Multinomial Theorem
The multinomial theorem describes how to expand the power of a sum of more than two terms. It is a generalization of the binomial theorem to polynomials with any number of terms. It expresses a power \( (x_1 + x_2 + \cdots + x_k)^n \) as a weighted sum of monomials of the form \( x_1^{b_1} x_2^{b_2} \cdots x_k^{b_k}, \) where the weights are given by generalizations of binomial coefficients called multinomial coefficients.
Contents
Multinomial Theorem Statement
The first important definition is the multinomial coefficient:
For non-negative integers \(b_1, b_2, \ldots, b_k\) such that \(\displaystyle \sum_{i=1}^{k} b_i = n,\) the multinomial coefficient is
\[\binom{n}{b_1, b_2,\ldots , \ b_k} = \frac{n!}{b_1!b_2!\cdots b_k!}. \]
Note that when \( k = 2\) this is the binomial coefficient: \[ \binom{n}{b_1,b_2} = \frac{n!}{b_1!b_2!} = \frac{n!}{b_1!(n-b_1)!} = \binom{n}{b_1}. \]
Now the multinomial theorem can be stated as follows:
Multinomial Theorem:
For a positive integer \(k\) and a non-negative integer \(n\),
\[\left(x_1 + x_2 + \cdots + x_k\right)^{n} = \sum_{b_1 + b_2 + \cdots +b_k = n} \binom{n}{b_1, b_2, b_3, \ldots, b_k} \prod_{j=1}^{k} x_j^{b_j}.\ _\square\]
The number of terms of this sum are given by a stars and bars argument: it is \( \binom{n+k-1}{n}\).
Proof of Multinomial Theorem
There are two proofs of the multinomial theorem, an algebraic proof by induction and a combinatorial proof by counting. The algebraic proof is presented first.
Proceed by induction on \(m.\)
When \(k = 1\) the result is true, and when \(k = 2\) the result is the binomial theorem. Assume that \(k \geq 3\) and that the result is true for \(k = p.\) When \(k = p+1,\)
\[\left(x_1 + x_2 + \cdots + x_p\right)^{n} = \left(x_1 + x_2 + \cdots +x_{p-1} + (x_p + x_{p+1})\right)^{n}. \]
Treating \(x_p + x_{p+1}\) as a single term and using the induction hypothesis,
\[ \sum_{b_1 + b_2 + \cdots +b_{p-1} + B = n} \binom{n}{b_1, b_2, b_3, \ldots, b_{p-1}, B} \prod_{j=1}^{p-1} x_j^{b_j} \times (x_p + x_{p+1})^B.\]
By the binomial theorem, this becomes
\[ \sum_{b_1 + b_2 + \cdots +b_{p-1} + B = n} \binom{n}{b_1, b_2, b_3, \ldots, b_{p-1}, B} \prod_{j=1}^{p-1} x_j^{b_j} \times \sum_{b_p + b_{p+1} = B}\binom{B}{b_p} x_p^{b_p} x_{p+1}^{b_{p+1}}.\]
Since \( \binom{n}{b_1, b_2, b_3, \ldots, b_{p-1}, B} \binom{B}{b_p} = \binom{n}{b_1, b_2, b_3, \ldots, b_{p+1}} \) this can be rewritten as:
\[ \sum_{b_1 + b_2 + \cdots +b_{p+1} = n} \binom{n}{b_1, b_2, b_3, \ldots, b_{p+1}} \prod_{j=1}^{k} x_j^{b_j}. \ \square \]
Here is the combinatorial proof, which relies on a fact from the multinomial coefficients wiki.
Consider a term in the expansion of \(\left(x_1 + x_2 + \cdots + x_k\right)^{n}\). It must be of the form \(\displaystyle \alpha\prod_{i=1}^{k} x_i^{\beta_i}\) for some integer \(\alpha\) and non-negative integers \(\beta_i.\) Since each term must come from choosing one summand from \(x_1 + x_2 + \cdots +x_k,\) we must have \(\beta_1 + \beta_2 + \cdots + \beta_k = n.\) The number of different ways that we can get this term will be the number of ways to choose \( \beta_1 \) copies of \( x_1\), \( \beta_2\) copies of \( x_2 \), and so on. By a result from the multinomial coefficients wiki, this is exactly \(\alpha = \binom{n}{\beta_1, \beta_2, \ldots, \beta_k}. \square \)
Multinomial Theorem Examples - Number of Terms
How many terms are in the expansion of \( (x_1+x_2+x_3)^4\)?
Solution: There is one term for each ordered triple \( (b_1,b_2,b_3) \) with \( b_1+b_2+b_3= 4\). One way to count these triples is to represent them as collections of \( 2 \) bars and \( 4 \) stars; for instance, \[ *|***| \] represents the triple \( (1,3,0) \). The number of such collections is \( \binom{4+2}{4} = 15 \), so there are \( 15 \) terms in the expansion.
Multinomial Theorem Examples - Specific Terms
Determine the coefficient of \(a^2b^4d\) in the expansion of the polynomial \( (3a + 5b -2c +d)^7.\)
A general term in the expansion of \( (3a + 5b -2c +d)^7\) will be of the form \(\binom{7}{b_1,b_2,b_3,b_4}(3a)^{b_1}(5b)^{b_2}(-2c)^{b_3}(d)^{b_4}.\) To have the term with \(a^2b^4d,\) we need \(b_1 = 2, b_2 = 4, b_3 = 0, b_4 = 1.\) This gives us
\[\begin{align}\binom{7}{2,4,1}(3a)^{2}(5b)^{4}(d)^{1} &= \frac{7!}{4!2!1!}(9a^2)(625b^4)(d) \\ &= 105(5625)a^2b^4d \\ &= 590625a^2b^4d,\end{align}\]
implying the answer is 590625. \(\square\)
Find the coefficient of \(t^{20}\) in the expansion of \(\left(t^3 - 3t^2 + 7t +1\right)^{11}.\)
A general term of the expansion has the form \(\binom{11}{b_1, b_2, b_3, b_4}\left(t^3\right)^{b_1}\left(-3t^2\right)^{b_2}(7t)^{b_3}(11)^{b_4}.\) In order to have a coefficient of \(t^{20}\) we must have \(b_1 + b_2 + b_3 +b_4 = 11\) and \(3b_1 + 2b_2 + b_3 = 20.\) We can write \(b_3 = 20 - 2b_2 - 3b_1\) and \(b_4 = 11 - b_1 - b_2 - b_3 = 2b_1 + b_2 - 9.\)
Thus, the coefficient will be the sum \(\displaystyle\sum_{b_1,b_2 \geq 0} \binom{11}{b_1,b_2,20-2b_2-3b_1,2b_1+b_2-9} \) such that \(2b_2+3b_1 \leq 20\) and \(2b_1+b_2 \geq 9.\)
We can evaluate this as -7654572342. \( _ \square \)
Find the term that does not contain the variable \(x\) in the complete expansion of \[\left(x^2+x+\frac{1}{x}\right)^{10}\]
To be clear, if an expression is completely expanded, then all like terms have been combined together, leaving unlike terms in the final answer. For example, \[(a+1)^4=a^4+4a^3+6a^2+4a+1\]