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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.
(b) [BB] Of the
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 Summer '10
 any
 Graph Theory

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