Discrete Mathematics with Graph Theory (3rd Edition) 212

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

This preview shows page 1. Sign up to view the full content.

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
This is the end of the preview. Sign up to access the rest of the document.

## 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.

Ask a homework question - tutors are online