How many 8-bit strings (that is, bit strings of length 8) are there which:Start with the sub-string 101? Have weight 5 (i.e., contain exactly five 1’s) and start with the sub-string 101? Either start with 101 or end with 11 (or both)? Have weight 5 and either start with 101 or end with 11?

Sure, let's break down each part of the question:

1. "How many 8-bit strings are there which start with the sub-string 101?"

An 8-bit string has 8 positions that can be either 0 or 1. If the string starts with '101', then the first three positions are already determined. This leaves 5 positions that can be either 0 or 1. Since each position has 2 possibilities (0 or 1), there are 2^5 = 32 such strings.

2. "Have weight 5 (i.e., contain exactly five 1’s) and start with the sub-string 101?"

If the string starts with '101', then two of the five 1's are already determined. This means we need to distribute the remaining three 1's in the remaining 5 positions. This is a combination problem, and can be solved using the binomial coefficient "5 choose 3", which equals 10. So, there are 10 such strings.

3. "Either start with 101 or end with 11 (or both)?"

If a string starts with '101', there are 2^5 = 32 possibilities, as we calculated in part 1. If a string ends with '11', there are 2^6 = 64 possibilities, because 6 positions can be either 0 or 1. However, we're double-counting the strings that both start with '101' and end with '11'. There are 2^3 = 8 such strings. So, the total number of strings that either start with '101', end with '11', or both, is 32 + 64 - 8 = 88.

4. "Have weight 5 and either start with 101 or end with 11?"

This is a more complex problem that combines the previous parts. If a string starts with '101' and has weight 5, there are 10 possibilities, as we calculated in part 2. If a string ends with '11' and has weight 5, we need to distribute the remaining three 1's in the remaining 6 positions, which is "6 choose 3" = 20. However, we're double-counting the strings that both start with '101', end with '11', and have weight 5. To find this, we need to distribute the remaining one 1 in the remaining 3 positions, which is "3 choose 1" = 3. So, the total number of strings that have weight 5 and either start with '101', end with '11', or both, is 10 + 20 - 3 = 27.