To solve this problem, we need to count how many bit strings of length 10 contain either 5 consecutive 0's or 5 consecutive 1's.

Step 1: Use the Inclusion-Exclusion Principle

We will approach this problem using the Inclusion-Exclusion Principle.

Let:

• $A$ be the set of bit strings that contain 5 consecutive 0's.
• $B$ be the set of bit strings that contain 5 consecutive 1's.

We are asked to find the number of bit strings that belong to either $A$ or $B$, which is given by:

$|A \cup B| = |A| + |B| - |A \cap B|$

Where:

• $|A|$ is the number of bit strings with 5 consecutive 0's,
• $|B|$ is the number of bit strings with 5 consecutive 1's,
• $|A \cap B|$ is the number of bit strings with both 5 consecutive 0's and 5 consecutive 1's.

Step 2: Count $|A|$ — Bit Strings with 5 Consecutive 0's

We need to count the number of bit strings of length 10 that contain at least one block of 5 consecutive 0's.

• If a block of 5 consecutive 0's appears, it can start at one of the positions 1 through 6 (because the block of 5 zeros must fit within the 10-bit string).
• Once we place this block, the remaining 5 bits can be filled freely with 0's or 1's.

Thus, for each starting position of the 5 consecutive 0's, there are $2^5$ ways to fill the remaining positions.

So, there are 6 possible positions for the 5 consecutive 0's, and for each position, there are $2^5 = 32$ ways to fill the remaining 5 bits.

Thus, the total number of bit strings with 5 consecutive 0's is: $|A| = 6 \times 32 = 192$

Step 3: Count $|B|$ — Bit Strings with 5 Consecutive 1's

This is similar to counting bit strings with 5 consecutive 0's, but now with 5 consecutive 1's. By the same reasoning, the number of bit strings with 5 consecutive 1's is also: $|B| = 6 \times 32 = 192$

Step 4: Count $|A \cap B|$ — Bit Strings with Both 5 Consecutive 0's and 5 Consecutive 1's

Now, we need to count how many bit strings of length 10 contain both 5 consecutive 0's and 5 consecutive 1's. These two blocks must overlap in some way.

There are only two ways for this to happen:

1. The 5 consecutive 0's and 5 consecutive 1's are arranged such that they do not overlap (e.g., one block of 5 0's and one block of 5 1's).
2. They must overlap in such a way that the remaining positions are filled in a manner that meets both conditions.

However, a bit string that contains both blocks must meet certain conditions to fulfill these patterns, and to keep things simple, this is difficult to resolve quickly with only basic guidance