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.




Contents






  • 1 Theorem


    • 1.1 Example


    • 1.2 Alternate expression


    • 1.3 Proof




  • 2 Multinomial coefficients


    • 2.1 Sum of all multinomial coefficients


    • 2.2 Number of multinomial coefficients


    • 2.3 Valuation of multinomial coefficients




  • 3 Interpretations


    • 3.1 Ways to put objects into bins


    • 3.2 Number of ways to select according to a distribution


    • 3.3 Number of unique permutations of words


    • 3.4 Generalized Pascal's triangle




  • 4 See also


  • 5 References





Theorem


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:


(x1+x2+⋯+xm)n=∑k1+k2+⋯+km=n(nk1,k2,…,km)∏t=1mxtkt,{displaystyle (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 _{t=1}^{m}x_{t}^{k_{t}},,}{displaystyle (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 _{t=1}^{m}x_{t}^{k_{t}},,}

where


(nk1,k2,…,km)=n!k1!k2!⋯km!{displaystyle {n choose k_{1},k_{2},ldots ,k_{m}}={frac {n!}{k_{1}!,k_{2}!cdots k_{m}!}}}{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


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


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




a2b0c1{displaystyle a^{2}b^{0}c^{1}}a^{2}b^{0}c^{1} has the coefficient (32,0,1)=3!2!⋅0!⋅1!=62⋅1⋅1=3.{displaystyle {3 choose 2,0,1}={frac {3!}{2!cdot 0!cdot 1!}}={frac {6}{2cdot 1cdot 1}}=3.}{displaystyle {3 choose 2,0,1}={frac {3!}{2!cdot 0!cdot 1!}}={frac {6}{2cdot 1cdot 1}}=3.}


a1b1c1{displaystyle a^{1}b^{1}c^{1}}a^{1}b^{1}c^{1} has the coefficient (31,1,1)=3!1!⋅1!⋅1!=61⋅1⋅1=6.{displaystyle {3 choose 1,1,1}={frac {3!}{1!cdot 1!cdot 1!}}={frac {6}{1cdot 1cdot 1}}=6.}{displaystyle {3 choose 1,1,1}={frac {3!}{1!cdot 1!cdot 1!}}={frac {6}{1cdot 1cdot 1}}=6.}



Alternate expression


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


(x1+⋯+xm)n=∑|α|=n(nα)xα{displaystyle (x_{1}+cdots +x_{m})^{n}=sum _{|alpha |=n}{n choose alpha }x^{alpha }}(x_{1}+cdots +x_{m})^{n}=sum _{{|alpha |=n}}{n choose alpha }x^{alpha }

where


α=(α1,α2,…,αm){displaystyle alpha =(alpha _{1},alpha _{2},dots ,alpha _{m})}{displaystyle alpha =(alpha _{1},alpha _{2},dots ,alpha _{m})}

and


xα=x1α1x2α2⋯xmαm{displaystyle x^{alpha }=x_{1}^{alpha _{1}}x_{2}^{alpha _{2}}cdots x_{m}^{alpha _{m}}}{displaystyle 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 x1n since there is only one term k1 = n in the sum. For the induction step, suppose the multinomial theorem holds for m. Then


(x1+x2+⋯+xm+xm+1)n=(x1+x2+⋯+(xm+xm+1))n=∑k1+k2+⋯+km−1+K=n(nk1,k2,…,km−1,K)x1k1x2k2⋯xm−1km−1(xm+xm+1)K{displaystyle {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}}}{displaystyle {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,



=∑k1+k2+⋯+km−1+K=n(nk1,k2,…,km−1,K)x1k1x2k2⋯xm−1km−1∑km+km+1=K(Kkm,km+1)xmkmxm+1km+1{displaystyle =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=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}}}}

=∑k1+k2+⋯+km−1+km+km+1=n(nk1,k2,…,km−1,km,km+1)x1k1x2k2⋯xm−1km−1xmkmxm+1km+1{displaystyle =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}}}=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


(nk1,k2,…,km−1,K)(Kkm,km+1)=(nk1,k2,…,km−1,km,km+1),{displaystyle {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}},}{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:


n!k1!k2!⋯km−1!K!K!km!km+1!=n!k1!k2!⋯km+1!.{displaystyle {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}!}}.}{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


