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 expands when raised to an arbitrary power:

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

The summation is taken over all sequences of nonnegative integer indices k₁ through k_m such that \sum_{i=1}^m {k_i} = n (for each term in the expansion, the exponents must add up to n). (As with the binomial theorem, quantities of the form 0⁰ which appear are taken to equal 1.)

The 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 nth power (e.g. coefficients={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)

Example

(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

We could have calculated each coefficient by first expanding (a+b+c)^2=a^2+b^2+c^2+2 a b+2 b c+2 a c, 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:

a^2 b^0 c^1 => {3 \choose 2, 0, 1} = \frac{3!}{2! 0! 1!} = \frac{6}{2*1*1} = 3

a^1 b^1 c^1 => {3 \choose 1, 1, 1} = \frac{3!}{1! 1! 1!} = \frac{6}{1*1*1} = 6

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

Interpretations

Ways to put objects into sized boxes

The multinomial coefficients have a direct combinatorial interpretation, as the number of ways of depositing n distinct objects in m bins, with k₁ objects in the first bin, k₂ objects in the second bin, and so on.

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 I's, 4 S's, and 2 P's is

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

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

(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 counted extra 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 to Pascal's pyramid or 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 x_1^n. 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

• Binomial theorem
• Multinomial distribution

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