The number of ways of arranging ${\text{ six }}^{\mathrm{\prime }}{+}^{\mathrm{\prime }}$ signs and

four $\text{ '-' }$ signs in a row, so that no two $"-"$signs occur

together, is 

The $6^{\mathrm{\prime }}{+}^{\mathrm{\prime }}$ signs can be arranged in a row in one

way. Between them before and after, there are 7 gaps

for $4^{\mathrm{\prime }}{-}^{\mathrm{\prime }}$ signs

$\therefore$ The desired number is $\left(\begin{array}{l}7\\ 4\end{array}\right)=35\text{. }$