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Unformatted text preview: HOMEWORK 1 SOLUTIONS (1) Let A = { 3 , 4 , 5 } ,B = { 3 , 4 } ,C = { 4 } . Find D = A 4 B 4 C . Solution: Recall from class that A 4 B = ( A \ B ) ∪ ( B \ A ). That is, A 4 B contains all elements that lie in A but not in B and all elements that lie in B but not in A . Note that we also have A 4 B = ( A ∪ B ) \ ( A ∩ B ). Thus ( A 4 B ) 4 C = h ( A 4 B ) \ C i ∪ h C \ ( A 4 B ) i = h ( A \ B ) ∪ ( B \ A ) \ C i ∪ h C \ ( A ∪ B ) \ ( A ∩ B ) i = A \ ( B ∪ C ) ∪ B \ ( A ∪ C ) ∪ C \ ( A ∪ B ) ∪ C ∩ A ∩ B . On the other hand A 4 ( B 4 C ) = h A \ ( B 4 C ) i ∪ h ( B 4 C ) \ A i = h A \ ( B ∪ C ) \ ( B ∩ C ) i ∪ h ( B \ C ) ∪ ( C \ B ) \ A i = A \ ( B ∪ C ) ∪ A ∩ B ∩ C ∪ B \ ( C ∪ A ) ∪ C \ ( B ∪ A ) . A close inspection shows that the last term in both of the above equations are the same. (2) Suppose 70% of Californians like cheese, 80% like apples and 10% like neither. What percentage of Californians like both cheese and apples? Solution: First let’s ask a different question. How many Californians like apples or like cheese? If we add the percentages together we get 70%+80% = 150%. This is clearly an overcounting. The problem is that we have counted those who like both apples and cheese twice. Since 10% of Californians like neither apples nor cheese, we know that 90% either like apples or like cheese (including those who like both). Hence, the difference between the percentages, 150%...
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This note was uploaded on 02/22/2011 for the course MATH 137 taught by Professor Adeboye during the Spring '11 term at UCSB.
 Spring '11
 ADEBOYE
 Graph Theory

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