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### Coefficients in a Trinomial Expansion

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Date: 04/24/2001 at 14:34:48
From: Mark Birkenfeld
Subject: Coefficient in trinomial expansion

In the expansion of (a+b+c)^6, what is the coefficient of a^2.b^2.c^2?

Without taking 10 or 15 minutes to multiply this out on paper, how can
I find the answer? I figure that the answer is related to the binomial
theorem and Pascal's triangle, but I cannot find any text to back up
my hypothesis.
```

```
Date: 04/24/2001 at 16:44:44
From: Doctor Rick
Subject: Re: Coefficient in trinomial expansion

Hi, Mark.

I would solve the problem using combinatorics. The question is
equivalent to this one: How many different six-letter "words" can be
formed using two a's, two b's, and two c's? To answer it, first
pretend the two a's are different (a1 and a2), and similarly for the
b's and the c's. Then there are 6! permutations of the six distinct
symbols. But any one of these "words" is in fact identical to the
"word" obtained by swapping the two a's; thus we must divide the total
by 2. Likewise we must divide by 2 again to account for the identical
b's, and again to account for the identical c's. The final answer is:

6!/2^3 = 6*5*4*3*2/(2*2*2) = 90

You can see the connection to the binomial theorem: for example, the
coefficient of a^3.b^3 in the expansion of (a+b)^6 is:

/6\      6!
(   ) = ----- = 6*5*4*3*2/(3*2*3*2) = 20
\3/    3! 3!

Here we again start with the permutations of six symbols. This time we
divide by the number of ways that the three a's can be permuted, and
the number of ways that the three b's can be permuted.

We're already very close to coming up with a general formula for the
coefficient of any term in the trinomial expansion. We just don't have
a single symbol for the answer, equivalent to the "combinations"
symbol.

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 04/24/2001 at 16:46:57
From: Doctor Greenie
Subject: Re: Coefficient in trinomial expansion

Hi, Mark --

Yes, you are right - this is very closely related to the binomial
theorem, Pascal's triangle, and "m choose n."

Let's look first at the more familiar binomial theorem, Pascal's
triangle, and the expansion of (a+b)^n.

As you know, the expansion of (a+b)^n is:

(a+b)^n = C(n,n)*a^n + C(n,n-1)*a^(n-1)b + ... + C(n,0)*b^n

The coefficient of the a^(n-k).b^k term in this expansion is

n!
C(n,n-k) = ---------
(n-k)!*k!

One place where this number C(n,n-k) comes up is as the number of
distinct ways to arrange (n-k) objects of one type and (k) objects of
another type - for example, the number of distinct arrangements of
the letters "aaaaabbb" is C(8,3) or C(8,5).

Let's try to draw a parallel between the expansion of (a+b)^n and the
arranging of a collection of two different kinds of objects.

Let's think of (a+b)^5 written out as the product of five binomials:

(a+b)^5 = (a+b)*(a+b)*(a+b)*(a+b)*(a+b)

Now let's think of looking at this multiplication and see if we can
determine the coefficient of the a^3.b^2 term in the expansion.

As an extension of the familiar FOIL method for multiplying two
binomials, the expansion of (a+b)^5 is found by summing all the
partial products formed by choosing one of the two terms in each of
the five binomial factors.

So in expanding (a+b)^5, we can write down all the partial products
that will contain three factors of a and two factors of b with the
following notation: ababa will be the partial product formed by
taking factors of a from the first, third, and fifth binomials and
factors of b from the second and fourth. Then our list of partial
products containing three factors of a and two factors of b is the
following:

aaabb
aabab
aabba
abaab
ababa
abbaa
baaab
baaba
babaa
bbaaa

There are ten such partial products. 10 is C(5,3); specifically, 10 is
the number of distinct arrangements of 3 a's and 2 b's. With our
notation for partial products as in the preceding list, there is a
one-to-one correspondence between the partial products a^3.b^2 and the
arrangements of three a's and two b's. So we have:

coefficient of (a^3b^2) term in (a+b)^5

= # partial products with 3 factors of 'a' and 2 factors of 'b'

= # arrangements of letters "aaabb"

5!
= ----
3!2!

= 10

Now we can expand this idea to the expansion of (a+b+c)^n, or
(a+b+c+d+e+f)^n, or to the expansion of any polynomial.

For your particular problem, you have (a+b+c)^6, which you can write
as:

(a+b+c)*(a+b+c)*(a+b+c)*(a+b+c)*(a+b+c)*(a+b+c)

If you think of using the extension of the FOIL method with this
multiplication, and if you want to find the coefficient of the
a^2.b^2.c^2 term, then you want to count the number of ways you can
form a partial product by choosing the a from two of the six
trinomial factors, the b from two of the other six, and the c from
the remaining two factors. But this number of ways is simply the
number of distinct arrangements possible of the letters aabbcc.

Here are a few examples of the general application of this idea (you
are already familiar with the first example, because of the binomial
theorem):

in the expansion      the coefficient of
of...               the... term                is...
-----------------     ------------------     --------------------

5!
(a+b)^5                a^3.b^2               ----  =  10
3!2!

6!
(a+b+c)^6                a^4.bc              ------  =  30
4!1!1!

10!
(a+b+c+d+e+f)^10        a.b^4.d^2.f^3         ------------  =  ?
1!4!0!2!0!3!

(Note the 0! factors in the denominator of the last example,
indicating 0 factors of c and 0 factors of e. This is one example of
why 0! is defined to have the value 1.)

I hope this helps. Write back if you have more questions.

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

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