Discrete Mathematics with Graph Theory (3rd Edition) 177

Discrete Mathematics with Graph Theory (3rd Edition) 177 -...

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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 possibilities-pf 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 one-to-one 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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