How many bit strings of length 10 contain either 5 consecutive 0’s or 5 consecutive 1’s

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

How many bit strings of length 10 contain either 5 consecutive 0’s or 5 consecutive 1’s?

Learn how to calculate the number of bit strings of length 10 that contain either 5 consecutive 0’s or 5 consecutive 1’s using the Inclusion-Exclusion Principle. The solution involves combinatorics and key counting principles, including how to handle overlapping conditions.

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