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Unformatted text preview: EECS 203: Homework 7 Solutions Section 5.1 1. (E) 24 a) Since the strings do not contain the same digit, there are 10 ways to choose the first digit, 9 ways to choose the second and so on. Therefore the answer is 10 9 8 7 = 5040 . b) There are 10 ways to choose each of the first three digits and 5 ways to choose the last digit. So the answer is 10 10 10 5 = 5000 . c) There are 4 ways to choose the position that is not 9 and 9 ways to choose the digit in that position. Therefore the answer is 4 9 = 36 . 2. (E) 30fgh f) The first two letters are fixed and we can choose 6 letters for the remaining 6 positions. Hence the answer is 26 6 = 308915776 . g) The first and last two letters are fixed and we can choose 4 letters for the remaining 4 positions. Thus the answer is 26 4 = 456976 . h) From part f, we have 26 6 strings that start with the letters BO. Similarly, it can be shown that there are 26 6 strings that end with the letters BO. By part g, we know that there are 26 4 strings that start and end with BO. So, by the inclusion exclusion principle, there are 26 6 + 26 6 26 4 strings that start or end with BO. The answer is 26 6 + 26 6 26 4 = 617374576 . 3. (E) 34 Each element in the domain { 1 , 2 , . . ., n } can map to either or 1 . Thus there are 2 n distinct functions from the set { 1 , 2 , . . ., n } to the set { , 1 } . The answer is 2 n . 4. (E) 40 a) We first need to choose 5 more people from the group except the bride and there are C (9 , 5) such ways. Moreover there are 6! ways to arrange the selected 5 people and the bride in a row. Thus, the answer is C (9 , 5) 6! = 90720 . b) We need to choose 4 more people except both the bride and the groom, giving us C (8 , 4) ways. Since we have 6! ways to arrange 6 people in a row, the answer is C (8 , 4) 6! = 50400 ....
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This note was uploaded on 02/12/2012 for the course EECS 203 taught by Professor Yaoyunshi during the Spring '07 term at University of Michigan.
 Spring '07
 YaoyunShi

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