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The number of different bit strings of length 7 can be calculated using the formula 2^n, where n is the length of the bit string. In this case, with a length of 7, there are 2^7 = 128 different bit strings possible. This is because each bit in the string can have 2 possible values (0 or 1), and there are 7 bits in total.
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In a bit string length 12 How many begin with 110?
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.
How many bit strings of length 8 are there which begin with a 0 and end with a 0?
-- 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.
How many bit strings of length not exceeding n consist entirely of 1s?
n+1 (counting the empty string)
How many bit strings of length 10 both begin and end with one?
Since there are 8 bits in between, and they can assume any of the two values (0 or 1), that results in a total of 28 different combinations.
How many bit strings are there of length six or less not counting the empty string?
That's just the highest number that can be counted with 6 bits ... 127 .
