The total number of ways in which six ′+′ and four ′−′ signs can be arranged in a line such that no two ′−′ signs occur together is ______.
Open in App
Solution
′+′ signs can be put in a row in one way creating seven gaps shown as arrows: Now 4′−′ signs must be kept in these gaps. So, no two ′−′ signs should be together. Out of these 7 gaps 4 can be chosen in 7C4 ways. Hence, required number of arrangements is 7C4=7C3=7×6×53×2×1=35.