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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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