My answer would certainly be: $2^6$. Because we know, that the little bit starts v $1$ and ends with $1$: which means, the $2$ that the size $8$ is used, also there room $6$ positions remaining, i beg your pardon is $2^6 = 64$.
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Yes the prize $2^6$ is correct. 2 bits you know are $1$ already, therefore there space $8-2=6$ bits left. Each bit deserve to either it is in a $0$ or a $1$, so you have actually $2$ possibilities because that each slot. Hence, as you finish correctly, you have $2 cdot 2 cdot 2 cdot 2 cdot 2 cdot 2 = 2^6$ possible combinations for your string.
Since there is one an option for the first position, one selection for the critical position, and two options for every of the 6 positions in between them, the variety of bit strings of size eight the begin and also end v a $1$ is $2^6$, therefore your systems is correct.
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offered a bit string $x$ of size $n$, how many strings space there that differ from $x$ in precisely $k$ positions?
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