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C__work_280sol5

# C__work_280sol5 - Solutions to Homework 5 4.2 3(a By the...

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Unformatted text preview: Solutions to Homework 5 March 30, 2006 4.2: 3 (a) By the product rule, the number of different meals is 5 × 6 × 20 × 3 = 1800 . (b) Here are two ways of doing this: one is to repeatedly apply the sum and product rules as shown below: • main dish only: 20 meals • main dish and soup only: 20 × 5 = 100 meals • main dish and appetizer only: 20 × 6 = 120 meals • main dish and dessert only: 20 × 3 = 60 meals • everything but soup: 20 × 6 × 3 = 360 meals • everything but an appetizer: 20 × 5 × 3 = 300 meals • everything but dessert: 20 × 5 × 6 = 600 meals • everything: 1800 meals (found in part a). Adding these up gives us a total of 3360 meals, the correct answer. A much slicker way of doing this is to think of soup, appetizers, and desserts as each having one more selection, namely the choice not to eat that item. By the product rule, we then have (5 + 1) × (6 + 1) × 20 × (3 + 1) = 3360 meals. 1 Comments: 1. In the alternate solution to (b), we don’t write (20+1) be- cause we must have a main dish. 2. The fact that the two solutions to (b) give the same answer is no coincidence. Indeed, if you expand out (5 + 1) × (6 + 1) × 20 × (3+1) by hand (not doing any of the computations), you get every term that appears in the first solution to (b). 4.2: 5 In terms of subject areas, there are three ways to choose one of History, Philosophy, or Religion; there are two ways to choose one of Math or Science; and there are three ways to choose one of Literature, Fine Arts, or Music. This gives a total of 3 × 2 × 3 = 18 different possibilities for the kinds of courses that a student can take. We must also consider the fact that for each subject area, the student can take one of three courses in that area. Since a schedule consists of three subject areas, the student has three choices for each subject and therefore 3 × 3 × 3 = 27 different schedules for each of the 18 possibilities we found above, giving a total of 18 × 27 = 486 different schedules. 4.2: 14 The computer can do one million checks per second and it costs 1 cent per second to do this, so for every 1 cent, it can do one million (= 10 6 ) checks. Therefore, for every dollar (= 100 cents), it can do 100 × 10 6 = 10 8 checks. Since we have \$10,000 (= \$10 4 ), the computer can do a maximum of 10 4 × 10 8 = 10 12 checks to stay within the budget constraints. The problem boils down to finding the (unique) value of n such that n ! ≤ 10 12 < ( n + 1)! Trying a few values on a calculator shows that n = 14 is the value of n satisfying this relationship. Therefore, we can visit 14 cities not including the hometown....
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C__work_280sol5 - Solutions to Homework 5 4.2 3(a By the...

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