Answer
$2^{n-2}$
Work Step by Step
We have to $n-2$ make choices as the string length is $n$ but the first and the last digits are already determined to be 1.
Each of the choice consists of choosing one of the 2 options 0 or 1.
Therefore, the number of bit strings of length 10 which both begin and end with a one = 2×2×...×2. [n-2 times]
=$2^{n-2}$
