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2. How many bit strings(strings of 1 or 0) of length 8 are there such that it contains no more nan 3 1s?

2. How many bit strings(strings of 1 or 0) of length 8 are there such that it contains no more nan 3 1s?

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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**Question:** How many bit strings (strings of 1 or 0) of length 8 are there such that it contains no more than 3 1s? **Explanation:** This problem asks about counting the number of bit strings with specific constraints. A bit string of length 8 means there are 8 positions, each of which can be either a 0 or a 1. The condition here is that there can be no more than 3 ones (1s) in the string. To solve this, you would calculate the sum of the possible combinations with 0, 1, 2, and 3 ones using combinations (binomial coefficients): 1. **0 ones:** C(8, 0) = 1 2. **1 one:** C(8, 1) = 8 3. **2 ones:** C(8, 2) = 28 4. **3 ones:** C(8, 3) = 56 **Total number of bit strings = C(8, 0) + C(8, 1) + C(8, 2) + C(8, 3) = 1 + 8 + 28 + 56 = 93** This solution leverages the concept of combinations where C(n, k) is the number of ways to choose k items from a set of n items without regard to order.
Transcribed Image Text:**Question:** How many bit strings (strings of 1 or 0) of length 8 are there such that it contains no more than 3 1s? **Explanation:** This problem asks about counting the number of bit strings with specific constraints. A bit string of length 8 means there are 8 positions, each of which can be either a 0 or a 1. The condition here is that there can be no more than 3 ones (1s) in the string. To solve this, you would calculate the sum of the possible combinations with 0, 1, 2, and 3 ones using combinations (binomial coefficients): 1. **0 ones:** C(8, 0) = 1 2. **1 one:** C(8, 1) = 8 3. **2 ones:** C(8, 2) = 28 4. **3 ones:** C(8, 3) = 56 **Total number of bit strings = C(8, 0) + C(8, 1) + C(8, 2) + C(8, 3) = 1 + 8 + 28 + 56 = 93** This solution leverages the concept of combinations where C(n, k) is the number of ways to choose k items from a set of n items without regard to order.
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