MATH 681
Problem Set #3
1.
(10 points)
Prove the following combinatorial identities:
(a)
(5 points)
Recall that
S
(
n, k
)
is equal to the number of ways to subdivide an
n
element set into
k
nonempty parts. Produce a combinatorial argument to show
that
S
(
n, k
) =
kS
(
n

1
, k
) +
S
(
n

1
, k

1)
.
Let us consider the ways to divide
{
1
,
2
, . . . , n
}
into
k
nonempty sets.
There
are two possibilities to be addressed: either
n
is in a set by itself or with other
numbers.
If
n
is in a set by itself, then removal of this set
{
n
}
leaves us with a partition of the
numbers
{
1
,
2
, . . . , n

1
}
into
k

1 sets. There are
S
(
n

1
, k

1) such partitions,
and each of them can be identified uniquely with a
k
partition of
{
1
,
2
, . . . , n

1
}
by adding in the extra set
{
n
}
.
On the other hand, if
n
is in a set with other numbers, removal of
n
from this
set leaves
k
sets behind partitioning the numbers
{
1
,
2
, . . . , n

1
}
.
There are
S
(
n

1
, k
) such possible partitions, but they are
not
uniquely identified with
k
partitions of
{
1
,
2
, . . . , n

1
}
, since the term
n
could have been removed from
(and can be reinserted into) any of the
k
sets. Thus, there are
k
different ways to
build a
k
partition of
{
1
,
2
, . . . , n
}
from a
k
partition of
{
1
,
2
, . . . , n

1
}
, so this
possibility accounts for
k
·
S
(
n

1
, k
) partitions.
Assembling these two cases, we see that
S
(
n, k
) =
S
(
n

1
, k

1) +
kS
(
n

1
, k
).
(b)
(5 points)
Prove that for any
m < n
,
∑
m
k
=0
(
n
k,m

k,n

m
)
= 2
m
(
n
m
)
.
This is actually very much like problem #1 from the previous problem set; it’s
actually a generalization, since that problem is equivalent to the
m
=
n

1 case
of this one.
The left side is fairly easy to interpret: it represents the number of ways to divide
an
n
element set into three distinct parts of size
k
,
m

k
, and
n

m
, for any value
of
k
. So, more properly, it represents the number of ways to divide
n
elements
into classes A, B, and C so that classes A and B together contain
m
elements,
and class C has
n

m
.
The right side can be interpreted as enumerating the same thing, in a different
way. We start by selecting the membership of classes A and B together; we want
to have
m
members, so there are
(
n
m
)
ways to select these, and the remaining
n

m
elements are consigned to class C. Now, we want to freely distribute these
m
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 Fall '09
 WILDSTROM
 Math, Pk, Recurrence relation, Fibonacci number, xk

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