Multinomial theorem explained

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 and any non-negative integer, the multinomial theorem describes how a sum with terms expands when raised to the th power:$$(x\_1\; +\; x\_2\; +\; \backslash cdots\; +\; x\_m)^n\; =\; \backslash sum\_\; x\_1^\; \backslash cdot\; x\_2^\; \backslash cdots\; x\_m^$$where$$=\; \backslash frac$$is a multinomial coefficient. The sum is taken over all combinations of nonnegative integer indices through such that the sum of all is . That is, for each term in the expansion, the exponents of the must add up to .

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

Example

The third power of the trinomial 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 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 } = (3 !)/(2 ! ⋅ 0 ! ⋅ 1 !) = (6)/(2 ⋅ 1 ⋅ 1) = 3 , a ^{1} b ^{1} c ^{1} has coefficient { 3 \choose 1 , 1 , 1 } = (3 !)/(1 ! ⋅ 1 ! ⋅ 1 !) = (6)/(1 ⋅ 1 ⋅ 1) = 6 , and so on.

Alternate expression

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

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

where

\alpha = (\alpha _{1,\alpha} _{2,...,\alpha} _{m)} andx ^{\alpha _{1}} _{1} x ^{\alpha _{2}} _{2} … x ^{\alpha _{m}} _{m}

Proof

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

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

\begin{align} & (x _{1+x} _{2+ … +x} _{m+x} _{m+1} ) ^{n} = (x _{1+x} _{2+ … +(x} _{m+x} _{m+1} )) ^{n} \\[ 6 pt] = {} & \sum _{k _{1+k} _{2+ … +k} _{m-1} + K = n} {n\choose k _{1,k} _{2,\ldots,k} _{m-1} ,K} x ^{k _{1}} _{1} x ^{k _{2}} _{2} … x ^{k _{m-1}} _{m -1} (x _{m+x} _{m+1} ) ^{K \end{align}}

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

= \sum _{k _{1+k} _{2+ … +k} _{m-1} + K = n} {n\choose k _{1,k} _{2,\ldots,k} _{m-1} ,K} x ^{k _{1}} _{1} x ^{k _{2}} _{2} … x ^{k _{m-1}} _{m -1} \sum _{k _{m+k} _{m+1} = K} {K\choose k _{m,k} _{m+1} }x_m^x_^= \sum _{k _{1+k} _{2+ … +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^x_2^\cdots x_^x_m^x_^

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} } =,as can easily be seen by writing the three coefficients using factorials as follows:(n!)/(k _{1!} k _{2!} … k _{m-1} !K!) (K!)/(= k _{m!} k _{m+1} !) (n!)/(k _{1!} k _{2!} … 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}} = (n!)/(k _{1!} k _{2!} … k _{m!}) = {k _{1\choose} k _{1}{k} _{1+k} _{2\choose} k _{2} … {k} _{1+k} _{2+ … +k} _{m\choose} k _{m}}

Sum of all multinomial coefficients

The substitution of for all into the multinomial theorem

\sum _{k _{1+k} _{2+ … +k} _{m=n}} {n \choose k _{1,} k _{2,} \ldots, k _{m}} x ^{k _{1}} _{1} x ^{k _{2}} _{2} … x ^{k _{m}} _{m} = (x _{1} + x _{2} + … + x ^{n} _{m)} gives immediately that \sum _{k _{1+k} _{2+ … +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,, is equal to the number of monomials of degree on the variables :

\# _{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 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, $$\backslash log\backslash binom\; =\; k\; n\; \backslash log(k)\; +\; \backslash frac\; \backslash left(\backslash log(k)\; -\; (k\; -\; 1)\; \backslash log(2\; \backslash pi\; n)\backslash right)\; -\; \backslash frac\; +\; \backslash frac\; -\; \backslash frac\; +\; O\backslash left(\backslash frac\backslash right)$$so for example,$$\backslash binom\; \backslash sim\; \backslash frac$$

Interpretations

Ways to put objects into bins

The multinomial coefficients have a direct combinatorial interpretation, as the number of ways of depositing distinct objects into distinct bins, with objects in the first bin, 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 on a set of total items, represents the number of items to be given the label . (In statistical mechanics is the label of the energy state.)

The number of arrangements is found by

• Choosing of the total to be labeled 1. This can be done

\tbinom{N}{n _{1}} ways.

• From the remaining items choose to label 2. This can be done

\tbinom{N - n _{1}{n} _{2}} ways.

• From the remaining items choose 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 } … = (N!)/((N - n _{1)!n} _{1!})} ⋅ ((N - n _{1)!})/((N - n _{1-n} _{2)!n} _{2!}) ⋅ ((N - n _{1-n} _{2)!})/((N - n _{1-n} _{2-n} _{3)!n} _{3!}) … .

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 elements, where is the multiplicity of each of the 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 } = (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.

See also

• Multinomial distribution
• Stars and bars (combinatorics)

Notes and References

1. Web site: NIST Digital Library of Mathematical Functions . National Institute of Standards and Technology . National Institute of Standards and Technology . May 11, 2010 . Section 26.4 . August 30, 2010.