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Unformatted text preview: Discrete Mathematics, Spring 2004 Homework 3 Sample Solutions 4.1 #52. How many integers from 5 to 200 (inclusive) are of the form xyz , where 0 negationslash = x < y and y > z ? Solution . Since x negationslash = 0, we must have x = 1; thus we are looking for integers of the form 1 yz , where 2 ≤ y ≤ 9 and 0 ≤ z ≤ y − 1. Given each choice of the digit y , we clearly have y choices for the digit z , and hence the number of integers with the desired form is 9 ∑ y =2 y = 2 + 3 + 4 + ... + 9 = 44 . 4.1 #53. (a) In how many ways can the months of the birthdays of five people be distinct? (b) How many possibilities are there for the months of the birthdays of five people? (c) In how many ways can at least two people among five have their birthdays in the same month? Solution . Denote the five people by A, B, C, D, E. (a) If the birthmonths of these people are distinct, we have 12 choices for the birthmonth of A. Then there are 11 remaining choices for the birthmonth of B, and then 10 remaining choices for the birthmonth of C, and so on. Hence there are 12 · 11 · 10 · 9 · 8 possibilities, by the Multiplication Principle. (b) Now there is no restriction on the birthmonths being distinct, so we have 12 choices for the birthmonths of each of A, B, C, D, and E. Hence there are 12 5 possibilities, again by the Multiplication Principle. (c) Any choice of the birthmonths for the people A, B, C, D, E satisfies either the condition in (a) or the condition in (c), but not both. Then by the Addition Principle, we findin (a) or the condition in (c), but not both....
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This note was uploaded on 06/12/2011 for the course MATH 103 taught by Professor Wouters during the Spring '08 term at Wisc Oshkosh.
 Spring '08
 WOUTERS
 Algebra, Integers

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