Difference between revisions of "Multinomial Theorem"
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− | The | + | The '''Multinomial Theorem''' states that |
+ | <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 <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>(a_1+a_2 | + | 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> | ||
− | + | == Proof == | |
+ | ===Proof by Induction=== | ||
+ | Proving the Multinomial Theorem by Induction | ||
− | <math>\binom{n}{ | + | 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> | ||
+ | |||
+ | <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> | ||
+ | <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 === | ||
+ | {{stub}} | ||
==Problems== | ==Problems== | ||
− | |||
===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 18: | Line 41: | ||
<math> \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</math> | <math> \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</math> | ||
− | + | (Source: [[2006_AMC_12A_Problems/Problem_24|2006 AMC 12A Problem 24]]) | |
===Olympiad=== | ===Olympiad=== | ||
− | |||
− | |||
− | |||
[[Category:Theorems]] | [[Category:Theorems]] | ||
− | [[Category: | + | [[Category:Combinatorics]] |
+ | [[Category:Multinomial Theorem]] |
Latest revision as of 19:37, 4 January 2023
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)