_{j = 0}^n \binom{n}{j} a_1^j a_2^{n - j}\]

Proof

Using induction and the Binomial Theorem

Template:Incomplete

Combinatorial proof

Template:Incomplete

Problems

Introductory

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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?

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

(Source: 2006 AMC 12A Problem 24)

Olympiad

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