Combinatorics Through Guided Discovery, 2004a

1.3. Some Applications of the Basic Principles 23
a The result we will derive is called the Chung-Feller Theorem; this approach is based of a
paper of Wen-jin Woan “Uniform Partitions of Lattice Paths and Chung-Feller Generalizations,”
American Mathematics Monthly 58 June/July 2001, p556.
1.3.2 The Binomial Theorem
◦ Problem 53. We know that (x + y) 2 = x 2 +2xy + y 2 . Multiply both sides
by (x + y) to get a formula for (x + y) 3 and repeat to get a formula for
(x + y) 4 . Do you see a pattern? If so, what is it? If not, repeat the process
to get a formula for (x + y) 5 and look back at Figure 1.2.4 to see the pattern.
Conjecture a formula for (x + y) n .
• Problem 54. When we apply the distributive law n times to (x + y) n ,we
get a sum of terms of the form x i y n−i for various values of the integer i. If
it is clear to you that each term of the form x i y n−i that we get comes from
choosing an x from i of the (x + y) factors and a y from the remaining n − i
of the factors and multiplying these choices together, then answer part a of
the problem and skip part b. In either case, be sure to answer part c.
(a) In how many ways can we choose an x from i terms and a y from n − i
terms?
(b) We can take this step-by-step and consider a small case to get started.
(i) Expand the product (x 1 + y 1 )(x 2 + y 2 )(x 3 + y 3 ).
(ii) What do you get when you substitute x for each x i and y for each
y i ?
(iii) Now imagine expanding
(x 1 + y 1 )(x 2 + y 2 ) ···(x n + y n ).
Once you apply the commutative law to the individual terms you
get, you will have a sum of terms of the form
x k1 x k2 ···x ki · y j1 y j2 ···y jn−i .
What is the set {k 1 , k 2 ,...,k i }∪{j 1 , j 2 ,...,j n−i }?
(iv) In how many ways can you choose the set {k 1 , k 2 ,...,k i }?
(v) Once you have chosen this set, how many choices do you have
for { j 1 , j 2 ,...,j n−i }?
(vi) If you substitute x for each x i and y for each y i , how many terms
of the form x i y n−i will you have in the expanded product
(x 1 + y 1 )(x 2 + y 2 ) ···(x n + y n )=(x + y) n ?