Answer
44
Work Step by Step
Answer = Number of bit strings of length seven which either begin with two 0s or end with three 1s
= Number of bit strings of length seven which begin with two 0s + Number of bit strings of length seven which end with three 1s - Number of bit strings of length seven which begin with two 0s AND end with three 1s.
=$2\times2\times2\times2\times2$ [7-2 decisions] + $2\times2\times2\times2$ [7-3 decisions] - $2\times2$ [7-(2+3) decisions] , each decision with 2 options, 0 or 1
= 32 + 16 - 4
= 44
[The last number (4) is subtracted because there will be 4 strings which will start with two 0s and end with three 1s and will hence be counted twice in the 32+16]
