sol.assig2 - N P i P j = 1 6 In the sum above we can hcoose...

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Solutions for Assignment 2 Discrete Mathematics II Macm 201 (Fall 2010) Section 3.7 Section 8.1:
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faculty of science department of mathematics MACM 201 - D100A SSIGNMENT #2 Solution to the instructor question (a) Let e i be the i -th employee and let P i be the property that e i is not assibned any job. In this problem we are counting the number of assignment that have none of the properties P 1 ,P 2 , or P 3 . Thus by our standard notation, we want N ( P 1 P 2 P 3 ) , where N ( P 1 P 2 P 3 )= N - ± 1 i 3 N ( P i )+ ± 1 i<j 3 N ( P i P j ) - N ( P 1 P 2 P 3 ) . N : is the total number of assignments of jobs, and we have N =3 6 . N ( P i ) : is the number of assignments where c i is not assigned any job. Hence N ( P i )=2 6 . In the sum above we can choose c i ( 3 1 ) ways. N ( P i P j ) : i sthe number of assignments where both c i and c j are not assigned job. Hence
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Unformatted text preview: N ( P i P j ) = 1 6 . In the sum above we can hcoose the pair c i , c j ( 3 2 ) ways. • N ( P 1 P 2 P 3 ) : is the number of assignments where none of employees is assigned a job; clearly there are no such assignments, so N ( P 1 P 2 P 3 ) = 0 . Therefore, N ( P 1 P 2 P 3 ) = 3 6-² 3 1 ³ 2 6 + ² 3 2 ³ 1 6 = 540 . (b) The problem in (a) is a special case of a problem of counting onto function. As we have seen it can be solved using inclusion/exclusion principle and by little observig we can generalize this to n m-² n 1 ³ ( n-1) m + ² n 2 ³ ( n-2) m-··· + (-1) n-1 ² n n-1 ³ 1 m . DR. L. STACHO, FALL 2010 1...
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sol.assig2 - N P i P j = 1 6 In the sum above we can hcoose...

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