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Problem 15 How many bit strings of length 10 have 1. Exactly three 0's? 2. More 0's than 1's? 3. At leat seven 1's? 4. At least three 1's?

Problem 15 How many bit strings of length 10 have 1. Exactly three 0's? 2. More 0's than 1's? 3. At leat seven 1's? 4. At least three 1's?

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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### Problem 15 **How many bit strings of length 10 have:** 1. Exactly three 0’s? 2. More 0’s than 1’s? 3. At least seven 1’s? 4. At least three 1’s? In the above problem, you are asked to determine the number of bit strings (sequences of 0s and 1s) of length 10 that meet various criteria. Let's break down each part to understand how to approach the problem. 1. **Exactly three 0’s:** - We need to find the number of bit strings of length 10 that contain exactly three 0’s. This can be calculated using combinations. The number of ways to choose 3 positions out of 10 for the 0's is given by the binomial coefficient \( \binom{10}{3} \). 2. **More 0’s than 1’s:** - For there to be more 0's than 1's in a bit string of length 10, there must be at least 6 zeros, because a 5-5 split would not meet the criteria. We need to count the number of bit strings with 6, 7, 8, 9, and 10 zeros respectively. 3. **At least seven 1’s:** - We need bit strings with at least seven 1's. This means 7, 8, 9, or 10 ones. Therefore, we need to sum the number of ways to have exactly 7 ones, 8 ones, 9 ones, and 10 ones in a sequence of 10 bits. 4. **At least three 1’s:** - For this part, we need the total number of bit strings with at least three 1's which is equivalent to the total number of bit strings minus the number of bit strings with fewer than three 1's. The cases with fewer than three 1's include 0, 1, and 2 ones. By understanding and applying combinatorial principles and binomial coefficients, one can determine the exact counts for each scenario described in the problem.
Transcribed Image Text:### Problem 15 **How many bit strings of length 10 have:** 1. Exactly three 0’s? 2. More 0’s than 1’s? 3. At least seven 1’s? 4. At least three 1’s? In the above problem, you are asked to determine the number of bit strings (sequences of 0s and 1s) of length 10 that meet various criteria. Let's break down each part to understand how to approach the problem. 1. **Exactly three 0’s:** - We need to find the number of bit strings of length 10 that contain exactly three 0’s. This can be calculated using combinations. The number of ways to choose 3 positions out of 10 for the 0's is given by the binomial coefficient \( \binom{10}{3} \). 2. **More 0’s than 1’s:** - For there to be more 0's than 1's in a bit string of length 10, there must be at least 6 zeros, because a 5-5 split would not meet the criteria. We need to count the number of bit strings with 6, 7, 8, 9, and 10 zeros respectively. 3. **At least seven 1’s:** - We need bit strings with at least seven 1's. This means 7, 8, 9, or 10 ones. Therefore, we need to sum the number of ways to have exactly 7 ones, 8 ones, 9 ones, and 10 ones in a sequence of 10 bits. 4. **At least three 1’s:** - For this part, we need the total number of bit strings with at least three 1's which is equivalent to the total number of bit strings minus the number of bit strings with fewer than three 1's. The cases with fewer than three 1's include 0, 1, and 2 ones. By understanding and applying combinatorial principles and binomial coefficients, one can determine the exact counts for each scenario described in the problem.
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