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Coefficients in a Trinomial Expansion

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" 

- Doctor Rick, The Math Forum   

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

     C(n,n-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 


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"

     = ----

     = 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 


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 

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

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

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

     (a+b+c+d+e+f)^10        a.b^4.d^2.f^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   
Associated Topics:
College Number Theory
High School Number Theory
High School Permutations and Combinations
High School Polynomials

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