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Unformatted text preview: with n elements. The number of subsets of S is
2n. The number of subsets with 0, 1, 2,..., n elements is given by
n
n
n
n
,
,
,...,
0
1
2
n Since subsets of S need to have between 0 and n elements, the claim
of the theorem follows.
31 Binomial Theorem
Theorem: Let x and y be variables. Let n be a
nonnegative integer. Then
n (x + y ) = n
k=0 n k n− k
xy
k Proof: Let us expand the left hand side. The terms of
the product in expanded form are xkynk for 0 <= k <= n.
To obtain the term xkynk one choose k x’s from
must
the n (x+y) terms. There are n ways to do that. k 32 Another Binomial Coefficient Identity
Corollary: Let n be a positive integer. Then
n
kn
=0
(−1)
k
k=0 Proof: We have 0 = 0n = (1+1)n. Expanding the right hand
side with the help of the binomial theorem, we obtain the
claim.
[This also says that the number of subsets with an even number
of elements is equal to the number of subsets with an odd number
of elements.]
33 Pascal’s Identity (1)
Theorem: Let n and k be positive integers with n>=k. Then n+1
k =
n
n
+
k−1
k Proof: We are going to prove this by counting the number of subsets
with k elements of a set T with n+1 elements in two different ways.
First way of counting:
The set T clearly contains n+1
k 34 subsets of size k. Pascal’s Identity (2)
Second way of counting:
Recall that T is a set with n+1 elements. Let us consider an element t
of T. We will count the subsets of T of size k that (a) contain the
element t, and (b) do not contain the element t.
(a) There are n
k−1 subsets of T that contain t, since t is already chosen, but the remaining k1 elements need to be chosen
from T{t}, a set of size n.
(b) There are
n
k subset of T not containing t, since one can choose any k elements from the set T{t} with n elements.
Since the two cases are exhaustive, C(n+1,k) = C(n,k1) + C(n,k). Pascal’s Identity (3) The first way of counting yields t...
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 Fall '11
 math

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