To determine the number of bit strings of length $n$ that start and end with 1s, let's analyze the problem step by step:

Explanation:

1. A bit string of length $n$: A string of 0s and 1s that has exactly $n$ characters.
2. Conditions: The first and last characters of the string must both be 1.

Derivation:

• The first and last positions are fixed as 1.
• This leaves $n - 2$ positions in between these fixed 1s.
• Each of these $n - 2$ positions can independently be either 0 or 1.

Thus, the total number of bit strings satisfying these conditions is: $2^{n-2}$ where $n \geq 2$ (since $n = 1$ doesn't allow both a starting and ending 1).

Example:

1. For $n = 3$: Possible bit strings are $101$, $111$. Total = $2^{3-2} = 2$.
2. For $n = 4$: Possible bit strings are $1001, 1011, 1101, 1111$. Total = $2^{4-2} = 4$.

Final Answer:

The number of bit strings of length $n$ that start and end with 1s is: $2^{n-2}, \quad \text{for } n \geq 2.$

If $n = 1$, the only string possible is $"1"$, and the count is $1$.

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Would you like a proof or further examples? Let me know!

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5 Related Questions:

1. How many bit strings of length $n$ start with 1 and end with 0?
2. How many bit strings of length $n$ have exactly $k$ ones?
3. What is the total number of bit strings of length $n$?
4. How many bit strings of length $n$ are palindromes?
5. How many bit strings of length $n$ start with a 0?

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Tip:

When solving problems involving bit strings, focus on fixing specific positions (like "start" and "end") and calculate the free positions separately.