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Unformatted text preview: 1. Consider a set of objects, each of which may satisfy one or more of four properties p 1 , p 2 , p 3 and p 4 . Write a formula for the number of objects that satisfy exactly one or exactly three of these properties (using only the terms of form N ( p i 1 ∩ p i 2 ∩ . . . ) that appear in the principle of inclusion and exclusion). There are several ways how to solve this exercise. One of them is to simply sum the expression for the number of objects satisfying exactly one property with the expression for the number of objects satisfy ing exactly three properties as given by the generalized Principle of Inclusion and Exclusion. Another, more elementary way follows the derivation of PIE: • In N 1 = N ( p 1 ) + N ( p 2 ) + N ( p 3 ) + N ( p 4 ), the object with one property is counted once, the one with two properties is counted 2 × , the one with three properties 3 × and the one with all four properties is counted 4 × . To shortens these descriptions, let us say that this expression counts the objects (1 , 2 , 3 , 4) × . • N 2 = ∑ 1 ≤ i<j ≤ 4 N ( p i ∩ p j ) counts the objects (0 , 1 , 3 , 6) × . As we do not want to count the objects with two properties, we consider the expression N 1 − 2 N 2 , that counts the objects (1 , , − 3 , − 8) × . • N 3 = ∑ 1 ≤ i<j<k ≤ 4 N ( p i ∩ p j ∩ p k ) counts the objects (0 , , 1 , 4) × , thus in order to count the objects with three properties once, we need to consider the expression N 1 − 2 N 2 + 4 N 3 , that counts the objects (1 , , 1 , 8) × ....
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 Spring '09
 MarniMishna
 Equals sign, 2 K, AirTrain Newark

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