there are 128 (2 to the power of 7) bit strings of length 7
-- There are 256 bit strings of length 8 . -- There are 4 bit strings of length 2, and you've restricted 2 of the 8 bits to 1 of those 4 . -- So you've restricted the whole byte to 1/4 of its possible values = 64 of them.
The question is exactly equivalent to: "What's the highest binary number with 6 bits ?There are 64, corresponding to the binary numbers from 0 to 63.
You are essentially asking how many different bit strings of length 9 are there, as the first 3 bits are fixed. The answer is 2^9 = 512.
n+1 (counting the empty string)
Every bit can either be a 0 or 1. So to find the amount of bit strings of length either, you do 2length to find the amount of bit strings there are of a given length.
1024
56 The number of triples of 1s on 8 bits
Basically, this is the same as finding the number of distinct ways of arranging seven 1s and 3 0s. That is (10!/7!3!) = (10*9*8)/(3*2*1) = 120. There are 120 bit strings of length 10 with exactly three 0s.
210=1024
there are 128 (2 to the power of 7) bit strings of length 7
Half of them.
There are no zero-length strings that start with 1 bit or end with 2 bits. In a zero-length string, there are no bits at all.
-- There are 256 bit strings of length 8 . -- There are 4 bit strings of length 2, and you've restricted 2 of the 8 bits to 1 of those 4 . -- So you've restricted the whole byte to 1/4 of its possible values = 64 of them.
There are 210.
1024 of them.
The question is exactly equivalent to: "What's the highest binary number with 6 bits ?There are 64, corresponding to the binary numbers from 0 to 63.