Discrete Mathematics, Spring 2004
Homework 4 Sample Solutions
4.2 #77.
Let
s
n,k
denote the number of ways to seat
n
persons at
k
round tables, with
at least one person at each table. (The numbers
s
n,k
are called
Stirling numbers of the first
kind
.) The ordering of the tables is
not
taken into account. The seating arrangement at a
table
is
taken into account, except for rotations.
(a) Show that
s
n,k
= 0 if
k > n
.
(b) Show that
s
n,n
= 1 for all
n
≥
1.
(c) Show that
s
n,
1
= (
n
−
1)! for all
n
≥
1.
(d) Show that
s
n,n

1
=
C
(
n,
2) for all
n
≥
2.
(e) Show that
s
n,
2
= (
n
−
1)!
parenleftbigg
1 +
1
2
+
1
3
+
· · ·
+
1
n
−
1
parenrightbigg
for all
n
≥
2.
(f) Show that
n
∑
k
=1
s
n,k
=
n
!
for all
n
≥
1
.
(g) Find a formula for
s
n,n

2
,
n
≥
3, and prove it.
Solution
.
(a) If
k < n
, there are no ways to place
n
people at
k
tables without at least one of the
tables being empty.
(b) There is exactly one way to place
n
people at
n
tables; each table has a single person
seated at it.
(c) Denote the individuals by
x
1
, . . . , x
n
; we wish to seat them at a single table.
Since
rotations are not taken into account, given any ordering of the
x
i
in a line with
x
1
in the first position, we can associate to it a unique seating around the table; simply
proceed counterclockwise around the table, starting with
x
1
.
Clearly we can reverse
this process, so that given a seating we can obtain an ordering of the
x
i
in a line with
x
1
in the first position.
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
But given that
x
1
must be in the first position of the line, there are exactly (
n
−
1)!
ways to order the remaining elements, and hence
s
n,
1
= (
n
−
1)!.
(d) With notation as in (c), the choice of any unordered pair
{
x
i
, x
j
}
determines a unique
seating of the
n
people at
n
−
1 tables; this pair sits at a single table (there is only one
way to seat them there), and each of the remaining tables has a single person seated
there. So the number of possible seatings is
C
(
n,
2).
(e) We determine a seating of
x
1
, . . . , x
n
at two tables in the following process:
(1) Select a
k
person subset of
{
x
2
, . . . , x
n
}
to sit at the same table with
x
1
.
The
remaining individuals will, of course, sit at the other table. (Notice that
k
cannot
equal
n
−
1, or else no one will be sitting at the other table.)
(2) Order the people sitting at the same table as
x
1
.
(3) Order the people at the other table.
For a fixed value of
k
, there are
C
(
n
−
1
, k
) ways to perform step 1.
By the same
argument as in (c), there are
k
! ways to perform step 2, and there are (
n
−
k
−
2)! ways
to perform step 3 (note that one performs this step after already having chosen the set
of people to sit at this table, so fix one of them and then use the same argument as in
(c)).
Summing over the possible values of
k
, it follows that the total number of ways to seat
x
1
, . . . , x
n
at two tables is given by
n

2
∑
k
=0
C
(
n
−
1
, k
)
·
k
!(
n
−
k
−
2)!
=
n

2
∑
k
=0
(
n
−
1)!
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 WOUTERS
 Math, Algebra, Natural number, ways, nonnegative

Click to edit the document details