There are 65, 19-bit sequences of 0s and 1s that start with a 0 and end with a 0 that exist.
Any subsequent 0 in the sequence must come exactly two or three spots down from the previous one.
We begin at point 1 and finish at position 19 in this instance.
That is, we advance along the line a whole of 18 positions.
In order to achieve 18, we must add all series consecutive 2s and 3s.
There are numerous methods for doing this:
Case 1: There is just one possible arrangement for nine 2s.
Case 2: Two 3s with six 2s - thus ⁸C₂ ways = 28 ways
Case 3: Four 3s with three 2s - thus ⁷C₄ ways = 35 ways
Case 4: Six 3s - Only 1 way to arrange.
Thus, the number of arrangements = 1 + 28 + 35 + 1 = 65 .
Therefore ,the total possible number of the sequence that cane be made is 65 .
To know more about the sequences, here
https://brainly.com/question/7882626
#SPJ4