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Unformatted text preview: EE 351K PROBABILITY & RANDOM PROCESSES FALL 2011 Instructor: Sujay Sanghavi [email protected] Homework 3 Solution Problem 1 Count the number of distinguishable ways in which you can arrange the letters in the words: (a) children (b) bookkeeper Sol : (a) The word “children” consists of 8 distinct letters, so the number of arrangements is the same as the number of possible permutations, namely 8!. (b) In the word “bookkeeper”, the letters “b”, “o”, “k”,“e”, “p”, and “r” appear 1, 2, 2, 3, 1, and 1 times respectively. Arguing exactly as in the text the number of distinguishable rearrangements is 10! 3! 2! 2! . Problem 2 In how many ways can 8 people be seated in a row if (a) there are no restrictions on the seating arrangement; (b) persons A and B must sit next to each other; (c) there are 4 men and 4 women and no 2 men or 2 women can sit next to each other; (d) there are 5 men and they must sit next to each other; (e) there are 4 married couples and each must sit together? Sol : (a) Any person can sit any seats. There are thus 8! possibilities. (b) First consider A and B as one person, arrange seating for 7 people without any restriction. Then permute A and B for every possible arrangement. The answer is 7!2!. (c) Similar to (b), pair one man and one woman as one person. Arrange seating for 4 virtual people first to get 4!. There are also 4! unique ways to pair a man and a woman, similar to arranging 4 men in 4 seats. There are two alternative man and woman seating for every arrangement and pairing, man onseats....
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This note was uploaded on 02/20/2012 for the course EE 351k taught by Professor Bard during the Spring '07 term at University of Texas.
 Spring '07
 BARD

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