Answer
352
Work Step by Step
Answer = Number of bit strings of length 10 which either begin with three 0s or end with two 0s
= Number of bit strings of length 10 which begin with three 0s + Number of bit strings of length 10 which end with two 0s - Number of bit strings of length 10 which begin with three 0s AND end with two 0s.
=$2\times2\times2\times2\times2\times2\times2$ [10-3 decisions] + $2\times2\times2\times2\times2\times2\times2\times2$ [10-2 decisions] - $2\times2\times2\times2\times2$ [10-(2+3) decisions] , each decision with 2 options, 0 or 1
= 128+256-32
= 352
[The last number (32) is subtracted because there will be 32 strings which will start with three 0s and end with two 0s and will hence be counted twice in the 128+256]
