Discrete Mathematics with Graph Theory (3rd Edition) 174

# Discrete Mathematics with Graph Theory (3rd Edition) 174 -...

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172 Solutions to Exercises using (3) of Section 2.2 and its extension in Exercise 19 of Section 5.1. Applying the induction hypothesis to both of the unions of k sets above, we obtain i<j<5.k i<j<£<5.k + (_l)k+IIAI n A2 n··· n Akl + IAk+II- 2: IAi n Ak+11 i<5.k + 2: I(Ai n Ak+I) n (Aj n Ak+I)1 + . .. i<j<5.k Note that (Ai n Ak+d n (Aj n Ak+d = Ai n Aj n Ak+1 with similar simplifications in other terms above. Hence, IAI U A2 u··· U Ak+11 = (2: IAil) + IAk+ll- 2: IAi n Ajl- 2: IAi n Ak+11 i<5.k i<j<5.k i<5.k This is the desired result for n = k + 1. By the Principle of Mathematical Induction, the Principle of Inclusion-Exclusion is true for all n 2: 1. Exercises 6.2 1. (a) [BB] There are three choices for the first course and four for the second. By the multiplication
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Unformatted text preview: rule, the answer is 3 x 4 = 12. (b) [BB] This time she must choose either a morning or an afternoon course. There are 4 + 3 = 7 choices. 2. (a) By the addition rule, there are 2 + 2 + 6 = 10 choices of molding. (b) By the multiplication rule, there are 2 x 4 = 8 choices of metal molding, 2 x 4 = 8 choices of wood molding and 6 x 4 = 24 choices of plastic molding. By the addition rule, the consumer can choose any of 8 + 8 + 24 = 40 moldings. 3. [BB] 10 x 9 x 8 = 720. 4. (a) There are nine choices for the first digit, nine for the second and eight for the third; in all 9 x 9 x 8 = 648 numbers. (b) There are five choices for the last digit, then eight choices for the first and eight choices for the middle; in all, 5 x 8 x 8 = 320 numbers....
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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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