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Unformatted text preview: Math331, Spring 2008 Instructor: David Anderson Section 2.2 Hw answers Homework: pgs. 44 and 45 #’s 1 , 3 , 5 , 14 , 15 , 17. 1. Let E i be the possibilities for the i th digit of a 6 digit number. Then E 1 = { 1 , 2 , . . . , 9 } consists of 9 elements and E i = { , 1 , . . ., 9 } for i ∈ { 2 , 3 , . . . , 6 } consists of 10 digits. There fore, by the generalized counting principle, there are 9 × 10 5 = 900 , 000 ways to construct a 6 digit number. To find how many of the 6 digit numbers contain the number 5 we first consider all those numbers that begin with a 5. That leaves the choices from E 2 , . . . , E 6 undecided upon. Therefore, there are 10 5 numbers that begin with a 5. Now we consider the second digit, but only count those without a 5 in the first digit, because those have already been counted. Therefore, the options for E 1 are { 1 , 2 , 3 , 4 , 6 , . . ., 9 } and the options are open for the others....
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This note was uploaded on 03/31/2008 for the course MATH 331 taught by Professor Anderson during the Spring '08 term at University of Wisconsin.
 Spring '08
 Anderson
 Math, Probability

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