Multinomial Theorem

Revision as of 13:44, 10 October 2017 by Shreyj (talk | contribs) (Proof)

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

Proof by Induction

Combinatorial proof

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Problems

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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