Question

How many bit strings of length eight contain either threeconsecutive os or four consecutive 1 s?

Answer 1

Number of strings with 3 consecutive zeros = 25+5×24=11225+5×24=112, because the 3 zeros can start at bit number 1, 2, 3, .., 6

Number of strings with 4 consecutive ones = 24+4×23=4824+4×23=48, I used the same reasoning