Discrete Mathematics with Graph Theory (3rd Edition) 202

Discrete Mathematics with Graph Theory (3rd Edition) 202 -...

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200 Solutions to Exercises (c) Let Al be the set of solutions to (*) with Xl ~ 8. Define A2, A3 and A4 similarly. The question asks for the number of elements in Al C n A2 c n A3 c n A4 c = (AI U A2 U A3 U A4)C (by one of the laws of De Morgan). By the Principle of Inclusion-Exclusion, IA l UA 2 UA 3 UA 4 1 = L IAil- L IAi n Ajl + L IAi n Aj n Akl-IAl n A2 n A3 n A41 i<j i<j<k By part (a), JAIl = 560. Similarly, each of the four terms in Ei IAil is 560. By part (b), IA l nA 2 1 = 56 and similarly, each of the @ = 6 terms in Ei<j IAinAjl is 56. Now AlnA2 nA3 is the set of solutions to (*) with Xl, X2, X3 ~ 8 and z ~ o. There are no such solutions! Thus, IAI U A2 U A3 U A41 = 4(560) - 6(56) = 1904. Since the number of solutions to (*) in nonnegative integers is (~f) = 2024, the number of solutions with each variable at most seven is 2024 - 1904 = 120. (d) We proceed as in part (c). Let Ai be the set of all solutions to the given equation in which variable Xi ~
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.

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