Answer
a)222
Work Step by Step
Step 1- Observe first 5 consecutive 0s.
The first 5 consecutive 0’s could start at position 1, 2, 3, 4, 5, or 6
-Start at first position
Other 5 bits can be anything: 2^5 = 32
-Start at second position
First bit must be a 1 because if first bit is 0 then we can have a string of the form 000000- - - - which is already counted in the previous case. Hence strings must be of the form 100000- - - - .
Remaining bits can be anything: 2^4 = 16
-Start at third position
Second bit must be a 1 due to above mentioned reason.
First bit and last 3 bits can be anything: 2^4 = 16
-Starting at 4,5 and 6 positions
Same as starting at positions 2 or 3: 16 each
Total = 32 + 16 + 16 + 16 + 16 + 16 = 112
Step 2-The five consecutive 1’s follow the same pattern, and have different like 112 possibilities.
Step 3- There would be two cases counted twice (that we thus need to exclude): 0000011111 and 1111100000
Hence by Inclusion Exclusion Principle.
Total = 112 + 112 – 2 = 222
