Multinomial Theorem: Difference between revisions

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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

Proving the Multinomial Theorem by Induction

For a positive integer $k$ and a non-negative integer $n$ , $(x_1 + x_2 + x_3 + ... + x_{k-1} + x_k)^n = \sum_{b_1 + b_2 + b_3 + ... + b_{k-1} + b_k= n}{\binom{n}{b_1, b_2, b_3, ..., b_{k-1}, b_k} \prod_{j=1}^{k}{x_j^{b_j}}}$

[b]Proof:[/b] When $k=1$ the result is true, and when $k=2$ the result is the binomial theorem. Assume that $k \ge 3$ and that the result is true for $k=p$ When $k=p+1$ $(x_1 + x_2 + x_3 + ... + x_{p-1} + x_p)^n = (x_1 + x_2 + x_3 + ... + x_{p-1} + (x_p +x_{p+1})^n$ Treating $x_p + x_{p+1}$ as a single term and using the induction hypothesis: $\sum_{b_1 + b_2 + b_3 + ... + b_{p-1} + B = n}{\binom{n}{b_1, b_2, b_3, ..., b_{p-1}, B} \cdot (x_p + x_{p+1})^B \cdot \prod_{j=1}^{p-1}{x_j^{b_j}}}$ By the Binomial Theorem, this becomes: $\sum_{b_1 + b_2 + b_3 + ... + b_{p-1} + B = n}{\binom{n}{b_1, b_2, b_3, ..., b_{p-1}, B}} (\prod_{j=1}^{p-1}{x_j^{b_j}}) \cdot \sum_{b_p + b_{p+1} = B}{\binom{B}{b_p} \cdot x_p^{b_p} x_{p+1}^{b_{p+1}}}$ Since $\binom{n}{b_1, b_2, b_3, ... ,b_p, B}\binom{B}{b_p} = \binom{n}{b_1, b_2, b_3, ... ,b_p, b_{p+1}}$ , this can be rewritten as: $\sum_{b_1 + b_2 + ... b_p + b_{p+1}= n}{\binom{n}{b_1, b_2, b_3, ..., b_p, b_{p+1}}\prod_{j=1}^{k}{x_j^{b_j}}}$

Combinatorial proof

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Problems

Intermediate

The expression

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

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

This problem has not been edited in. Help us out by adding it.

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

@@ Line 12: / Line 12: @@
 == Proof ==
-==Proof by Induction==
+===Proof by Induction===
+Proving the Multinomial Theorem by Induction
+For a positive integer <math>k</math> and a non-negative integer <math>n</math>,
+<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>
+[b]Proof:[/b]
+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>
+Treating <math>x_p + x_{p+1}</math> as a single term and using the induction hypothesis:
+<cmath>\sum_{b_1 + b_2 + b_3 + ... + b_{p-1} + B = n}{\binom{n}{b_1, b_2, b_3, ..., b_{p-1}, B} \cdot (x_p + x_{p+1})^B \cdot \prod_{j=1}^{p-1}{x_j^{b_j}}}</cmath>
+By the Binomial Theorem, this becomes:
+<cmath>\sum_{b_1 + b_2 + b_3 + ... + b_{p-1} + B = n}{\binom{n}{b_1, b_2, b_3, ..., b_{p-1}, B}} (\prod_{j=1}^{p-1}{x_j^{b_j}}) \cdot \sum_{b_p + b_{p+1} = B}{\binom{B}{b_p} \cdot x_p^{b_p} x_{p+1}^{b_{p+1}}}</cmath>
+Since <math>\binom{n}{b_1, b_2, b_3, ... ,b_p, B}\binom{B}{b_p} = \binom{n}{b_1, b_2, b_3, ... ,b_p, b_{p+1}}</math>, this can be rewritten as:
+<cmath>\sum_{b_1 + b_2 + ...  b_p + b_{p+1}= n}{\binom{n}{b_1, b_2, b_3, ..., b_p, b_{p+1}}\prod_{j=1}^{k}{x_j^{b_j}}}</cmath>
 === Combinatorial proof ===
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