mdm12Section5_3_OddSolutionsFinal

# mdm12Section5_3_OddSolutionsFinal - Q1-7 Q21 Q9-19 CHAPTER...

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Q1-7 Q9-19 Q21 CHAPTER 5 Section 5.3, pp. 286–288 Solutions to Odd Number Problems 1. For each coin, you can choose to include it or choose not to include it. Assuming that you will include at least one coin, number of sums = 2 4 – 1 = 15 3. a) n = 2 2 = 4 b) n = 2 4 = 16 c) n = 2 7 = 128 5. a) Combinations; order is unimportant. 12 C 3 = 220 b) Permutations; order is important. 12 P 3 = 1320 c) Combinations; order is unimportant. ( 3 C 1 ) ( 4 C 1 ) ( 2 C 1 ) = 24 d) Permutations; order is important. 4 P 2 = 12 7. There are 3 chocolate-chip cookies, 2 peanut-butter cookies, 1 lemon cookie, 1 almond cookie, and 5 raisin cookies in the jar. You may choose all, some, or none of each type of cookie. Assume that you will take at least one cookie. a) n = (4)(3)(2)(2)(6) – 1 = 287 b) (3)(3)(2)(2)(6) = 216 Top

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9. There are 24 cards to choose from. a) There are 12 red cards. n (5 red cards) = 12 C 5 = 792 b) n (at least two red cards) = n (no restrictions) – n (no red cards) – n (1 red card) = 24 C 5 – ( 12 C 0 )( 12 C 5 ) – ( 12 C 1 )( 12 C 4 ) = 35 772 c) n (at most 2 red cards) = ( 12 C 0 )( 12 C 5 ) + ( 12 C 1 )( 12 C 4 ) + ( 12 C 2 )( 12 C 3 ) = 21 252 11. 5880 = (2 3 )(7 2 )(3)(5) a) Total number of divisors including 1 is (4)(3)(2)(2) = 48.
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## This note was uploaded on 01/23/2012 for the course MATHS MDM-01 taught by Professor Mr.m during the Spring '10 term at Seneca.

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mdm12Section5_3_OddSolutionsFinal - Q1-7 Q21 Q9-19 CHAPTER...

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