Difference between revisions of "Multinomial Theorem"
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The '''Multinomial Theorem''' states that | The '''Multinomial Theorem''' states that | ||
| − | + | <cmath> | |
| − | <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 \\ | + | (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> | + | \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>. | 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>. | ||
Note that this is a direct generalization of the [[Binomial Theorem]]: when <math>k = 2</math> it simplifies to | 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> |
| + | (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=== | ||
| + | {{problem}} | ||
| + | |||
===Intermediate=== | ===Intermediate=== | ||
*The [[expression]] | *The [[expression]] | ||
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<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=== | ||
| − | + | {{problem}} | |
{{stub}} | {{stub}} | ||
[[Category:Theorems]] | [[Category:Theorems]] | ||
[[Category:Combinatorics]] | [[Category:Combinatorics]] | ||
Revision as of 18:11, 29 April 2008
The Multinomial Theorem states that
![\[(a_1+a_2+\cdots+a_k)^n=\sum_{\substack{j_1,j_2,\ldots,j_k \\ 0 \leq j_i \leq n \textrm{ for each } i \\ \textrm{and } j_1 + \ldots + j_k = n}}\binom{n}{j_1; j_2; \ldots ; j_k}a_1^{j_1}a_2^{j_2}\cdots a_k^{j_k}\]](http://latex.artofproblemsolving.com/4/5/c/45cb6228eda6d98ac694f0fa5bf19c95eb0edff4.png)
where 
is the multinomial coefficient 
.
Note that this is a direct generalization of the Binomial Theorem: when 
it simplifies to
![\[(a_1 + a_2)^n = \sum_{\substack{0\leq j_1, j_2 \leq n \\ j_1 + j_2 = n}} \binom{n}{j_1; j_2} a_1^{j_1}a_2^{j_2} = \sum_{j = 0}^n \binom{n}{j} a_1^j a_2^{n - j}\]](http://latex.artofproblemsolving.com/1/c/5/1c5dc64793d2f619276e8443c44d83a30551fdd2.png)
Problems
Introductory
This problem has not been edited in. If you know this problem, please help us out by adding it.
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)
Olympiad
This problem has not been edited in. If you know this problem, please help us out by adding it.
This article is a stub. Help us out by expanding it.
