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HW2_sol

# HW2_sol - CSE21 FA11 Homework#2 In this homework we will...

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CSE21 FA11 Homework #2 (10/3/11) In this homework, we will consider “ordinary” decks of playing cards which have 52 cards, with 13 of each of the four suits (Hearts, Spades, Diamonds and Clubs) with each suit have the 13 ranks (Ace, 2, 3, . . . , Jack, Queen, King). (See the text if you need more details). 2.1. (a) How many different rearrangements of all the letters in UNSCRUPULOUS are there? (b) How many different rearrangements of all the letters in supercalifragilisticexpialidocious are there? Solution: (a) Since there are 1 ‘C’, 1 ‘L’, 1 ‘N’, 1 ‘O’, 1 ‘P’, 1 ‘R’, 2 ‘S’, and 4 ‘U’, the number of different rearrangements is 12 1 , 1 , 1 , 1 , 1 , 1 , 2 , 4 . (b) Since there are 3 ‘a’, 3 ‘c’, 1 ‘d’, 2 ‘e’, 1 ‘f’, 1 ‘g’, 7 ‘i’, 3 ‘l’, 2 ‘o’, 2 ‘p’, 2 ‘r’, 3 ‘s’, 1 ‘t’, 2 ‘u’ and 1 ‘x’, the number of different rearrangements is 34 3 , 3 , 1 , 2 , 1 , 1 , 7 , 3 , 2 , 2 , 2 , 3 , 1 , 2 , 1 . 2.2. A 20-member club must have a president, vice-president, secretary, and a treasurer, as well as a three-person nominating committee. If the officers must be different people, and if no officer may be on the nominating committee, in how many ways could the officers and nominating committee be chosen? Answer the same question if officers may be on the nominating committee. Solution: 1

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(a) Note that there are 20 4 ways to choose 4 members out of 20 members as president, vice-president, secretary, and treasurer respectively. Then, for the left 16 members, there are ( 16 3 )
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