Difference between revisions of "Multinomial Theorem"
Dinoking93 (talk | contribs) m |
Dinoking93 (talk | contribs) m |
||
Line 6: | Line 6: | ||
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> | <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} | (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} |
Latest revision as of 18:37, 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
[hide]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)