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Trinomial Expansion

Date: 02/25/2003 at 09:45:26
From: Sureyya Hikmet
Subject: Trinomial expansion

How can I find the number of terms of (a+b+c)^140 ?


Date: 02/26/2003 at 00:14:24
From: Doctor Greenie
Subject: Re: Trinomial expansion

Hello, Sureyya -

Let's look at a much more simple example and then extend the analysis 
to your particular case.

Let's find the number of terms in (a+b+c)^3.  We will first list them 
all; then we will find a way to count the number of terms without 
listing them.

The different terms in the expansion of (a+b+c)^3 are the following:

  a^3  b^3  c^3  (3)
  a^2b  a^2c  b^2a  b^2c  c^2a  c^2b  (6)
  abc (1)

By listing them and counting, we see that there are 10 terms in the 
expansion of (a+b+c)^3.

How do we find this number without listing the terms? We know that 
each term must contain 3 variable factors, each of which is either a, 
b, or c. In any term, if x = number of factors of a, y = number of 
factors of b, and z = number of factors of c, then we must have

  x+y+z = 3

So the number of terms in (a+b+c)^3 is equal to the number of ordered 
triples (x,y,z) of non-negative integers with x+y+z = 3.  This in 
turn is equal to the number of ways of partitioning a set of 3 
elements into 3 disjoint subsets.

The number of ways of partitioning a set of 3 elements into 3 disjoint 
subsets can be modeled using 3 markers as the elements of the set and 
2 markers as the 'dividers' between the subsets.  If we use * for an 
element and | for a divider, then we have, for example,

  |**|*  corresponds to the b^2c term  (no elements in the 1st
           subset; two in the 2nd; one in the 3rd - i.e., no
           factors of a, two factors of b, and one factor of c)

  *||**  corresponds to the ac^2 term (one element in the 1st
           subset; none in the 2nd; two in the 3rd - i.e., one
           factor of a, no factors of b, and two factors of c)

So now we see that the number of terms in the expansion of (a+b+c)^3 
is equal to the number of distinct arrangements of the characters

   ***||

This number of arrangements, as you probably know, is equal to

      5!       120
  -------- = ------ = 10
  (3!)(2!)   (6)(2)


Having analyzed the relatively simple case of (a+b+c)^3, we can 
extend our reasoning to say that the number of terms of (a+b+c)^140 
is equal to

     142!      (142)(141)
  ---------- = ---------- = (71)(141) = 10011
  (140!)(2!)        2

I hope all 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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