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Suppose that in a certain state, all automobile license plates have three uppercase letters followed by three digits. Use the method illustrated in Example 9.2.2 to answer the following questions. (a) How many different license plates are possible? To answer this question, think of creating a license plate as a 6-step process, where steps 1-3 are to choose the uppercase letters to put in positions 1-3 and the remaining steps are to choose the digits to put in the remaining positions. There are 17576 ways to perform steps 1-3, and there are 1000 111ways to perform the remaining steps. Thus, the number of license plates is 17576000 (b) How many license plates could begin with A and end in 0? In this case, the number of ways to place the A in Step 1 is 1 and the number of ways to place the 0 in the final step is 1 Thus, the answer is 67600 (c) How many license plates could begin with HVI? In this case, the number of ways to perform steps 1-3 is 1 Thus, the answer is 1000 (d) How many license plates are possible in which all the letters and digits are distinct? 11232000 (e) How many license plates could begin with AB and have all letters and digits distinct? 17260

Suppose that in a certain state, all automobile license plates have three uppercase letters followed by three digits. Use the method illustrated in Example 9.2.2 to answer the following questions. (a) How many different license plates are possible? To answer this question, think of creating a license plate as a 6-step process, where steps 1-3 are to choose the uppercase letters to put in positions 1-3 and the remaining steps are to choose the digits to put in the remaining positions. There are 17576 ways to perform steps 1-3, and there are 1000 111ways to perform the remaining steps. Thus, the number of license plates is 17576000 (b) How many license plates could begin with A and end in 0? In this case, the number of ways to place the A in Step 1 is 1 and the number of ways to place the 0 in the final step is 1 Thus, the answer is 67600 (c) How many license plates could begin with HVI? In this case, the number of ways to perform steps 1-3 is 1 Thus, the answer is 1000 (d) How many license plates are possible in which all the letters and digits are distinct? 11232000 (e) How many license plates could begin with AB and have all letters and digits distinct? 17260

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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**Title: Understanding License Plate Combinations** Suppose that in a certain state, all automobile license plates have three uppercase letters followed by three digits. Use the method illustrated in Example 9.2.2 to answer the following questions. **(a) How many different license plates are possible?** To answer this question, think of creating a license plate as a 6-step process, where steps 1-3 are to choose the uppercase letters to put in positions 1-3 and the remaining steps are to choose the digits to put in the remaining positions. There are \(17576\) ways to perform steps 1-3, and there are \(1000\) ways to perform the remaining steps. Thus, the number of license plates is \(17576000\). **(b) How many license plates could begin with A and end in 0?** In this case, the number of ways to place the A in Step 1 is \(1\), and the number of ways to place the 0 in the final step is \(1\). Thus, the answer is \(676000\). **(c) How many license plates could begin with HV7?** In this case, the number of ways to perform steps 1-3 is \(1\). Thus, the answer is \(1000\). **(d) How many license plates are possible in which all the letters and digits are distinct?** \(11232000\) **(e) How many license plates could begin with AB and have all letters and digits distinct?** \(17280\)
Transcribed Image Text:**Title: Understanding License Plate Combinations** Suppose that in a certain state, all automobile license plates have three uppercase letters followed by three digits. Use the method illustrated in Example 9.2.2 to answer the following questions. **(a) How many different license plates are possible?** To answer this question, think of creating a license plate as a 6-step process, where steps 1-3 are to choose the uppercase letters to put in positions 1-3 and the remaining steps are to choose the digits to put in the remaining positions. There are \(17576\) ways to perform steps 1-3, and there are \(1000\) ways to perform the remaining steps. Thus, the number of license plates is \(17576000\). **(b) How many license plates could begin with A and end in 0?** In this case, the number of ways to place the A in Step 1 is \(1\), and the number of ways to place the 0 in the final step is \(1\). Thus, the answer is \(676000\). **(c) How many license plates could begin with HV7?** In this case, the number of ways to perform steps 1-3 is \(1\). Thus, the answer is \(1000\). **(d) How many license plates are possible in which all the letters and digits are distinct?** \(11232000\) **(e) How many license plates could begin with AB and have all letters and digits distinct?** \(17280\)
**Example 9.2.2: Counting and Probability of PINs** **Problem:** - **b.** How many different PINs are possible if repetition of symbols is not allowed? - **c.** What is the probability that a PIN does not have a repeated symbol assuming that all PINs are equally likely? **Solution:** **a.** Some possible PINs are RCAE, 3387, B92B, and so forth. You can think of forming a PIN as a four-step operation where each step involves placing a symbol into one of four positions, as shown below. **Diagram Explanation:** - The pool of available symbols includes the alphabets A-Z and digits 0-9. - For each of the four positions (1, 2, 3, and 4), there are 36 choices. **Steps:** - **Step 1:** Choose a symbol to place in position 1. - **Step 2:** Choose a symbol to place in position 2. - **Step 3:** Choose a symbol to place in position 3. - **Step 4:** Choose a symbol to place in position 4. There are a fixed number of ways to perform each step, namely 36, regardless of how preceding steps were performed. By the multiplication rule: \(36 \times 36 \times 36 \times 36 = 36^4 = 1,679,616\) PINs in all. **b.** Again, think of forming a PIN as a four-step operation: Choose the first symbol, then the second, then the third, and then the fourth. There are: - 36 ways to choose the first symbol, - 35 ways to choose the second, - 34 ways to choose the third, - 33 ways to choose the fourth. The multiplication rule can be applied: \(36 \times 35 \times 34 \times 33 = 1,413,720\) different PINs with no repeated symbols. **c.** By part (b) there are 1,413,720 PINs with no repeated symbol, and by part (a) there are 1,679,616 PINs in all. Thus the probability that a PIN chosen at random contains no repeated symbol is \(\frac{1,413,720}{1,679,616} = 0.8417\). This means
Transcribed Image Text:**Example 9.2.2: Counting and Probability of PINs** **Problem:** - **b.** How many different PINs are possible if repetition of symbols is not allowed? - **c.** What is the probability that a PIN does not have a repeated symbol assuming that all PINs are equally likely? **Solution:** **a.** Some possible PINs are RCAE, 3387, B92B, and so forth. You can think of forming a PIN as a four-step operation where each step involves placing a symbol into one of four positions, as shown below. **Diagram Explanation:** - The pool of available symbols includes the alphabets A-Z and digits 0-9. - For each of the four positions (1, 2, 3, and 4), there are 36 choices. **Steps:** - **Step 1:** Choose a symbol to place in position 1. - **Step 2:** Choose a symbol to place in position 2. - **Step 3:** Choose a symbol to place in position 3. - **Step 4:** Choose a symbol to place in position 4. There are a fixed number of ways to perform each step, namely 36, regardless of how preceding steps were performed. By the multiplication rule: \(36 \times 36 \times 36 \times 36 = 36^4 = 1,679,616\) PINs in all. **b.** Again, think of forming a PIN as a four-step operation: Choose the first symbol, then the second, then the third, and then the fourth. There are: - 36 ways to choose the first symbol, - 35 ways to choose the second, - 34 ways to choose the third, - 33 ways to choose the fourth. The multiplication rule can be applied: \(36 \times 35 \times 34 \times 33 = 1,413,720\) different PINs with no repeated symbols. **c.** By part (b) there are 1,413,720 PINs with no repeated symbol, and by part (a) there are 1,679,616 PINs in all. Thus the probability that a PIN chosen at random contains no repeated symbol is \(\frac{1,413,720}{1,679,616} = 0.8417\). This means
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