Multinomial theorem

From Wikipedia, the free encyclopedia
Jump to: navigation, search

In mathematics, the multinomial theorem says 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 to polynomials.

Theorem[edit]

For any positive integer m and any nonnegative integer n, the multinomial formula tells us how a sum 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!}

is a multinomial coefficient. The sum is taken over all combinations of nonnegative integer indices k1 through km such that the sum of all ki is n. That is, for each term in the expansion, the exponents of the xi 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).

In the case m = 2, this statement reduces to that of the binomial theorem.

Example[edit]

The third power of the trinomial a + b + c is given by

(a+b+c)^3 = a^3 + b^3 + c^3 + 3 a^2 b + 3 a^2 c + 3 b^2 a + 3 b^2 c + 3 c^2 a + 3 c^2 b + 6 a b c.

This can be computed by hand using the distributive property of multiplication over addition, but it can also be done (perhaps more easily) with the multinomial theorem, which gives us a simple formula 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:

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

Alternate expression[edit]

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 α = (α12,…,αm) and xα = x1α1x2α2xmαm.

Proof[edit]

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}!}.

Multinomial coefficients[edit]

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 nth 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 nth power (e.g. (a + b + c)3)

Sum of all multinomial coefficients[edit]

The substitution of xi = 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\,.

Number of multinomial coefficients[edit]

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


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

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

Central multinomial coefficients[edit]

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.

Interpretations[edit]

Ways to put objects into boxes[edit]

The multinomial coefficients have a direct combinatorial interpretation, as the number of ways of depositing n distinct objects into m distinct bins, 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[edit]

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 {ni} on a set of N total items, ni 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[edit]

The multinomial coefficient is also the number of distinct ways to permute a multiset of n elements, and ki 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[edit]

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.

See also[edit]

References[edit]

External links[edit]