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MATH 681
Problem Set #2 Solutions
1.
(10 points)
Prove the following identity combinatorially:
n
X
i
=0
i
±
n
i
²
=
n
2
n

1
Both sides can easily be interpreted as enumerating the ways to select an element
x
of
{
1
,
2
,
3
,...,n
}
, and then choose a subset
S
from among the remaining
n

1 elements.
The right side of this equation clearly enumerates the set of ways to do the above:
there are
n
choices for
x
, and, regardless of the choice of
x
,
S
is a subset of a set of
size
n

1; thus there are 2
n

1
choices of
S
.
To calculate the left side of the equation, we begin by choosing a set
T
which will
contain both
S
and
x
, then we select
x
as an element of
T
and deﬁne
S
=
T
 {
x
}
.
Thus, one may uniquely determine
x
and
S
by, instead of choosing
x
then a set
S
not
containing
x
, choosing a set
T
and an element thereof to serve as
x
. For any value of
i
, there are
(
n
i
)
possible choices of
T
of size
i
; then, having schosen a set
T
of size
i
,
any of its
i
elements can serve as a value of
x
, so there are
i
(
n
i
)
selections of
T
and
x
given

T

=
i
. Ranging over all possible sizes of
T
, we see that there are
∑
n
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This note was uploaded on 01/12/2012 for the course MATH 681 taught by Professor Wildstrom during the Fall '09 term at University of Louisville.
 Fall '09
 WILDSTROM
 Math

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