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Unformatted text preview: AMS 301.2 (Spring 2011) Homework Set # 7 – Solution Notes # 14, 5.1: Here we must select in which of the 6 places we should place the “2”, and then in which of the 5 remaining places we should place the “3”. The 4 “1”s will be placed in the remaining 4 spots. So the answer is 6 · 5 = 30. This is equivalent to looking at how many different “spellings” there are using the characters “111123”: 6! 4!1!1! # 18, 5.1: (a). The letters can appear first or last, 2 choices. The number of ways of chooosing 3 letters, repetition allowed and order is important is 26 3 , similarly 3 numbers 10 3 , so the answer is 2 · 26 3 · 10 3 . (b). The number of choices for the letters are (26 + 26 2 + 26 3 ). If there is a single digit, the letters can appear either before it or after, so there are 2 · 10 choices. If there are 2 digits, then the placement of the letters has 3 choices, and there are 100 possibilities for the digits. The final answer is: (26 + 26 2 + 26 3 )(2 · 10 + 3 ·...
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This note was uploaded on 05/28/2011 for the course AMS 301 taught by Professor Estie during the Spring '11 term at University of Florida.
 Spring '11
 Estie

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