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Unformatted text preview: AMS 301 Final Solution Ning SUN August 20, 2009 1. (20 pt) Derive a recurrence relation for a n , the number of ways to roll a die n times, with no consecutive 4's. Write down a n and compute a 3 . If the rst result is 1, 2, 3, 5 or 6 (5 possibilities), the game restarts after the rst trial. If the rst result is 4, the second result cannot be 4 (it can be 1, 2, 3, 5 or 6, 5 possibilites), and the game restarts after the rst 2 trials. So a n = 5 a n 1 + 5 a n 2 ; a = 1 , a 1 = 6 . a 2 = 5(1 + 6) = 35 , a 3 = 5(6 + 35) = 205 . 2. (20 pt) Among 150 people on a picnic, 90 bring salads or sandwiches, 80 bring sandwiches or cheese, 100 bring salads or cheese, 60 bring at least two foods. (a) How many people bring just salads? De ne A: people who bring salads, D: people who bring sadwiches, C: people who bring cheese. So N = 150 , N ( A ∪ D ) = N 1 + N 2 + ( N 4 + N 5 + N 6 + N 7 ) = 90 , N ( D ∪ C ) = N 2 + N 3 +( N 4 + N 5 + N 6 + N 7 ) = 80 , N ( A ∪ C ) = N 1 + N 3 +( N 4 + N 5 + N 6 + N 7 ) = 100 , N (( A ∩ C ) ∪ ( A ∩ D ) ∪ ( C ∩ D )) = N 4 + N 5 + N 6 + N 7 = 60 . The requested is N ( A ∩ ¯ C ∩ ¯ D ) = N 1 . From the conditions, N 1 + N 2 = 90 60 = 30 , N 2 + N 3 = 80 60 = 20 , N 1 + N 3 = 100 60 = 40 . So N 1 = 25 , N 2 = 5 , N 3 = 15 . So 25 people bring just salads. (b) How many people bring nothing? The requested is N 8 = 150 N 1 N 2 N 3 ( N 4 + N 5 + N 6 + N 7 ) = 150 25 20 60 = 45 . So 45 people bring nothing. 3. (20 pt) How many permutations of the 26 letters are there that contain NONE of the sequences RUNS, FROM or JOE? Constraint: (no RUNS ) AND (no FROM ) AND (no JOE ). So we formulate it as an intersection of ¯ A 1 , ¯ A 2 and ¯ A 3 , where U (universe): all arrangements of the 26 letters; A 1 : all arrangements of the 26 letters containing RUNS ; A 2 : all arrangements of the 26 letters containing FROM ; A 2 : all arrangements of the 26 letters containing JOE . The requested is N ( ¯ A 1 ¯ A 2 ¯ A 3 ) = N N ( A 1 ) N (...
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 Spring '08
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