The number of bit strings of length 10 that contain either five consecutive 0’s or five consecutive 1’s is

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NIMCET 2015 Official Paper

Answer 1

Calculation:

Observe first 5 consecutive 0s

Summation rule: the first 5 consecutive 0’s could start at position 1, 2, 3, 4, 5, or 6

Start at the first position

Other 5 bits can be anything: 2⁵ = 32

Start at the second position

The first bit must be a 1

There are various possibilities that can be included in it.

Remaining bits can be anything: 2⁴ = 16

Start at third position

The second bit must be a 1 due to above-mentioned reason.

First bit and last 3 bits can be anything: 2⁴ = 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

The five consecutive 1’s ensue the same pattern and have different like 112 possibilities

There would be two cases counted twice (that we thus need to exclude): 0000011111 and 1111100000
Total = 112 + 112 - 2 = 222