Section 6.3
175
24. For a particular
i,
the entire term
pfi
must be a factor
of
s
or
of
t
(since gcd(s,
t)
=
1). Hence, for
each
i,
there are two
possibilitiespf
i
is in
s
or in
t.
Thus, there are
2T
ways
of
dividing the powers
between
sand
t.
In this scheme, however, a given pair
s,
t
is counted twice, once in the order
s,
t
and
once
as
t,
s.
So the answer is
~(2T)
=
2
T

l
.
25.
(a) [BB] The easiest way to see that there are
2
n
functions from
A
to
B
is to note that
{ai
I
(ai'O)
E
f}
+4
f
is a onetoone correspondence between the
(2n)
subsets
of
A
and the set
of
functions
f:
A+B.
Alternatively, one can argue as follows. There are two choices for
j
such that
(aI,
j)
is part
of
a function. For each
of
these choices, there are two possibilities for
j
such that
(a2,
j)
is part
of
a function. Thus, there are four possibilities for both the pairs
(aI,
jl),
(a2,
j2)
to be part
of
a
function. Continuing this way, we see that there are
2
n
functions from
A
to
B.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Summer '10
 any
 Set Theory, Graph Theory, Recursion, Natural number, Structural induction

Click to edit the document details