mdm12Section5_3_OddSolutionsFinal

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 3

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online