Chapter 6 - Section 6.3 - Permutations and Combinations - Exercises - Page 414: 36

126 strings

We have 5 zeros and 14 ones $0$ must be immediately followed by $11$ So we have five $011$ and four $1$ to arrange. The arrangements for $n$ objects with $r$ similar objects is given by $\frac{n!}{r!}$ So in our case we have five $011$ and four $1$ to arrange i.e total 9 objects. Hence it is given by $\frac{9!}{4!.5!}$ = 126