A little string is a limited sequence that \$0\$’s and \$1\$’s. How plenty of bit strings of size \$8\$ begin and also end v a \$1\$?

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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Is this the exactly answer?

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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