Multinomial theorem

In mathematics, the multinomial theorem says how to write a power of a sum in terms of powers of the terms in that sum. It is the generalization of the binomial theorem to polynomials.

Theorem

For any positive integer m and any nonnegative integer n, the multinomial formula tells us how a polynomial with m terms expands when raised to an arbitrary power n:

$(x_1 + x_2 + \cdots + x_m)^n = \sum_{k_1+k_2+\cdots+k_m=n} {n \choose k_1, k_2, \ldots, k_m} \prod_{1\le t\le m}x_{t}^{k_{t}}\,,$

where

${n \choose k_1, k_2, \ldots, k_m} = \frac{n!}{k_1!\, k_2! \cdots k_m!}$

as described below. The sum is taken over all combinations of nonnegative integer indices k₁ through k_m such that the sum of all k_i is n. That is, for each term in the expansion, the exponents must add up to n. Also, as with the binomial theorem, quantities of the form x⁰ that appear are taken to equal 1 (even when x equals zero). Alternatively, this can be written concisely using multiindices as

$(x_1+\cdots+x_m)^n = \sum_{|\alpha|=n}{n \choose \alpha}x^\alpha$

where α = (α₁,α₂,…,α_m) and x^α = x₁^α₁x₂^α₂⋯x_m^α_m.

Number of multinomial coefficients

The number of terms in multinomial sum, #_n,m, is equal to the number of monomials of degree n on the variables x₁, …, x_m:

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

The count can be easily performed using the Stars and bars (combinatorics) method.

Multinomial coefficients

The numbers

${n \choose k_1, k_2, \ldots, k_m} = \frac{n!}{k_1!\, k_2! \cdots k_m!}$

(which can also be written as:)

$= {k_1\choose k_1}{k_1+k_2\choose k_2}\cdots{k_1+k_2+\cdots+k_m\choose k_m} = \prod_{i=1}^m {\sum_{j=1}^i k_j \choose k_i}$

are the multinomial coefficients. Just like "n choose k" are the coefficients when you raise a binomial to the n^th power (e.g. the coefficients are 1,3,3,1 for (a + b)³, where n = 3), the multinomial coefficients appear when one raises a multinomial to the n^th power (e.g. (a + b + c)³)

Sum of all multinomial coefficients

The substitution of x_i = 1 for all i into:

$\sum_{k_1+k_2+\cdots+k_m=n} {n \choose k_1, k_2, \ldots, k_m} x_1^{k_1} x_2^{k_2} \cdots x_m^{k_m} = (x_1 + x_2 + \cdots + x_m)^n\,,$

gives immediately that

$\sum_{k_1+k_2+\cdots+k_m=n} {n \choose k_1, k_2, \ldots, k_m} = m^n\,.$

Central multinomial coefficients

All of the multinomial coefficients for which the following holds true:

$\left\lfloor\frac{n}{m}\right\rfloor \le k_i \le \left\lceil\frac{n}{m}\right\rceil,\ \sum_{i=1}^m{k_i} = n,$

are central multinomial coefficients: the greatest ones and all of equal size.

A special case for m = 2 is central binomial coefficient.

Example multinomial coefficients

(a + b + c)³ = a³ + b³ + c³ + 3a²b + 3a²c + 3b²a + 3b²c + 3c²a + 3c²b + 6abc.

We could have calculated each coefficient by first expanding (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac, then self-multiplying it again to get (a + b + c)³ (and then if we were raising it to higher powers, we'd multiply it by itself even some more). However this process is slow, and can be avoided by using the multinomial theorem. The multinomial theorem "solves" this process by giving us the closed form for any coefficient we might want. It is possible to "read off" the multinomial coefficients from the terms by using the multinomial coefficient formula. For example:

a²b⁰c¹ has the coefficient ${3 \choose 2, 0, 1} = \frac{3!}{2!\cdot 0!\cdot 1!} = \frac{6}{2 \cdot 1 \cdot 1} = 3$

a¹b¹c¹ has the coefficient ${3 \choose 1, 1, 1} = \frac{3!}{1!\cdot 1!\cdot 1!} = \frac{6}{1 \cdot 1 \cdot 1} = 6$.

We could have also had a 'd' variable, or even more variables—hence the multinomial theorem.

The binomial theorem and binomial coefficients are special cases, for m = 2, of the multinomial theorem and multinomial coefficients, respectively.

Interpretations

Ways to put objects into boxes

The multinomial coefficients have a direct combinatorial interpretation, as the number of ways of depositing n distinct objects into m distinct bins, such that no bin remains empty, with k₁ objects in the first bin, k₂ objects in the second bin, and so on.^[1]

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 n₁ of the total N to be labeled 1. This can be done $N\choose n_1$ ways.
• From the remaining N − n₁ items choose n₂ to label 2. This can be done $N-n_1 \choose n_2$ ways.
• From the remaining N − n₁ − n₂ items choose n₃ to label 3. Again, this can be done $N-n_1-n_2 \choose n_3$ ways.

Multiplying the number of choices at each step results in:

${N \choose n_1}{N-n_1\choose n_2}{N-n_1-n_2\choose n_3}...=\frac{N!}{(N-n_1)!n_1!}\frac{(N-n_1)!}{(N-n_1-n_2)!n_2!}\frac{(N-n_1-n_2)!}{(N-n_1-n_2-n_3)!n_3!}....$

Upon cancellation, we arrive at the formula given in the introduction.

Number of unique permutations of words

In addition, the multinomial coefficient is also the number of distinct ways to permute a multiset of n elements, and k_i are the multiplicities of each of the distinct elements. 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

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

(This is just like saying that there are 11! ways to permute the letters—the common interpretation of factorial as the number of unique permutations. However, we created duplicate permutations, due to the fact that some letters are the same, and must divide to correct our answer.)

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.

The case of n = 3 can be easily drawn by hand. The case of n = 4 can be drawn with effort as a series of growing pyramids.

Proof

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

First, for m = 1, both sides equal x₁ⁿ since there is only one term k₁ = n in the sum. For the induction step, suppose the multinomial theorem holds for m. Then

$(x_1+x_2+\cdots+x_m+x_{m+1})^n = (x_1+x_2+\cdots+(x_m+x_{m+1}))^n$

$= \sum_{k_1+k_2+\cdots+k_{m-1}+K=n}{n\choose k_1,k_2,\ldots,k_{m-1},K} x_1^{k_1}x_2^{k_2}\cdots x_{m-1}^{k_{m-1}}(x_m+x_{m+1})^K$

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:

$\frac{n!}{k_1! k_2! \cdots k_{m-1}!K!} \frac{K!}{k_m! k_{m+1}!}=\frac{n!}{k_1! k_2! \cdots k_{m+1}!}.$

See also

• Multinomial distribution
• Stars and bars (combinatorics)

References

1. ^ National Institute of Standards and Technology (May, 11, 2010). "NIST Digital Library of Mathematical Functions". Section 26.4. http://dlmf.nist.gov/. Retrieved August 30, 2010.