The total number of ways in which six '+' and four '–' signs can be arranged in aline such that no two '–' signs occur together
Question:
The total number of ways in which six '+' and four '–' signs can be arranged in aline such that no two '–' signs occur together is ___________.
Solution:
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
$=\frac{7 !}{4 ! 3 !}=\frac{7 \times 6 \times 5}{2 \times 3}=35$