Math 6a  Solution Set 5
Problem # 1
Show that
S
(
n,
2) = 2
n

1

1 and
S
(
n,
3) =
1
2
(3
n

1

2
n
+ 1) for
n
≥
1
.
Solution.
Note that
S
(
n,
2) counts the number of partitions of an
n
element set
S
into 2
nonempty subsets. So if we call the sets
A
and
B,
each element in
S
can either be placed
in
A
or
B,
yielding 2
n
partitions. However, this counts the partition where
A
is empty and
B
=
S
and the partition where
B
is empty and
A
=
S,
both of which are not partitions into
two nonempty subsets (all other partitions have both
A
and
B
nonempty). Further, each
partition will be counted twice, as we can relabel
A
and
B
and still have the same partition.
Thus
S
(
n,
2) =
1
2
(2
n

2) = 2
n

1

1
.
We prove the second result by induction. Note that if
n
= 1
,
we clearly cannot partition an
n
element set into 3 nonempty subsets, so
S
(
n,
3) = 0 =
1
2
(3
n

1

2
n
+ 1)
.
So assume that it
is true for
n
=
m.
Then consider the case
n
=
m
+ 1
.
Note that we have the formula (from
class)
S
(
n, k
) =
kS
(
n

1
, k
) +
S
(
n

1
, k

1)
so, by induction and our result in part (a),
S
(
m
+ 1
,
3) = 3
S
(
m,
3) +
S
(
m,
2) = 3(
1
2
(3
m

1

2
m
+ 1)) + 2
m

1

1
S
(
m
+ 1
,
3) =
1
2
(3
m
)

3(2
m

1
) + 2
m

1
+
3
2

1 =
1
2
(3
m
)

2
m
+
1
2
S
(
m
+ 1) =
1
2
(3
m
+1

1

2
m
+1
+ 1)
which finishes the induction and shows, for all
n
≥
1
,
that
S
(
n,
3) =
1
2
(3
n

1

2
n
+ 1)
.
Problem # 2
A partition has
type
1
k
1
2
k
2
. . . r
k
r
if it has
k
i
blocks of size
i
for
i
= 1
, . . . , r.
Show that the
number of partitions of an
n
element set of type 1
k
1
2
k
2
. . . r
k
r
is
n
!
Q
r
m
=1
(
m
!)
k
m
(
k
m
)!
.
1
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Solution.
We count the number of pairs (
S, T
) where
S
is a permutation of
{
1
,
2
, . . . , n
}
written as a sequence (
a
1
, a
2
, . . . , a
n
) of distinct elements of
{
1
,
2
, . . . , n
}
and
T
is the parti
tion of
{
1
,
2
, . . . , n
}
of type 1
k
1
2
k
2
. . . r
k
r
formed by taking the first
k
1
of the
a
i
as singleton
sets, the next 2
k
2
of the
a
i
consecutively as pairs of the form
{
a
i
, a
i
+1
}
,
the next 3
K
3
of the
a
i
consecutively in triples of the form
a
i
, a
i
+1
, a
i
+2
,
etc (in general we take the next
jk
j
of
the
a
i
consecutively in
j
element sets of the form
{
a
i
, a
i
+1
, . . . , a
i
+
j

1
}
). So, for example, if
n
= 10 and if
k
1
= 3
, k
2
= 2
,
and
k
3
= 1
,
if
S
= (4
,
1
,
3
,
2
,
5
,
8
,
6
,
9
,
7
,
10) we will have
T
=
{{
4
}
,
{
1
}
,
{
3
}
,
{
2
,
5
}
,
{
8
,
6
}
,
{
9
,
7
,
10
}}
.
Thus, if
N
is the number of all such pairs (
S, T
) it follows that, since for each
S
we have a
unique
T,
and there are
n
! such permutations
S
we must have
N
=
n
!
.
So consider any
T
of type 1
k
1
2
k
2
. . . r
k
r
.
To get
S,
we first choose an order on each of the
elements of each block. A block with
m
elements can be ordered in
m
! ways, and there are
k
m
such blocks for each
m,
so the number of ways to order the elements in each block is
r
Y
m
=1
(
m
!)
k
m
.
Further, we must order the blocks so that the blocks of size 1 come first, followed by the
blocks of size 2
,
etc. However, we can order each of the blocks of size
m
in any order we
choose, and there are
k
m
such blocks, giving (
k
m
)! ways to order these blocks, thus yielding
r
Y
m
=1
(
k
m
)!
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 Fall '07
 Wilson
 Math, Sets, Order theory, Partially ordered set

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