D-Math-W05-Ch6

D-Math-W05-Ch6 - SHEN'S CLASS NOTES-191 Chapter Six...

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SHEN’S CLASS NOTES-191 1 Chapter Six Counting Methods and The Pigeonhole Principle 6.1 Basic Principles Example 1 Suppose Kay’s Quick Lunch has the following lunch menu. APPETIZERS Nachos ………………… 2.15 Salad ……………………… 1.90 Hamburger ………………… 3.15 MAIN COURSES Cheeseburger ……………. . 3.65 Fish Files …………………. 3.15 BEVERAGES Tea ………………………. 0.70 Milk ………………………. . 0.85 Cola ………………………. 0.75 Root Beer ………………. 3.15 We would like to know (1) How many different dinners that consist of one main course and one beverage?
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SHEN’S CLASS NOTES-191 2 (2) How many different dinners that consist of one appetizer, one main course and one beverage? (3) How many different choices do we have if we order a single item, either one of the appetizers, or one of main courses, or one of the beverages? The answer to (1) is 12. It can be seen by listing all of them: HT, HM, HC, HR, CT, CM, CC, CR, FT, FM, FC, FR. Another way to find the number is to consider each category as a set and look at the Cartesian product of M a i n C o u r s e s × Beverages. Tea Milk Cola Root Beer Hamburger Cheeseburger Fish Filet Obviously, there are 12 different combinations. The answer to (2) is obviously equal to 2 × 3 × 4 = 24. The answer to (3) is obviously equal to 2 + 3 + 4 = 9.
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SHEN’S CLASS NOTES-191 3 Multiplication Principle If an activity can be constructed in t successive steps. Moreover, step 1 can be done in n 1 ways, step 2 can be done in n 2 ways, step t can be done in n t ways, then the number of different possible activities is n 1 n 2 n t . Example 2 (6.1.1) How many different dinners are available from the Kay’s Quick Lunch consisting of one main course and an optional beverage? Answer: We can construct a dinner in two steps. Step 1 is to choose a main course. Obviously, n 1 = 3. Step 2 is to choose a beverage, or not choose it at all. Because there are 4 choices of beverage, n 2 = 4 + 1 = 5. Therefore, there are n 1 n 2 = 3 5 = 15 possible dinners. Example 3 (6.1.2) Left for students.
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SHEN’S CLASS NOTES-191 4 Example 4 (6.1.3) (a) How many strings of length 4 can be formed using the letters ABCDE if repetition are not allowed. (b) How many strings of part (a) begin with letter B? (c) How many strings of (a) do not begin with letter B? Answer: (a) We construct the string of length 4 in 4 steps. For the first letter of the string, we have 5 choices, so n 1 = 5. When we construct the second letter, we have n 2 = 4 choices because we cannot use the same letter that is used for the first letter. Similarly, n 3 = 3, and n 4 = 2. By the multiplication principle, the answer to (a) is 5 4 3 2 = 120. (b) We construct the string of length 4 in 4 steps. For the first step, we must choose B, so n 1 = 1. Obviously, we have n 2 = 4, n 3 = 3, and n 4 = 2. By the multiplication principle, the answer to (b) is 1 4 3 2 = 24.
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D-Math-W05-Ch6 - SHEN'S CLASS NOTES-191 Chapter Six...

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