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Multinomial Theorem and Coefficients of Polynomial Expansions

Date: 06/24/2007 at 19:23:36
From: Siddhant
Subject: Multinomial Theorem

Could you explain the Multinomial Theorem to me?  For example, find 
(x + y + z + a + b)^4?  I looked up the Multinomial Theorem, but can't
quite understand it.




Date: 06/25/2007 at 02:12:12
From: Doctor Greenie
Subject: Re: Multinomial Theorem

Hello, Siddhant --

I found one page in the Dr. Math archives where this topic is very 
briefly discussed:

  Multinomial Coefficients
    http://mathforum.org/library/drmath/view/64616.html 

Let me go into much greater detail for you on your question....

Let's start with the binomial theorem.  I assume if you are asking 
questions about the multinomial theorem that you are familiar with 
the binomial theorem.

The binomial theorem says that the coefficient of the

  (x^m) * (y^(n-m))

term in the expansion of (x+y)^n is "n choose m", denoted variously 
as nCm or C(n,m) or as "n" above "m" in a single large set of
parentheses.  (I can't type that one.)

The number nCm has the value

      n!
  ----------
  (m!)(n-m)!

By a well-known principle of combinatorics, this same expression is 
the number of distinct ways of arranging "n" objects of which "m" 
are identical and the remaining "n-m" are identical.  Let's use this 
interpretation of this expression to see why the binomial coefficients 
are these nCm numbers.

Suppose, for a specific example, we want to determine the coefficient 
of the x^2 * y^3 term in the expansion of (x+y)^5.  Let's think of the 
expansion as the product of five factors of (x+y):

  (x+y)(x+y)(x+y)(x+y)(x+y)

The product of any group of polynomials is the sum of the partial 
products formed by selecting one of the terms from each factor.  In 
this example, we can pick either an x term or a y term from each of 
the five factors.  If we want to count the number of partial products 
which are x^2 * y^3, then we need to select the "x" term from 2 of the 
5 factors and the "y" term from 3 of the 5 factors.

We could find the number of these partial products by using basic 
combinatoric principles; the number of these partial products is "5 
choose 2", because we need to choose 2 of the 5 factors to be the ones 
from which we choose the "x" term.  But using the "a choose b" concept 
will not help us when we move on to examine the multinomial 
coefficients.

So let's count the number of partial products of the form x^2 * y^3 
in a different way.  We will write out all the ways we can select the 
"x" terms from 2 of the 5 factors.  If we use the string of characters 
"xxyyy" to indicate selecting the x terms from the first two factors 
and the y terms from the last three factors, then the complete list of 
ways we can get a partial product of x^2 * y^3 is

  xxyyy
  xyxyy
  xyyxy
  xyyyx
  yxxyy
  yxyxy
  yxyyx
  yyxxy
  yyxyx
  yyyxx

This is clearly the number of distinct arrangements of the letters 
"xxxyy", which we know is

     5!
  -------- = 10
  (2!)(3!)

So the number of partial products of the form x^2 * y^3--and therefore 
the coefficient of the x^2 * y^3 term in the expansion of (x+y)^5--is 
the number of distinct ways of arranging 2 x's and 3 y's.

The conclusion in the last paragraph makes the extension of our 
investigation to multinomial coefficients rather easy.  Suppose we 
take your example to look at the question of multinomial coefficients:

  (x+y+z+a+b)^4

If we want to know the coefficient of the x^3 * a^1 term in this 
expansion, we know the number of partial products of that form is the 
number of distinct ways of arranging the letters "xxxa".  And that 
number of ways is

     4!
  -------- = 4
  (3!)(1!)

Of course, we don't need the "1!" in that computation; I simply showed 
it in this case for clarity.

And similarly the coefficient of the x^1 * y^2 * b^1 in this expansion 
is

  4!
  -- = 12
  2!

And the coefficient of the abxy term in the expansion is just

         4!
  ---------------- = 24
  (1!)(1!)(1!)(1!)

Just a couple of random examples to make the method for determining 
multinomial coefficients more clear....

The coefficient of the a^2 * b^3 * c^4 term in the expansion of
(a+b+c)^9 is the number of distinct arrangements of the letters 
"aabbbcccc", which is

       9!
  ------------ = 1260
  (2!)(3!)(4!) 

The coefficient of the a^4 * b^6 * c^2 * d^7 * e^3 term in the 
expansion of (a+b=c+d+e)^22 is the number of distinct arrangements of 
the letters "aaaabbbbbbccdddddddeee", which is

           22!
  -------------------- = ???
  (4!)(6!)(2!)(7!)(3!)

I hope this helps.  Please write back if you have any further 
questions about any of this.

- Doctor Greenie, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Permutations and Combinations
High School Polynomials

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