Multinomial theorem

In mathematics, the multinomial theorem describes how to expand a power of a sum in terms of powers of the terms in that sum. It is the generalization of the binomial theorem from binomials to multinomials.

Theorem

For any positive integer $m$ and any non-negative integer $n$ , the multinomial theorem describes how a sum with $m$ terms expands when raised to the $n$ th power: $(x_{1}+x_{2}+\cdots +x_{m})^{n}=\sum _{\begin{array}{c}k_{1}+k_{2}+\cdots +k_{m}=n\\k_{1},k_{2},\cdots ,k_{m}\geq 0\end{array}}{n \choose k_{1},k_{2},\ldots ,k_{m}}x_{1}^{k_{1}}\cdot x_{2}^{k_{2}}\cdots x_{m}^{k_{m}}$ where ${n \choose k_{1},k_{2},\ldots ,k_{m}}={\frac {n!}{k_{1}!\,k_{2}!\cdots k_{m}!}}$ is a multinomial coefficient. This can be proved by the slider method. The sum is taken over all combinations of nonnegative integer indices $k 1$ through $k m$ such that the sum of all $k i$ is $n$ . That is, for each term in the expansion, the exponents of the $x i$ must add up to $n$ .^[1]^[a]

In the case $m = 2$ , this statement reduces to that of the binomial theorem.^[1]

Example

The third power of the trinomial $a + b + c$ is given by $(a+b+c)^{3}=a^{3}+b^{3}+c^{3}+3a^{2}b+3a^{2}c+3b^{2}a+3b^{2}c+3c^{2}a+3c^{2}b+6abc.$ This can be computed by hand using the distributive property of multiplication over addition and combining like terms, but it can also be done (perhaps more easily) with the multinomial theorem. It is possible to "read off" the multinomial coefficients from the terms by using the multinomial coefficient formula. For example,

$a^{2}b^{0}c^{1}$ has coefficient ${3 \choose 2,0,1}={\frac {3!}{2!\cdot 0!\cdot 1!}}={\frac {6}{2\cdot 1\cdot 1}}=3$ , $a^{1}b^{1}c^{1}$ has coefficient ${3 \choose 1,1,1}={\frac {3!}{1!\cdot 1!\cdot 1!}}={\frac {6}{1\cdot 1\cdot 1}}=6$ , and so on.

Alternate expression

The statement of the theorem can be written concisely using multiindices:

(x_{1}+\cdots +x_{m})^{n}=\sum _{|\alpha |=n}{n \choose \alpha }x^{\alpha }

where

\alpha =(\alpha _{1},\alpha _{2},\dots ,\alpha _{m})

and

x^{\alpha }=x_{1}^{\alpha _{1}}x_{2}^{\alpha _{2}}\cdots x_{m}^{\alpha _{m}}

Proof

This proof of the multinomial theorem uses the binomial theorem and induction on $m$ .

First, for $m = 1$ , both sides equal $x 1 n$ since there is only one term $k 1 = n$ in the sum. For the induction step, suppose the multinomial theorem holds for $m$ . Then

{\begin{aligned}&(x_{1}+x_{2}+\cdots +x_{m}+x_{m+1})^{n}=(x_{1}+x_{2}+\cdots +(x_{m}+x_{m+1}))^{n}\\[6pt]={}&\sum _{k_{1}+k_{2}+\cdots +k_{m-1}+K=n}{n \choose k_{1},k_{2},\ldots ,k_{m-1},K}x_{1}^{k_{1}}x_{2}^{k_{2}}\cdots x_{m-1}^{k_{m-1}}(x_{m}+x_{m+1})^{K}\end{aligned}}

by the induction hypothesis. Applying the binomial theorem to the last factor,

=\sum _{k_{1}+k_{2}+\cdots +k_{m-1}+K=n}{n \choose k_{1},k_{2},\ldots ,k_{m-1},K}x_{1}^{k_{1}}x_{2}^{k_{2}}\cdots x_{m-1}^{k_{m-1}}\sum _{k_{m}+k_{m+1}=K}{K \choose k_{m},k_{m+1}}x_{m}^{k_{m}}x_{m+1}^{k_{m+1}}

=\sum _{k_{1}+k_{2}+\cdots +k_{m-1}+k_{m}+k_{m+1}=n}{n \choose k_{1},k_{2},\ldots ,k_{m-1},k_{m},k_{m+1}}x_{1}^{k_{1}}x_{2}^{k_{2}}\cdots x_{m-1}^{k_{m-1}}x_{m}^{k_{m}}x_{m+1}^{k_{m+1}}

which completes the induction. The last step follows because

{n \choose k_{1},k_{2},\ldots ,k_{m-1},K}{K \choose k_{m},k_{m+1}}={n \choose k_{1},k_{2},\ldots ,k_{m-1},k_{m},k_{m+1}},

as can easily be seen by writing the three coefficients using factorials as follows:

{\frac {n!}{k_{1}!k_{2}!\cdots k_{m-1}!K!}}{\frac {K!}{k_{m}!k_{m+1}!}}={\frac {n!}{k_{1}!k_{2}!\cdots k_{m+1}!}}.

