This theorem applies to the expansion of the expression (x+y)
n
.
Oddly enough, this expansion is tied to the counting (in
particular combinations), that we have been doing.
Consider multiplying this expression out long hand:
(x+y)(x+y).
....(x+y)
Now, each term from this expansion is a product of terms,
where each term is chosen from each binomial. One such term
is x
n
, which is created by choosing the x in each binomial.
Another term is x
n1
y. This can be created by choosing the y in
the first binomial and the x in all of the following binomials.
But, you could also get that same term by picking x from the
first binomial, y from the second, and x from the rest. In fact,
you can get that term in exactly n ways.
Thus, the expansion of
(x+y)
n
starts out as x
n
+ nx
n1
y. It seems
tedious to work this entire expansion out in this manner. We
know that each term is going to be of the form x
k
y
nk
, for
integers k that range from 0 to n, inclusive.
So now, the question becomes, can we get a general formula for
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 Spring '09
 nCk

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