HW7-Solutions

# HW7-Solutions - EECS 203 DISCRETE MATHEMATICS Homework 7...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

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

Unformatted text preview: EECS 203: DISCRETE MATHEMATICS Homework 7 Solutions 1. (4 points) There are 15 gold coins, all of which are identical except for one counterfeit coin, which is either slightly heavier or slightly lighter than all the rest. You have at your disposal a balance scale and wish to identify the counterfeit coin and decide whether it is heavier or lighter. Prove, using the pigeonhole principle, that it is not possible to always identify the coin (& whether it is heavier or lighter) in 3 weighings. 1 Solution : There are three outcomes to the first weighing (balanced, tilt left, tilt right), and three outcomes for the 2nd and 3rd, thus 3 3 = 27 outcomes in total. There are 30 possibilities for the correct answer since each of the fifteen coins can be either heavier or lighter than the rest. Thus, there is no assignment of correct answers (pigeons) to outcomes of the 3 weighings (pigeonholes) that doesn’t assign two pigeons to the same hole. In particular there are at least 30- 27 = 3 possible correct answers that can never be guessed. (This question is similar to the “30 questions” game discussed in class.) 2. (4 points) You select n + 1 distinct numbers from the set { 1 , 2 ,..., 3 n } . Prove, using the pigeonhole principle, that there must be two chosen numbers that differ by at most 2. Solution : Define the pigeonholes to be the following n sets of 3 numbers: { 1 , 2 , 3 } , { 4 , 5 , 6 } ,..., { 3 n- 2 , 3 n- 1 , 3 n } . The n +1 numbers that are chosen are the pigeons. There must be two chosen numbers that appear in the same set of three consecutive numbers. These two numbers will differ by at most 2. 3. (4 points) A palindrome is a sequence of letters that reads the same forward & backward. How many 7-letter palindromes are there that (i) contain at least two vowels and (ii) contain no more than two consecutive vowels? (There are 26 letters, 5 of which are vowels.) Solution: Given the two criteria there are 9 possible patterns for vowels (V) and consonants (x): VxxxxxV, xxVxVxx, xVxxxVx, VxxVxxV, xVxVxVx, VxVxVxV, VVxxxVV, xVVxVVx, VVxVxVV....
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

HW7-Solutions - EECS 203 DISCRETE MATHEMATICS Homework 7...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online