Math 15B
Final Exam Solutions
6 December 2001
1. In each case,
give an example or explain why none exists
.
(a) A permutation
f
of
{
1
,
2
,
3
,
4
,
5
}
such that, for
some
x
∈{
1
,
2
,
3
,
4
,
5
}
,
f
20
(
x
)
6
=
x
.
A.
In cycle form, choose
f
to have a 3cycle and
x
to belong to the cycle. (There
are 120 correct answers.)
(b) A permutation
f
of
{
1
,
2
,
3
,
4
,
5
}
such that, for
every
x
1
,
2
,
3
,
4
,
5
}
,
f
20
(
x
)
6
=
x
.
A.
f
20
(
x
)=
x
if and only if
x
belongs to a cycle of length
k
for some divisor
k
of 20.
Since the domain has 5 elements,
k
=1
,
2
,
4, or 5. Thus,
f
20
(
x
)
6
=
x
implies that
x
belongs to a 3cycle. Hence
every
x
1
,
2
,
3
,
4
,
5
}
must belong to a 3cycle,
which is impossible.
(c) A tree with exactly 10 vertices and exactly 10 edges.
A.
A tree with
n
vertices has
n

1 edges, so this is impossible.
2. In each case,
give an example or explain why none exists
.
(a) A function
f
(
n
) such that
f
(
n
)isin
O
(
n
2
) but
f
(
n
) is not Θ(
n
2
).
A.
Any function which grows slower than
n
2
will work; for example,
f
(
n
n
.
(b) A function
f
(
n
) such that
f
(
n
)is
O
(
n
log
n
) but
f
(
n
) is not
O
(
n
2
).
A.
Impossible, since anything that doesn’t grow faster than
n
log
n
does not grow
faster than
n
2
. (Alternatively, you could note that (
n
log
n
)
/n
2
is bounded for
large
n
.)
(c) A probability space (
U, P
) and two subsets
S
and
T
of
U
such that
P
(
S
P
(
T
)=2
/
3 and
S
6
=
T
.
A. There are many possible answers.
For example,
U
=
{
a, b, c
}
,
P
is uniform,
S
=
{
a, b
}
and
T
=
{
b, c
}
. Another:
S
=
U
=
{
a, b
}
,
T
=
{
a
}
and
P
(
a
) is chosen
greater than 1
/
2.