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Unformatted text preview: Chapter 1. P1 (a) 26 2 Â· 10 5 (b) 26 Â· 25 Â· 10 Â· 9 Â· 8 Â· 7 Â· 6 P4 (If each of the boys can play all four instruments) 4! = 24 (If Jay and Jack can each play only piano and drums) 2 Â· 1 Â· 2 Â· 1 = 4 ((two options for Jay) Ã— (one remaining option for Jack) Ã— (two remaining options for John) Ã— (one remaining instrument for Jim)) P7 (a) (3 + 3)! = 720 (b) 2 Â· 3! Â· 3! = 72 ((boys before girls or girls before boys) Ã— (the # of all permutations of boys) Ã— (the # of all permutations of girls)) (c) (1 + 3)! Â· 3! = 144 ((the # of all permutations of 4 objects: boys treated as single object, and 3 girls each treated as a separate object) Ã— (the # of all permutations of boys)) (d) 2 Â· 3! Â· 3! = 72 ((two options for the sex of the person sitting first - this determines the sex of people in all remaining sits) Ã— (the # of all permutations of boys) Ã— (the # of all permutations of girls)) P8 (a) 5! (c) 11!...
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This note was uploaded on 10/27/2010 for the course MATH 351 taught by Professor Moumen,f during the Spring '08 term at George Mason.
- Spring '08