(nk1,k2,…,km){displaystyle {n choose k_{1},k_{2},ldots ,k_{m}}}{displaystyle {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:


(nk1,k2,…,km)=n!k1!k2!⋯km!=(k1k1)(k1+k2k2)⋯(k1+k2+⋯+kmkm){displaystyle {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}}}{displaystyle {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 xi = 1 for all i into the multinomial theorem


∑k1+k2+⋯+km=n(nk1,k2,…,km)x1k1x2k2⋯xmkm=(x1+x2+⋯+xm)n{displaystyle 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}}{displaystyle 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


∑k1+k2+⋯+km=n(nk1,k2,…,km)=mn.{displaystyle sum _{k_{1}+k_{2}+cdots +k_{m}=n}{n choose k_{1},k_{2},ldots ,k_{m}}=m^{n}.}{displaystyle 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 x1, …, xm:


#n,m=(n+m−1m−1).{displaystyle #_{n,m}={n+m-1 choose m-1}.}{displaystyle #_{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{displaystyle p}p that divides a multinomial coefficient may be computed using a generalization of Kummer's theorem.



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 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 {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 (Nn1){displaystyle N choose n_{1}}{N choose n_{1}} ways.

  • From the remaining N − n1 items choose n2 to label 2. This can be done (N−n1n2){displaystyle N-n_{1} choose n_{2}}{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−n1−n2n3){displaystyle N-n_{1}-n_{2} choose n_{3}}{N-n_{1}-n_{2} choose n_{3}} ways.


Multiplying the number of choices at each step results in:


(Nn1)(N−n1n2)(N−n1−n2n3)⋯=N!(N−n1)!n1!⋅(N−n1)!(N−n1−n2)!n2!⋅(N−n1−n2)!(N−n1−n2−n3)!n3!⋯.{displaystyle {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 .}{displaystyle {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 .}

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



Number of unique permutations of words


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


(111,4,4,2)=11!1!4!4!2!=34650.{displaystyle {11 choose 1,4,4,2}={frac {11!}{1!,4!,4!,2!}}=34650.}{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, because 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.



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. Retrieved August 30, 2010..mw-parser-output cite.citation{font-style:inherit}.mw-parser-output .citation q{quotes:"""""""'""'"}.mw-parser-output .citation .cs1-lock-free a{background:url("//upload.wikimedia.org/wikipedia/commons/thumb/6/65/Lock-green.svg/9px-Lock-green.svg.png")no-repeat;background-position:right .1em center}.mw-parser-output .citation .cs1-lock-limited a,.mw-parser-output .citation .cs1-lock-registration a{background:url("//upload.wikimedia.org/wikipedia/commons/thumb/d/d6/Lock-gray-alt-2.svg/9px-Lock-gray-alt-2.svg.png")no-repeat;background-position:right .1em center}.mw-parser-output .citation .cs1-lock-subscription a{background:url("//upload.wikimedia.org/wikipedia/commons/thumb/a/aa/Lock-red-alt-2.svg/9px-Lock-red-alt-2.svg.png")no-repeat;background-position:right .1em center}.mw-parser-output .cs1-subscription,.mw-parser-output .cs1-registration{color:#555}.mw-parser-output .cs1-subscription span,.mw-parser-output .cs1-registration span{border-bottom:1px dotted;cursor:help}.mw-parser-output .cs1-ws-icon a{background:url("//upload.wikimedia.org/wikipedia/commons/thumb/4/4c/Wikisource-logo.svg/12px-Wikisource-logo.svg.png")no-repeat;background-position:right .1em center}.mw-parser-output code.cs1-code{color:inherit;background:inherit;border:inherit;padding:inherit}.mw-parser-output .cs1-hidden-error{display:none;font-size:100%}.mw-parser-output .cs1-visible-error{font-size:100%}.mw-parser-output .cs1-maint{display:none;color:#33aa33;margin-left:0.3em}.mw-parser-output .cs1-subscription,.mw-parser-output .cs1-registration,.mw-parser-output .cs1-format{font-size:95%}.mw-parser-output .cs1-kern-left,.mw-parser-output .cs1-kern-wl-left{padding-left:0.2em}.mw-parser-output .cs1-kern-right,.mw-parser-output .cs1-kern-wl-right{padding-right:0.2em}









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