Multinomial coefficients

The numbers

{n \choose k_{1},k_{2},\ldots ,k_{m}}

appearing in the theorem are the multinomial coefficients. They can be expressed in numerous ways, including as a product of binomial coefficients or of factorials:

{n \choose k_{1},k_{2},\ldots ,k_{m}}={\frac {n!}{k_{1}!\,k_{2}!\cdots k_{m}!}}={k_{1} \choose k_{1}}{k_{1}+k_{2} \choose k_{2}}\cdots {k_{1}+k_{2}+\cdots +k_{m} \choose k_{m}}

Sum of all multinomial coefficients

The substitution of $x i = 1$ for all $i$ into the multinomial theorem

\sum _{k_{1}+k_{2}+\cdots +k_{m}=n}{n \choose k_{1},k_{2},\ldots ,k_{m}}x_{1}^{k_{1}}x_{2}^{k_{2}}\cdots x_{m}^{k_{m}}=(x_{1}+x_{2}+\cdots +x_{m})^{n}

gives immediately that

\sum _{k_{1}+k_{2}+\cdots +k_{m}=n}{n \choose k_{1},k_{2},\ldots ,k_{m}}=m^{n}.

Number of multinomial coefficients

The number of terms in a multinomial sum, $# n, m$ , is equal to the number of monomials of degree $n$ on the variables $x 1, \dots, x m$ :

\#_{n,m}={n+m-1 \choose m-1}.

The count can be performed easily using the method of stars and bars.

Valuation of multinomial coefficients

The largest power of a prime $p$ that divides a multinomial coefficient may be computed using a generalization of Kummer's theorem.

Asymptotics

By Stirling's approximation, or equivalently the log-gamma function's asymptotic expansion, $\log {\binom {kn}{n,n,\cdots ,n}}=kn\log(k)+{\frac {1}{2}}\left(\log(k)-(k-1)\log(2\pi n)\right)-{\frac {k^{2}-1}{12kn}}+{\frac {k^{4}-1}{360k^{3}n^{3}}}-{\frac {k^{6}-1}{1260k^{5}n^{5}}}+O\left({\frac {1}{n^{6}}}\right)$ so for example, ${\binom {2n}{n}}\sim {\frac {2^{2n}}{\sqrt {n\pi }}}$

Interpretations

Ways to put objects into bins

The multinomial coefficients have a direct combinatorial interpretation, as the number of ways of depositing $n$ distinct objects into $m$ distinct bins, with $k 1$ objects in the first bin, $k 2$ objects in the second bin, and so on.^[2]

Number of ways to select according to a distribution

In statistical mechanics and combinatorics, if one has a number distribution of labels, then the multinomial coefficients naturally arise from the binomial coefficients. Given a number distribution ${n i}$ on a set of $N$ total items, $n i$ represents the number of items to be given the label $i$ . (In statistical mechanics $i$ is the label of the energy state.)

The number of arrangements is found by

Choosing $n 1$ of the total $N$ to be labeled 1. This can be done ${\tbinom {N}{n_{1}}}$ ways.
From the remaining $N - n 1$ items choose $n 2$ to label 2. This can be done ${\tbinom {N-n_{1}}{n_{2}}}$ ways.
From the remaining $N - n 1 - n 2$ items choose $n 3$ to label 3. Again, this can be done ${\tbinom {N-n_{1}-n_{2}}{n_{3}}}$ ways.

Multiplying the number of choices at each step results in:

{N \choose n_{1}}{N-n_{1} \choose n_{2}}{N-n_{1}-n_{2} \choose n_{3}}\cdots ={\frac {N!}{(N-n_{1})!n_{1}!}}\cdot {\frac {(N-n_{1})!}{(N-n_{1}-n_{2})!n_{2}!}}\cdot {\frac {(N-n_{1}-n_{2})!}{(N-n_{1}-n_{2}-n_{3})!n_{3}!}}\cdots .

Cancellation results in the formula given above.

Number of unique permutations of words

The multinomial coefficient

{\binom {n}{k_{1},\ldots ,k_{m}}}

is also the number of distinct ways to permute a multiset of $n$ elements, where $k i$ is the multiplicity of each of the $i$ th element. For example, the number of distinct permutations of the letters of the word MISSISSIPPI, which has 1 M, 4 Is, 4 Ss, and 2 Ps, is

{11 \choose 1,4,4,2}={\frac {11!}{1!\,4!\,4!\,2!}}=34650.

Generalized Pascal's triangle

One can use the multinomial theorem to generalize Pascal's triangle or Pascal's pyramid to Pascal's simplex. This provides a quick way to generate a lookup table for multinomial coefficients.

References

^ As with the binomial theorem, quantities of the form $x 0$ that appear are taken to equal 1, even when $x$ equals zero.

^ ^a ^b Stanley, Richard (2012), Enumerative Combinatorics, vol. 1 (2 ed.), Cambridge University Press, §1.2
^ National Institute of Standards and Technology (May 11, 2010). "NIST Digital Library of Mathematical Functions". Section 26.4. Retrieved August 30, 2010.

[2] As with the binomial theorem, quantities of the form $x 0$ that appear are taken to equal 1, even when $x$ equals zero.

[EC1-1] Stanley, Richard (2012), Enumerative Combinatorics, vol. 1 (2 ed.), Cambridge University Press, §1.2

[3] National Institute of Standards and Technology (May 11, 2010). "NIST Digital Library of Mathematical Functions". Section 26.4. Retrieved August 30, 2010.

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