Chapter 6 - Section 6.6 - Generating Permutations and Combinations - Supplementary Exercises - Page 441: 4

184 strings

The number of strings beginning with 000 is $2^{7}$ since every bit after 000 can be either 0 or 1. By similar logic, we get the number of strings that end with 1111 i.e $2^{6}$ The number of strings that start with 000 and end with 1111 is $2^{3}$ as there are only 3 bits that have to be filled. So the solution is $2^{7}$+$2^{6}$-$2^{3}$ = 184