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Concrete mathematics : a foundation for computer science

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The empty set<br />

is pointless.<br />

A ANSWERS TO EXERCISES 547<br />

7.44 Each partition into k nonempty subsets can be ordered in k! ways, so<br />

bk = k!. Thus Q(Z) = ,&k>O {E}k!zn/n! = tkao(eZ - l)k = l/(2 - e’).<br />

And this is the geometric series tkao ekZ/2k+‘, hence ok = 1 /2kf’. Finally,<br />

ck = 2k; consider all permutations when the x’s are distinct, change each ‘>’<br />

between subscripts to ‘

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