Discrete Mathematics with Graph Theory (3rd Edition) 212

Discrete Mathematics with Graph Theory (3rd Edition) 212 -...

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210 Solutions to Exercises But (~) + (n ~ 1) + . .. + (i ~~) + (i) (n-l) [(n-2) (n-2)] [( ~) (~)] (~-1) ~ [(~ ~ 1)++ (n ~ 2) : .. ~ (~~6l++ [(n-i 2 )++ (n-~13) : .. ~+-(:~:) 1 = an-l + an-2, by the induction hypothesis = an, as before. In either case, we obtain an = an-l + an-2, the recurrence relation which we know defines the Fibonacci numbers. By the Principle of Mathematical Induction we conclude that this is true for all n2':1. 24. (a) There is just one way to select 0 books from n. (b) We prove this by contradiction. So assume the result is not true. Then we can find k 2': ~ + 1 such that [~] =I- 0; that is, a selection of the required type can be made. Label the books 1,2, ... ,n from left to right and first assume n is even. If we partition the books into the ~ sets {I, 2}, {3, 4}, ... , {n - 1, n}, the Pigeon-Hole Principle immediately tells us that two of the k books selected must be in the same set, contradicting the fact that no two adjacent books are to be chosen. Now assume
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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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