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Unformatted text preview: 1 Question 1 (10 points) (a) (2 points) State the principle of inclusion and exclusion for three prop erties p 1 , p 2 and p 3 , and explain the meaning of the terms. Answer : N ( p 1 ∪ p 2 ∪ p 3 ) = N ( p 1 ) + N ( p 2 ) + N ( p 3 ) N ( p 1 ∩ p 2 ) N ( p 1 ∩ p 3 ) N ( p 2 ∩ p 3 ) + N ( p 1 ∩ p 2 ∩ p 3 ), where N ( p 1 ∪ p 2 ∪ p 3 ) is the number of objects that have at least one of the properties p 1 , p 2 and p 3 , N ( p i ) (for i = 1 , 2 , 3) is the number of objects that have property p i , N ( p i ∩ p j ) (for 1 ≤ i < j ≤ 3) is the number of objects that have properties p i and p j , and N ( p 1 ∩ p 2 ∩ p 3 ) is the number of objects that have all three properties p 1 , p 2 and p 3 . (b) (4 points) You have 16 balls. 8 of them are red, 8 of them are made from glass, 7 of them are new, 5 are both red and made from glass, 4 are both new and made from glass, 2 of them are both red and new, and 2 of them are new, red and made from glass. How many of the balls are neither red, new, nor made from glass? How many of the glass balls are not new? Answer : By PIE, there are 16 8 8 7 + 5 + 4 + 2 2 = 2 balls that are neither red, new, nor made from glass. There are 8 glass balls and 4 balls that are both glass and new, hence there are 4 balls that are made from glass, but are not new. (c) (4 points) Consider an object that has properties p 2 , p 4 and p 5 , but not the properties p 1 and p 3 . How many times is such an object counted in the expression N ( p 1 ∩ p 4 ∩ p 5 ) + 2 N ( p 2 ) N ( p 2 ∩ p 3 ∩ p 4 ∩ p 5 ) + summationdisplay 1 ≤ i<j ≤ 5 N ( p i ∩ p j )? Answer : The object is not counted in the first and in the third term, as it does not have properties p 1 and p 3 , respectively. It is counted in the second term (twice), and in N ( p 2 ∩ p 4 ), N ( p 2 ∩ p 5 ) and N ( p 4 ∩ p 5 ) in the last term. Therefore, it is counted 5 times. 2 Question 2 (12 points) Consider the integers 1, 2, . . . , 99999. (a) (3 points) How many of these numbers are divisible by 3 or 5 or 7?...
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This note was uploaded on 07/15/2010 for the course MACM 201 taught by Professor Marnimishna during the Spring '09 term at Simon Fraser.
 Spring '09
 MarniMishna

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