The total number of ways in which six '+' and four '–' signs can be arranged in aline such that no two '–' signs occur together

Number of  '+' sign = 6 and Number of  '−' sign = 4

After placing 6 '+' sign, there are 7 places for '−' sign.

Now, we need to select 4 sign out of 7 places

∴ The total number of ways of arranging signs

Such that no two '−' are together = 7C4

$=\frac{7 !}{4 ! 3 !}=\frac{7 \times 6 \times 5}{2 \times 3}=35$